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Algebra and Logic
For Statistics
Lectures, Exercises, Worksheets,
and Sample Tests
by
Deborah H. White
© 2016
Table of Contents
Page
Part I Topics in Arithmetic
Lecture #1: Using Algorithms in Arithmetic
1
Lecture #2: Fractions, Decimals, and Percents
11
Lecture #3: Graphing Fractions, Decimals, and Percents
21
Lecture #4: Rounding Numbers
27
Lecture #5: Rates and Ratios
33
Lecture #6: Solving Percent-Base-Amount Problems
37
Lecture #7: Unit Measure and Conversion
43
Sample Test on Topics in Arithmetic
51
Part II Topics in Algebra and Logic
Lecture #8: The Arithmetic of Signed Numbers
55
Lecture #9: Order of Operations and the Syntax of Algebra
63
Lecture #10: Evaluating and Simplifying Algebraic Expressions
71
Lecture #11: Solving Linear Equations in One Variable
79
Lecture #12: Modeling Situations with Linear Equations
89
Lecture #13: Linear Inequalities in One Variable
99
Lecture #14: Equations and Inequalities with Absolute Value
109
Lecture #15: More about Exponents
115
Lecture #16: Roots, Radicals, and Rational Exponents
125
Sample Test on Topics in Algebra
133
Lecture #17: Linear Equations in Two Variables
136
Lecture #18: The Rectangular Coordinate System
143
Lecture #19: Functions and Function Notation
157
Lecture #20: Types of Functions
163
Lecture #21: Formulas
177
Lecture #22: Symbolic Logic
193
Sample Test on Algebra and Logic
203
i
Part III Assorted Topics in Mathematics and Statistics
Lecture #23: Summation
209
Lecture #24: What Is Calculus?
212
Lecture #25: Quantitative Language
218
Lecture #26: Variables in Statistics
220
Lecture #27: Visual Display of Data
223
Index
229
ii
Lecture #1: Using Algorithms in Arithmetic
The word algorithm is one we’ll be using all semester in this course. You can find
various definitions of it online. Here’s a good one from The Free Dictionary: “A step-by-step
problem-solving procedure, especially an established, recursive computational procedure for
solving a problem in a finite number of steps.” (http://www.thefreedictionary.com/algorithm)
The important words here are “step-by-step,” “recursive,” and “finite.” “Step-by-step” is
pretty obvious -- you follow instructions in the exact order they’re given. “Recursive” means
you might have to repeat a process, kind of a loop, for some number of times before you move
on. “Finite” means that the procedure will come to an end – you’re not stuck in a loop forever,
and you will get an answer.
What a strange-looking word, though! What is its etymology? Etymology is the study of
where words come from. It’s a long story, but basically the word algorithm takes its name from
Al-Khwarizmi, a Persian Islamic mathematician who lived from about 790 to 850 AD and wrote
the first book about algebra. It was called Hisab al-jabr w'al-muqabala, and our word “algebra”
is taken from the title. This is one fascinating guy!
Algorithms can be made for almost anything you do, both in daily life and in math.
Here’s one for daily life:
Sorting Laundry before Putting in Washer:
1) Pick up item and observe its color.
a) If it’s white, put it in the white pile.
b) Otherwise, put it in the colored pile.
2) Repeat (1) until no items are left to sort.
Pretty mindless, right? But still, it has all the characteristics of an algorithm: it’s stepby-step, it’s recursive, and it’s finite (unless you have an infinite amount of laundry, which you
may think you do….).
Obviously we’re more interested in applying these techniques to mathematics than to
household tasks, but now you know the basics of the processes. How about an algorithm for
reducing a fraction? (The process is really called reducing a fraction to lowest terms, the
terms being the number on top, the numerator, and the number on the bottom, the
denominator. If you say you’re reducing a fraction, it sounds like you’re making it smaller,
which isn’t true – you’re keeping its value the same, just making the top and bottom smaller.)
So here’s the algorithm for reducing a fraction to lowest terms:
1) Find a whole number bigger than 1 that is a divisor (or factor) of both the numerator and
denominator.
1
a) If there isn’t one, the fraction is already reduced to lowest terms.
b) If there is one,
i)
Divide it into the numerator and let the quotient be the new numerator;
ii)
Divide it into the denominator and let the quotient be the new denominator.
2) Repeat step (1).
Let’s perform this algorithm on an actual fraction,
60
.
75
1) The number 5 is a divisor of both 60 and 75.
b) i) The quotient of 60 and 5 is 12.
ii) The quotient of 75 and 5 is 15.
12
So the new fraction is
.
15
12
2) Repeat step (1) using
.
15
1) The number 3 is a divisor of both 12 and 15.
b) i) The quotient of 12 and 3 is 4.
ii) The quotient of 15 and 3 is 5.
4
So the new fraction is .
5
4
2) Repeat step (1) using .
5
1) There is no whole number bigger than 1 which is a divisor of both 4 and 5.
4
.
5
I know this is a terribly tedious process, and if you can do it more easily (for instance, we
could have started with 15 as the number to divide into the numerator and denominator and
finished in one step) you should. But if you can’t, then the algorithm will get you to the correct
answer if performed correctly.
a) The fraction is reduced to lowest terms and is
Notice that the algorithm assumes that you know the meaning of certain terms: whole
number (0, 1, 2, 3, etc.), divisor (divides evenly into, in other words leaves no remainder),
factor (another word for divisor), numerator and denominator, which I already defined, and
quotient (the answer when dividing). And of course you have to know the meaning of the nonmathematical words. But that’s how all knowledge is: it builds upon previous knowledge.
As you can see, math has a specialized vocabulary, and knowing what the terms mean is
key to understanding the processes involved. For instance, we say that 3 is a factor or divisor of
12, meaning that the quotient of 12 and 3 is a whole number. Looking at the situation from the
opposite perspective, we say that 12 is a multiple of 3, meaning that there is a whole number (4
in this case) which when multiplied by 3 has the product 12.
2
Here are some other algorithms for the arithmetic of fractions.
Raising a Fraction to Higher Terms
1) Divide the old denominator into the new denominator, and call the quotient the “building
factor”.
2) Multiply the old numerator by the building factor, and replace the old numerator by the
product.
(The product is the answer when you multiply.)
2
Example: Change
to a fraction with denominator 12.
3
1) The quotient of 12 and 3 is 4. 4 is the building factor.
8
2) The product of 4 and 2 is 8. The new fraction is
.
12
Notice the word order surrounding the concepts of quotients and dividing. Here are three ways
of saying the same thing: 3 divided into 12 is 4. 12 divided by 3 is 4. The quotient of 12 and
3 is 4. (The quotient of 3 and 12 is something else entirely – ¼ or 0.25.)
Changing a Mixed Number to an Improper Fraction
2
A mixed number is a number like 3 made up of a whole number and a fraction. It
5
2
really means 3  . You have three whole things (pies, periods of times, distances, whatever)
5
2
and
of another. You could split each of the whole things into five parts, and you would have
5
2 17 17
15 parts. Those plus the other two would give you 17. So 3  .
is called an improper
5 5
5
fraction (not because it’s immoral!), which is a fraction whose numerator is at least as big as its
denominator. Here’s the algorithm:
1) Multiply the whole number part by the denominator.
2) Take the sum of the product in (1) and the numerator.
3) Make the sum in (2) the numerator of the improper fraction and the denominator its
denominator.
Changing an Improper Fraction into a Mixed Number
17
Here we reverse the steps. If we’re starting with
we’re seeing how many groups of
5
five there are in 17, and how many fifths are left over. Here’s the algorithm:
3
1) Divide the denominator into the numerator and note the quotient and remainder.
2) Make the quotient the whole number part, the remainder the numerator of the fraction
part, and the denominator its denominator. If the remainder is 0 there is no fraction part.
3) Reduce the fraction part if necessary.
33
Examples:
8
1) The quotient of 33 and 8 is 4, with a remainder of 1
2) 4
1
8
Adding (and Subtracting) Fractions
1) If the fractions have the same denominator,
A) Let that denominator be the denominator of the answer.
B) Add (or subtract) the numerators and let the sum (or difference) be the numerator of
the answer.
C) Reduce the new fraction to lowest terms.
2) If the fractions don’t have the same denominator, find a number that both denominators
are factors of (divide evenly into). Raise both fractions to higher terms using the number
you found as the new denominator. Then do step (1).
There are some pretty important things to notice here. First, the sum is the answer when
you add, and the difference is the answer when you subtract. Together with the product and
quotient, these are among the big words of arithmetic, because they are names for the
answers to the four basic operations. When someone asks what the product of 5 and 6 is, you
don’t have to inquire what you’re supposed to do with the 5 and the 6 – the word ‘product’
tells you.
Second, the algorithm refers to two previous algorithms – reducing fractions to lowest
terms and raising fractions to higher terms. The instructions for doing these procedures do
not need to be repeated in this algorithm. See how wonderfully economical this system is!
We build up a set of algorithms and can use them as soon as they’ve been introduced.
They’re kind of like sub-routines in a computer program.
Examples:
1 3

8 8
4
1) A) The denominators are the same, so the denominator of the sum is 8.
4
B) 1 + 3 = 4, so the numerator of the sum is 4. The sum is .
8
44 1
1
C) 4 goes evenly into both 4 and 8.
.
 . The answer is
84 2
2
5 7

6 10
2) The denominators aren’t the same. 30 is a number that both 6 and 10 go into evenly.
(30 is called the least common denominator, not because it’s the most unusual -least common – but because it’s the smallest common denominator, i.e. the smallest
number that both 6 and 10 go into evenly. It’s perfectly all right to find a larger
common denominator, like 60 or 120 or 48,000; it just means there will be more work
in the reducing process. In fact, if it takes you a while to find the smallest common
denominator, you’d be better off just using the product of the denominators (60 in this
case) as the new denominator. That always works.) Raise both fractions to higher
terms:
5  5 25
3  7 21


5  6 30
3  10 30
25 21 4
1) A) and B)


30 30 30
C)
42
2
2
. The answer is
.

30  2 15
15
And here’s one last algorithm:
Multiplying Fractions
1) Use the product of the numerators as the numerator of the answer.
2) Use the product of the denominators as the denominator of the answer.
3) Reduce the result of (1) and (2) to lowest terms.
Example:
3 2

4 9
1) and 2)
3 2 6
 
4 9 36
5
3)
66 1
1
 . The answer is .
36  6 6
6
(You may well know another way of doing this problem, using a procedure often referred to as
“cross-cancelling.” For the purposes of learning to use algorithms, do them according to this
algorithm instead.)
Note how much simpler a process this is than adding or subtracting fractions. Different
denominators? No problem; just multiply them. No need to raise the fractions to higher terms
and obtain a common denominator. Just remember that this process applies to multiplication,
not addition and subtraction. Otherwise you might find yourself making the following mistake:
1 1 11 2 2  2 1
 
 
 . Not!!!!! Let’s see….you have half a dollar and then you get
2 2 22 4 42 2
another half dollar, and now you have…..half a dollar. Not really:
1 1 11 2 2  2 1
 
 
  1 . That’s better – you have a dollar.
2 2
2
2 22 1
6
Worksheet for Lecture #1
Work in pairs to complete the following problems using the algorithms from the lecture.
1. Write in lowest terms:
6
12
2. Write in lowest terms:
28
36
3. Change
1
to a fraction with denominator 18.
2
4. Change
6
to a fraction with denominator 100.
25
5. Change 7
6. Change
7. Add:
5
to an improper fraction.
6
31
to a mixed number.
4
5 1

12 12
8. Subtract:
5 3

6 4
9. Multiply:
1 8

4 9
10. Multiply:
4 5

9 6
7
8
Assignment for Lecture #1: Using Algorithms in Arithmetic
1) Pick something you do in daily life
(not sorting laundry) and write an
algorithm for the process.
5
to a fraction with
8
denominator 72.
9
F) Change
to a fraction with
14
denominator 84.
E) Change
2) Using the algorithm for reducing
fractions to lowest terms, reduce
these fractions to lowest terms:
36
A)
81
45
B)
60
84
C)
200
27
D)
441
125
E)
2,000
F)
4) Use the algorithm for changing a
mixed number to an improper
fraction on these mixed numbers:
3
A) 2
4
3
B) 4
10
5) Use the algorithm for changing an
improper fraction to a mixed number
on these improper fractions:
45
A)
2
93
B)
10
6,000
42,000
3) Using the algorithm for raising a
fraction to higher terms, do these
problems:
1
A) Change to a fraction with
5
denominator 75.
1
B) Change to a fraction with
4
denominator 20.
2
C) Change to a fraction with
3
denominator 66.
3
D) Change to a fraction with
4
denominator 600.
6) Using the algorithm for adding or
subtracting fractions, do these
problems:
A)
B)
C)
D)
E)
9
5 3

14 14
5 1

9 9
3 1

10 10
1 7

5 10
3 1

4 6
F)
8 2

25 15
8) Write an algorithm for dividing
fractions.
7) Using the algorithm for multiplying
fractions, do these problems:
5 8
A) 
6 25
2 4
B) 
3 5
7 1
C)

10 21
10 3
D)

21 5
3 2
E) 
4 9
9 7
F)

14 15
9) Use your algorithm to do these
problems:
2 4

9 15
1 1
ii) 
3 2
3 5
iii) 
4 6
iv) A problem of your own.
i)
10
Lecture #2: Fractions, Decimals, and Percents
As you know, we have three different systems of naming and writing numbers, especially
when they’re between 0 and 1: fractions, decimals, and percents. You might wonder why we
need three different systems. The answer is that mathematics is a language, and languages have
many ways of expressing the same concepts. Sometimes there’s a shade of meaning in one word
that makes it more appropriate for one context, while another word works better in a different
situation. If you’re a parent, you might refer to those people whose parent you are as “my kids,”
in an informal conversation, as “my children” in a more formal one, even as “my offspring” if
you’re emphasizing that they are descended from you, or in any number of other ways. So let’s
not begrudge mathematics its multiple systems.
Fractions are the most basic and simplest form of the concept. You know how it goes:
3
means that you took something – a pie, a period of time, an amount of money, the length of a
4
journey, you name it – and split it into four equal parts and are now referring to three of those
four parts. Maybe you’re eating them, in the case of the pie, or spending them, in the case of
time and money, or traveling them, in the case of the journey. The important thing is that the
pieces of whatever it is you’re splitting be equal. This is not one-half of a pie:
Decimals (more correctly called “decimal fractions”) are a special way to write fractions,
and they have two advantages: They are easier to calculate with than fractions, and they can
easily be estimated or approximated, unlike fractions. To illustrate this second point, look at
5
.
13
It means that you took whatever it is, split it into 13 equal parts (good luck with that!), and
selected exactly five of them. But if you look at the decimal representation of
0.384615384615.... , you can call it 0.4 (
5
, which is
13
4
38
) or 0.38 (
) and gain a better picture of it
10
100
without losing too much detail.
And percents – well, percents are a special kind of decimals that we like to use when
talking about such business concepts as discounts and taxes, and about increases and decreases in
populations and other measures, like unemployment. We could certainly get along without them,
but why would we want to? They’re so expressive! “He gave 100%,” said the football coach,
sounding a lot better than, “He gave 1.”
11
The important thing is that you understand how to go back and forth among the three
systems easily and accurately; otherwise you’ll be the victim of all kinds of costly mistakes.
Changing One Form into Another
Let’s say we want to change a fraction into a percent, or a decimal into a fraction, or –
well, how many different changes are there among the three systems? This is something covered
in statistics, in the probability part of the course. The general term is “rules of counting,” which
means figuring out how many different ways there are to do something. Here we’ll use the
multiplication rule of counting, which tells us that if we know how many options there are for
the first step, for the second step, and so on, we simply multiply the number of options to find the
total number of ways there are to do the process.
In this case, first we pick the system we want to translate from. We can choose fractions,
decimals, or percents – three places to start. Then we pick the system we want to translate to.
Notice that we have only two options now, because we don’t want just to translate the system
into itself. So, since 3  2  6 , there are six different translations we want to be able to make.
Let’s start with the easiest ones.
Changing a Decimal to a Percent
First there are a couple of facts to note that you probably already know. In writing a
decimal which doesn’t have a whole number part, you can write a zero to the left of the decimal
point or not; it makes no difference. For instance, .25 and 0.25 mean the same thing. Why write
the 0, then? Because it draws attention to the decimal point. Second, if a number has no decimal
point, you can put one to the right of the last digit. So 34. means the same thing as 34 does. To
the right of the final digit is the default placement of the decimal point. The decimal point
declares that the digit to its immediate left is in the ones’ place.
So here’s the algorithm:
1)
2)
3)
4)
5)
Move the decimal point two places to the right.
If in doing (1) there were any empty spaces, fill them in with zeroes.
Place a percent sign to the right of the final digit.
Drop any zeroes to the left of the first non-zero digit.
If the number ends with a decimal point, delete it.
12
Examples:
0.36
1) 036.
3) 036.%
5) 36%
2.1
1) 21__ .
2)
3)
4)
5)
210.
210.%
5.%
210%
0.05
1) 005.
1 (which is 1.)
1) 1__ __ .
3) 005.%
2) 100.
3) 100.%
5) 5%
5) 100%
0.005
1) 000.5
3) 000.5%4) 36.%
4) .5% (though 0.5%
is also correct)
Changing a Percent to a Decimal
1) Delete the percent sign.
2) Move the decimal point two places to the left.
3) If in doing (2) there were any empty spaces, fill them in with zeroes.
Examples:
In the first three, remember the default placement of the decimal point to the right of the final
digit.
42%
6%
250%
9.5%
.02%
1) 42.
2) .42
1) 6.
2) . __ 6
3) .06
or 0.06
1) 250.
2) 2.50
1) 9.5
2) . __ 95
3) .095
or 0.095
1) .02
2) . __ __ 02
3) .0002
or 0.0002
Changing a Fraction to a Decimal
Now we’re at the tricky part. Not the way you’re going to do it, which will be on your
calculator, using this very simple algorithm:
1) Divide the numerator by the denominator.
But I also want you to understand the reasoning behind it. You see, a decimal is just a
special sort of fraction, with a denominator of 1 followed by a certain number of zeroes. How
many? Well, that’s determined by how many digits there are to the right of the decimal point.
For 0.4, for instance, the denominator is 10, and the numerator is 4:
denominator is 100, and the numerator is 31:
4
. For 0.31, the
10
31
. For 0.028, the denominator is 1000, and the
100
13
28
. So basically the denominator has as many zeroes after the 1 as there are
1000
digits to the right of the decimal point, and the numerator is all the digits to the right of the
decimal point, except that you can delete zeroes which are to the left of the first non-zero digit.
numerator is 28:
But that’s about translating a decimal to a fraction. Hopefully you can see that going the
other way involves building up a fraction to have one of the decimal denominators: 10, 100,
1000, etc.
1)
2)
3)
4)
Here’s the algorithm:
Find a number of the form 1 followed by any number of zeroes of which the
denominator is a divisor.
Raise the fraction to higher terms so that its denominator is the number you found in (1).
Write a decimal point followed by as many spaces as the number in (1) has zeroes.
Place the new numerator after the decimal point so that its right-most digit is in the rightmost space, its next digit to the left is in the next space to the left, and so on, and fill in
any empty spaces to the left with zeroes.
Examples:
3
5
7
20
1
8
2
25
1) 10
1) 100
1) 1000
1) 100
3 2 6

2)
5  2 10
3) . __
4) .6 or 0.6
7  5 35

2)
20  5 100
3) . __ __
4) .35 or 0.35
1  125 125

2)
8  125 1000
3) . __ __ __
4) .125 or 0.125
2 4
8

25  4 100
3) . __ __
4) .08 or 0.08
2)
But wait a minute – what if you couldn’t find a number in (1), because the denominator
of your fraction is not a divisor of any number of the form 1 followed by zero(es)? These
denominators include 3, 6, 7, 9, 11, 12, 13, 14, and 15, just to list the first few. And how can you
tell that there isn’t such a number? What if you just haven’t found it, but it’s out there?
Well, here’s how to find out. If a number’s only divisors are 2’s and 5’s, it is the divisor
of one of the decimal denominators. If not, it isn’t. How can you determine that? You could
keep dividing your number by 2 until it is no longer a whole number, then take the last whole
number you got and keep dividing that by 5 until it is no longer a whole number, and then look at
the last whole number you got. It is either 1, in which case your denominator is a divisor of one
of the decimal denominators, or it isn’t 1, in which case your denominator isn’t a divisor of one
of the decimal denominators. Another way is to find the prime factorization of your
denominator. If the prime factorization consists only of 2’s and 5’s, it’s a divisor of one of the
decimal denominators; if not, it isn’t. (The prime factorization of a whole number expresses it as
14
the product of only prime numbers, which can’t be broken down into smaller whole number
factors. For example, 50  2  5 2 , so 50 is a divisor of a decimal denominator, but 24  2 3  3 , so
24 is not a divisor of a decimal denominator.)
Or you could simply look at your denominator and try to find a divisor which isn’t 2 or 5.
For instance, 13 has 13 as a divisor, 14 has 7, and 15 has 3.
So how do you change this kind of fraction to a decimal? Well, you could use your
calculator of course, but I want to show how the process looks when you use long division.
1
. The calculator gives you 0.3333333333. It stops after ten decimal places. This is what
3
it looks like using long division:
Take
Hopefully you know how to do long division – putting a decimal point after the one, placing a
decimal point above the line directly over the one in the dividend (the number being divided
into), and adding zeroes as needed in the dividend. Note all the ones. This is why all the digits
in the quotient are 3’s, because each time you have to divide 3 into 10, and 10  9  1.
To show that the quotient has 3’s forever, we use a repeater bar, and the number itself is
1
 0. 3 , we mean 0.333… But the calculator
3
doesn’t have the capability of using a repeater bar, so it just repeats the 3’s until it runs out of
room.
2
But notice that when you use the calculator to change
to a decimal, you get
3
0.6666666667. That’s because the calculator doesn’t just stop when it runs out of room – it
2
considers the next digit and rounds the number accordingly. Using the repeater bar,  0. 6 .
3
Check out the long division version and you’ll see that the remainder is always 2, because you
get 6 when you divide 3 into 20, and 20  18  2 .
called a repeating decimal. When we write
15
Not all fractions which are repeating decimals produce the same remainder at each step of
the long division process. Some have two or more. Look at
17
:
33
In this case, the remainders alternate between 17 and 5, so the repeated part has two digits.
Using the repeater bar,
17
 0.51 .
33
There can be any number of digits in the cycle under the repeater bar, but there’s a limit
for each denominator: the number has to be smaller than the denominator itself. You can see
why. Otherwise the denominator would go evenly into one of the remainders, which would be
itself.
A denominator like 33 is very modest in its cycle, having only two digits, the 7 goes all
the way, requiring six digits before repeating. The 7th’s are particularly neat:
1  0.142857
7
2  0.285714
7
3  0.428571
7
4  0.571428
7
5  0.714285
7
6  0.857142
7
It’s the same six digits in the same order, just starting at a different part of the cycle for each
fraction.
Sometimes the repetition doesn’t begin at the first digit to the right of the decimal point
(the 10th’s place). The repeater bar shows this by being placed only over the repeating part:
8  0.533333....  0.53
14  0.186666....  0.186
15
75
Repeater bars aren’t terribly useful in the real world, but they’re a lot of fun!
16
I haven’t mentioned changing mixed numbers to decimals yet, because they’re not very
popular in algebra. I’ll explain why later in the course. But they present no problem to our
conversion process: just put the whole number part to the left of the decimal point and proceed
1
5
Changing a Decimal to a Fraction
as before. For example, 4  4.2 .
Don’t worry – this explanation is going to be much shorter than the one for turning
fractions into decimals. Here’s the algorithm:
1) Give the fraction a denominator consisting of a 1 followed by as many zeroes as there
are places to the right of the decimal point.
2) Give the fraction a numerator consisting of the digits to the right of the decimal point.
3) If the numerator begins with one or more zeroes, delete them.
4) Reduce the fraction.
Examples:
0.4
0.04
0.004
0.45
0.00056
1)
1)
1)
1)
1)
10
4
2)
10
4)
2
5
100
04
2)
100
4
3)
100
1
4)
25
100
45
2)
100
1000
004
2)
1000
4
3)
1000
1
4)
250
4)
9
20
100000
00056
2)
100000
56
3)
100000
7
4)
12500
But what about repeating decimals? You can hardly put an infinite number of zeroes
after the 1 in the denominator! Well, there are a couple of ways of dealing with them, but we
won’t go into them. It’s enough if you know that 0.333… is
1
2
and that 0.666… is . It might
3
3
1
2
, that 0.222… is , etc. In short, the numerator in the case of
9
9
th
9 ’s is repeated infinitely in the decimal version. Note that this even works for 0.333…and
1 3
2 6
9
0.666…, because  and  . It also means that 0.999… is , or 1 – something that
3 9
9
3 9
baffles a lot of people.
help to know that 0.111… is
The decimal-to-fraction algorithm makes it clear why zeroes to the right of the decimal
point but to the left of the first non-zero digit are important, while those after the last non-zero
17
digit aren’t. Look at 0.4, 0.04, and 0.004. The zeroes in bold type determine the denominator by
making more places. Whereas you could say the same thing for 0.4, 0.40, and 0.400, in fact step
(4) of the algorithm reduces all three fractions (
lowest terms,
4 40
400
,
, and
) to the same fraction in
10 100
1000
2
.
5
Changing a Fraction to a Percent
If you’re dreading another long explanation, fear not! We’re going to go via decimals.
Here’s the algorithm:
1) Change the fraction to a decimal.
2) Change the decimal to a percent.
Example:
2
25
1) .08
2) 8%
There is one slight complication, which involves the fractions which turns into repeating
decimals, but we’ll worry about it only when the repeating part is 3’s or 6’s. We know that
1
is 0.33333…, so when we move the decimal point two places to the right and tack on the
3
percent sign we get 33.333…%. In that case, we replace the .333… by its fractional
1
1
2
2
equivalent, , so the percent version is written as 33 %. Likewise,
becomes 66 %.
3
3
3
3
Changing a Percent to a Fraction
1) Change the percent to a decimal.
2) Change the decimal to a fraction.
18
Worksheet for Lecture #2
Fill in the table.
Fraction
Decimal
Percent
3/5
0.35
80%
3 9/20
6.3%
7.5
19
Assignment for Lecture #2: Fractions, Decimals, and Percents
1) With fractions, decimals, and percents, we saw that there were 6 different conversions
among the three systems. But what if there was another system to represent numbers?
Let’s call it “perfractimals.” So now there are four systems. How many different
conversions would this new arrangement have? (Use the multiplication rule of
counting to determine the answer.)
2) Define the word tithe. Give three versions of the definition, one for each system –
fractions, decimals, and percents.
3) Complete the worksheet for conversions of fractions, decimals, and percents.
Fraction
Decimal
Percent
2/5
Fraction
Decimal
Percent
3 1/5
0.5
72%
30%
7/10
0.06
4
4%
2.6
0.75
1%
1 1/4
7/8
100%
0.1
2.2
55%
7/100
0.01
0.05
88.10%
150%
0.85
3 3/4
4/5
2 1/2
3/100
0.3%
2.7
1/8
8%
20
Lecture #3: Graphing Fractions, Decimals, and Percents on a Number Line
In arithmetic, we tend to think of numbers as amounts, or quantities: there are 15 books,
6.3 ounces, 1½ hours, etc. But in algebra, we’re more likely to think of numbers as locations, or
addresses. We do this by using a number line.
For now, our number line will begin at a point which we label 0 and continue directly to
the right, with an arrow at the end to show that it goes on forever:
Now we choose some other point on the line, make a hash mark to indicate it, and label it 1:
And now we’re done – not in the sense that we don’t have to label any more points, but in the
sense that we no longer have any choice about those points. Their location is determined once
we place the 1. For instance, 2 has to be located exactly as far to the right of 1 as 1 is to the right
of 0. Then 3….well, you get the idea. We can’t go messing with the scale once we’ve started.
Now let’s concentrate on the part of the line between 0 and 1. We’ll enlarge it:
1
(or 0.5, or 50%). You divide the line segment between 0 and 1 into
2
two parts of equal length, and you do that by placing one hash mark exactly halfway from 0 to 1:
Say you want to locate
1
2
and ? Now you have to split the segment into three parts of equal length, and
3
3
1
you do this by placing two hash marks at equal intervals. The first one is the location of , the
3
2
second of :
3
How about
21
And so on and so forth. Remember: to form n parts of equal length, you have to place n – 1
equally-spaced hash marks. (This is algebra-speak. Instead of saying that for four equal parts
you need three hash marks, for five equal parts you need four hash marks, etc., etc., we just call
the number of equal parts n and say that you need one fewer hash mark, or n – 1. A very, very
efficient system!!)
If the number you’re trying to locate on the number line is bigger than 1 (a mixed
number, an improper fraction, a decimal with at least one non-zero digit to the left of the decimal
point, or more than 100 percent), you find which two whole numbers the number is between, and
then locate the fractional part between those two numbers. If the number is an improper fraction,
change it to a mixed number.
If we wish to graph (locate)
11
on the number line, we first change it to the mixed
4
3
. We divide the segment between 2 and 3 into four equal parts (by placing three
4
hash marks, of course), and count over to the third of them. I placed the number’s name, 11/4,
above the line only so that it wouldn’t run into the 3 below.
number 2
Approximate and Relative Placement
Sometimes you don’t have to be so careful about exact placement of a number, but you
might want to find out which of two numbers is bigger, and thus has to be placed further to the
right. This isn’t too difficult if the whole-number parts of the numbers are different. (If the
numbers are improper fractions, change them to mixed numbers to find out.) But if the wholenumber parts are the same, follow this algorithm:
1) Change fractions and percents to decimals if they aren’t already.
2) Compare the digits in the 10th s place (the first place to the right of the decimal point).
3) If one number’s digit is larger than the other, it is the larger number and goes to the
right of the other one.
4) If the digits in step (2) are equal, compare the digits in the next place to the right.
5) Perform step (3) on these new digits. Continue to the right as needed until you reach
unequal digits.
So, if asked to graph
14
7
and
,
33
15
22
1) Change them to decimals:
14
 0.424242... , and
33
7
 0.466666...
15
2 & 3) The 10th s digits are equal (both 4’s).
4) The next digits (in the 100th s place) are different, and since 6 is bigger than 2,
3)
7
14
is the larger number and goes to the right of
.
15
33
7
you’d
15
have to make 14 equally-spaced hash marks between 0 and 1 and count over to the 7th one, and
14
for
it would be even worse), where should you put them?
33
There are a couple of “quick and dirty” ways to do this. First, look at the decimal
1
versions of the numbers. They’re both less than , or 0.5. So they go closer to the 0 than to the
2
1
1, but not much closer. On the other hand, they’re bigger than , or 0.3333…. This tells you
3
that they are more than a third of the way from 0 to 1. It’s easy to visually divide a line segment
into three more or less equals parts. Our two numbers are more than one-third of the way from 0
7
14
to 1 but less than half of the way, and don’t forget that
is to the right of
. Here’s an
15
33
approximate graph:
But if you’re going to locate them approximately (a good idea, because in the case of
I know this is a lot of effort to go to, considering how unlikely you are ever to be asked to
graph numbers like this (except in this course….), but it brings up ways of looking at numbers,
their locations, and the words that describe their relative positions.
For instance, there’s the word “between,” which we’ll encounter frequently in doing
14
7
and
”, we mean those that are bigger than (to
33
15
14
7
the right of)
and smaller than (to the left of)
. Do we ever use the “between….and”
33
15
math. If we refer to the numbers “between
23
phrase and mention the bigger number first? It’s hard to tell using such obscure numbers as our
fractions; let’s try something simpler: “I’m expecting to have between 40 and 20 papers to grade
this weekend.” While understandable, it certainly sounds a little jarring to me, whereas
“between 20 and 40” would attract no attention at all.
And one final note about comparing fractions. If your number sense is at all active, you
might have noticed that one of the numerators (14) is twice the other (7). But the denominator of
the fraction whose numerator is twice the other’s is more than twice the denominator of the
other fraction (33 vs. 15). So even though we took twice as many pieces of the pie in
14
than
33
7
, the pieces themselves were less than half as big, so we wound up with
15
less of the pie. If your reaction to this explanation is, “Say what?” just stick with the algorithm.
we did of the pie in
Finding the Whole Numbers between Two Numbers
In statistics it is sometimes necessary to list all the whole numbers between two numbers
which you have calculated. Let’s say the two numbers are 0.72 and 3.45. It might be obvious to
you that the whole numbers between them are 1, 2, and 3. If not, placing 0.72 and 3.45 on a
number line might help:
24
Worksheet for Lecture #3
1. For reach of these numbers, construct a number line from 0 to 5, and graph the number as
accurately as possible. Use a separate number line for each number.
(a)
7
3
(b) 0.75
(c) 325%
2. Use the algorithm and your calculator to determine which is larger,
Justify your response.
3. Use the algorithm to determine which is larger, 0.823 or 1.04.
Justify your response.
4. List all whole numbers between 0.98 and 3.52.
25
19
6
or .
34
11
Assignment for Lecture #3: Graphing Fractions, Decimals, and
Percents on a Number Line
Locate these numbers on a number line. Use a separate number line for each number.
3
1) 4
2
2) 5
2
3
5
4
1
4
3)
4)
5) 3.4
6) 1.5
7) 175%
8) Which is larger,
3
5
or
? Use the algorithm and show your work.
14
23
9) Which is larger, 3
7
5
or 4 ? Use the algorithm and show your work.
6
8
1
1
or
? Use the algorithm and show your work. If you can, explain
57
56
how you could have found the answer without converting the fractions to decimals.
10) Which is larger,
26
Lecture #4: Rounding Numbers
First, why would you want to round a number? There are a couple of reasons. A large
number with lots of non-zero digits can be hard to comprehend. Isn’t 570,000,000 easier to take
in than 567,319,835? True, some detail is lost, but the improved understanding might more than
compensate for the loss. Also, the further to the left a digit is, the more important it is and the
more weight it has. For instance, the 8 is only 8 hundred, but the 5 is 5 hundred million!
A second reason is that sometimes a computation will result in a number which gives the
impression of being more accurate than it really is. Let’s say you want to find the quotient of 82
and 17. The calculator tells us it’s 4.823529412, and that’s not even the whole thing (of course
it’s a repeating decimal since 17 is by no means composed solely of factors of 2 and 5 – in fact it
doesn’t even have a 2 or a 5 as a factor). We’d be way overstating our case if we use all ten
digits, since 82 and 17 are just whole numbers. I’d say we could stretch it to 4.8, but an even
more modest and realistic result might be “about 5.”
Before I give the algorithm for rounding numbers, you need to know the names of the
places, or place values. For numbers without a decimal point, they are as follows:
We start at the right-most digit and work our way left. The commas are inserted every three
digits, and they divide the number into what are called periods – the ones period, the thousands
period, the millions period, etc. Each period gets its name from its right-most place. The
commas function as visual dividers, making it easier to read the number. Each period is read in
itself, with the name of the period added: Two hundred ten billion, one hundred twenty-three
million, four hundred fifty-six thousand, seven hundred eighty-nine. We don’t add the period
name “ones.”
The ingenious system of place values allows us to write very compactly a number with a
lot of complexity. 210,123,456,789 is actually short for
200,000,000,000  10,000,000,000  100,000,000  20,000,000
 3,000,000  400,000  50,000  6,000  700  80  9
Wouldn’t you just hate to have to write this out? This long form is called expanded notation.
Of course numbers without a decimal point (whole numbers) have one by default at their
extreme right. For numbers with a decimal point, the ones place is occupied by the first digit to
27
its left, the tens place by the second digit to its left, and so on. The first place to the right of the
decimal point is called the tenths place (all the places to the right of the decimal point have
names ending in “ths”). Here’s a number with digits to the right of the decimal point:
One kind of confusing thing is that the tens place is the second one to the left of the
decimal point, but the tenths place is the first one to the right of the decimal point. In reality,
the ones place is the center of the number, and the decimal point just serves to give notice that
the digit to its left is in the ones place. The key is that the digits to the right of the decimal point
are the numerators of fractions with denominators consisting of a one followed by as many
zeroes as the position of the digit. We could write the part of the number to the right of the
decimal point in expanded notation as
2
3
4
5
6




10 100 1,000 10,000 100,000
or, using decimal notation, as
0.2  0.03  0.004  0.0005  0.00006
To pronounce the decimal part of our number, first read it as though it were just 23456, or
23,456, and then add the name of the final place: Twenty-three thousand, four hundred fifty-six
hundred thousandths. To put the two parts of the number together, connect them with the word
“and.”
Algorithm for Rounding Numbers
Now that we’ve covered the names of the places, here’s the algorithm for rounding. You
are asked to round a number to the nearest __________ (name of place value).
1) Locate the digit in the specified place.
2) Locate the digit in the place directly to the right of the specified place. It is called the
decider.
3) If the decider is 5, 6, 7, 8, or 9 (at least 5), change the digit in the specified place to the
next higher digit.
4) If the decider is 0, 1, 2, 3, or 4 (less than 5), leave the digit in the specified place as is.
5) Leave all digits to the left of the specified place as is.
6) A) If the specified place is to the left of the decimal point, replace all digits to the right
of the specified place (including the decider) by zeroes. Omit the decimal point and all
digits to its right.
B) If the specified place is to the right of the decimal point, drop all digits to its right.
28
Examples:
Round 16,835.279 to the nearest hundred.
1) The digit in the hundreds place is 8.
2) The decider is 3.
4) Since the decider is less than 5, leave the 8 as is.
5) Leave the 1 and the 6 as is.
6) A) Replace the 3 and 5 by zeroes, drop the decimal point and all digits to its right: the
answer is 16,800.
Round 16,835.279 to the nearest tenth.
1) The digit in the tenths place is 2.
2) The decider is 7.
3) Since the decider is at least 5, change the 2 to a 3.
5) Leave the 1, 6, 8, 3, and 5 as is.
6) B) Drop all digits to the right: the answer is 16,835.3.
That’s pretty much the whole story, except where the digit in the place to which you’re
rounding is a 9 and the decider is at least 5, because then you can’t do step (3) – there is no next
higher digit. Let’s insert a new step into the algorithm under step (3):
3) A) If the digit in the specified place is a 9, change it to a 0 and replace the digit to its
left by the next higher digit. If that digit in turn is a 9, keep doing this until you reach a
digit which is less than 9. If all digits to the left of the specified place are 9’s, put a 1 to
the left of the last digit you changed to a 0.
Examples:
Round 439,621 to the nearest thousand.
1) The digit in the specified place is 9.
2) The decider is 6.
3) A) Since the decider is 6 and the digit in the specified place is 9, change the 9 to a 0
and the 3 to a 4.
6) A) Replace the decider and all digits to its right by zeroes: The answer is 440,000.
29
Round 59.9998 to the nearest thousandth.
1) The last 9 on the right is in the specified place.
2) The decider is 8.
3) A) Since 8 is at least 5, change the 9 in the thousandths place to a 0. Keep replacing the
9’s by zeroes until you reach the 5, and change it to a 6.
6) B) Drop all digits to the right of the specified place: The answer is 60.000.
Here’s why you don’t drop all of the zeroes to the right of the decimal point: You need to show
that you’ve rounded the number to the nearest thousandth. If you wrote 60 for your answer, it
would appear that you had rounded the number to the nearest one.
30
Worksheet for Lecture #4
Round 42,695,273.8926 to the nearest
1) Ten Million
2) Million
3) Hundred Thousand
4) Ten Thousand
5) Thousand
6) Hundred
7) Ten
8) One
9) Tenth
10) Hundredth
11) Thousandth
31
Assignment for Lecture #4: Rounding Numbers
Round these numbers to the indicated place. Example below.
The Number
374.6837
Tens
370
Ones
375
Tenths
374.7
Hundredth
374.68
Thousandth
374.684
1,704.6039
67.5539
145.9341
99.4507
801.7777
5,286.0356
29.9468
179.9946
9999.9997
1) In your own words, what is it about rounding numbers like 429.97 to the nearest tenth
that makes it trickier than, for instance, rounding numbers like 429.87 to the nearest
tenth?
2) Some people believe in the “domino theory” – as I call it – of rounding numbers. Asked
to round 37.46 to the nearest whole number, they would say 38. Can you explain what
they do to get this answer? Why is it incorrect? [Hint: What is the number exactly
halfway between 37 and 38? Which side of this number does 37.46 lie on?] How does
the algorithm prevent you from using the “domino theory”?
3) Explain in your own words why when asked to round 25.997 to the nearest hundredth we
write 26.00, but to the nearest whole number we write 26?
32
Lecture #5: Rates and Ratios
Ratios
A ratio is a useful way of showing the size relation of two quantities, expressed either as
a fraction or as two numbers separated by a colon.
For example, suppose a class has 18 women and 12 men. Whereas it is true that there are
six more women than men, a more important fact is that the ratio of women to men is
3
, or
2
2
, or 2:3. The quantity after the
3
word “of” is the numerator of the fraction; the quantity after the word “to” is the denominator.
Word order is very important in the translation.)
3:2. (We could also say that the ratio of men to women is
3
18
? This is because the fraction
,
12
2
having the number of women as its numerator and the number of men as its denominator, when
3
reduced to lowest terms, equals . You could split the class into groups of five (3 + 2), in
2
which each group has three women and two men. There would be six such groups ( 6  5  30 ).
Did you see that ratio of women to men is in fact
3
gives you a good idea of the nature of the class. There are
2
definitely more women than men, but not overwhelmingly more.
Knowing that the ratio is
4
.
3
How many women and men are in this class? There’s an easy way to figure this out, using the
idea of making groups of women and men. In this case you would make groups of seven people
(4 + 3), each with four women and three men. How many such groups could you make? Five,
since 35  7  5 . So there would be 20 women ( 5 4 ) and 15 men ( 5 3 ). Note that this checks,
20
4
because reduced to lowest terms the fraction
equals .
15
3
Make sure you understand this technique by determining the number of women and men
in a class of 30 in which the ratio of women to men is 2 to 1.
Let’s look at a different class, one with 35 students and a ratio of women to men of
Rates
A rate is a special kind of ratio, because it compares two measurements or quantities
which have different units. It usually contains the word “per” (from the Latin word for “for” or
“for each”) and often has a specific name. You are familiar with lots of these: miles per hour
(speed), miles per gallon (fuel efficiency), pounds per square inch (pressure), and cents per
33
ounce (unit price), to name a few. These and other similar terms are evidence of the lushness of
language, which provides multiple ways of expressing the same idea.
By reducing the fractions in the rates to lowest terms, you can turn a specific situation
into a general statement. For instance, if you go 110 miles in 2 hours, the fraction
110
becomes
2
55
(we don’t usually leave a 1 in the denominator, but I left it for emphasis), and your average
1
speed was 55 miles per hour. This means that if you had actually traveled a steady 55 mph for
two hours (which is pretty impossible), you would have gone the same distance, 110 miles, as
you actually went.
Let’s look at a fuel-efficiency example. Your car uses up 8 gallons of gas in going 256
miles. (You can determine this by starting the trip with a full tank of gas and then refilling it at
the end of the journey.) Since the fraction
256
in lowest terms (still keeping the 1 in the
8
32
, you’ve driven 32 miles per gallon of gas on the average. You haven’t
1
steadily used one gallon of gas for each 32 miles driven, of course, since perhaps sometimes you
were coasting down hills or going up them, and sometimes you were in city traffic and
sometimes on the highway. But the rate of 32 miles per gallon is a useful way to estimate fuel
needs and cost for a trip.
denominator) is
Unit price is a very important concept for consumers, and it does and should influence
buying habits. If you’re buying tomato paste, and a 5-oz can costs $2.15 while an 8-oz can costs
$2.15
$0.43
$2.88, the cost per ounce of the smaller can is
, or
, 43¢ per ounce. For the larger
1 oz
5 oz
$2.88
$0.36
can, the figures are
, or
, 36¢ per ounce. Unless the extra 3 ounces are going to go
8 oz
1oz
to waste, or you just don’t have enough money for the larger can, you are better off buying the
larger can.
A little later in the course we’ll discuss how to change the units in rates to answer
questions like this: how many feet per second do you go if your speed is 55 miles per hour?
34
Worksheet for Lecture #5
Express the following ratios in lowest terms:
1) 6 dogs to 24 cats
2) 60 men to 75 women
3) If a club has 72 people, and the ratio of women to men is
4
, how many women and how
5
many men are in the group?
4) If you travel 135 miles in 4 hours, what is your average speed in miles per hour, to the
nearest whole number?
5) If you go 413 miles on 20 gallons of gas, what is your fuel efficiency in miles per gallon,
to the nearest tenth?
6) If a 14-ounce box of cereal costs $4.99, and an 8-ounce box of the same cereal costs
$2.99, which is the better deal? Explain fully.
35
Assignment for Lecture #5: Rates and Ratios
Express in lowest terms the following ratios:
1) 24 women to 14 men
2) 16 men to 32 women
3) 85 women to 39 men
4) 75 men to 100 women
5) If a group has 36 people, and the ratio of women to men is
5
, how many women and
1
how many men are in the group?
6) If a group has 36 people, and the ratio of men to women is
4
, how many women and
5
how many men are in the group?
7) If you travel 147 miles in 4 hours, what is your average speed in miles per hour, to the
nearest whole number?
8) If you go 175 miles on 6 gallons of gas, what is your fuel efficiency in miles per gallon,
to the nearest tenth?
9) If a force of 230 pounds is exerted over an area of 8 square inches, what is the pressure in
pounds per square inch, to the nearest whole number?
10) If a 6-ounce can of pineapple costs $2.99, and a 10-ounce can costs $5.19, which is the
better deal? Explain fully.
11) In your own words, what is the difference between a ratio and a rate?
36
Lecture #6: Solving Percent-Base-Amount Problems
The first thing you need to know is what a proportion is and how to solve for a missing
part. A proportion is an equation which says that two fractions are equal to each other. Here’s
2 8
8
2
 . This is a true statement, since
reduced to lowest terms is . But what we’re
3 12
12
3
really interested in here is when one of the four numbers (called terms) is missing, and you have
to find the number which, when substituted for the missing term, would make the proportion
true.
one:
4 x
 . What value of x will
20 25
make the proportion true? The easiest way to find out is to follow this algorithm:
We’ll call the missing term x. Consider this proportion:
1) Take the number in the same fraction as the missing term and the number in the other
fraction directly opposite the missing term, and find their product.
2) Find the quotient of this product and the number diagonally across from the missing term
in the other fraction. This is the right value for the missing term.
Here’s a little diagram to help you understand and remember this process:
1) The number in the same fraction as the missing term is 25, and the number directly
opposite is 4. The product of 25 and 4 is 100.
2) The number diagonally across from the missing term is 20. The quotient of 100 and 20 is
5. The missing term is 5.
4
5
and
to decimals, you get 0.2 in both cases, so the two fractions are equal.
20
25
1
You could also reduce both to lowest terms, and in each case the answer would be .
5
20 25
 , you still get 5
Notice that if you turn both fractions upside down, which gives
4 x
for your answer, because the algorithm tells you to multiply 25 by 4, and then divide by 20, the
same process as in the first version of the proportion. You would also get the same answer by
changing the proportion to have one fraction consist of the two numerators and the other fraction
4 20
consist of the two denominators as long as you maintain their order: 
. Again, the
x 25
algorithm tells us to multiply 4 by 25, and then divide by 20. It’s true that now we’re
If you change
37
multiplying 4 by 25, and before we were multiplying 25 by 4, but it’s no secret that these both
give a product of 100.
4 25
 .
x 20
Here the algorithm would have us multiplying 4 by 20, and then dividing by 25, which gives 3.2,
not 5.
The best way to see what I mean by maintaining their order is to change it:
Percent-Base-Amount Problems
Proportions can be used to solve all kinds of math questions that occur in daily life. You
set up a proportion with a missing term and solve for it. But we’re going to cover only one kind
here, the percent kind.
As you know, percents are a specialized kind of decimal, but they are also a ratio (note
the per and the cent, which means hundred). So 20% is the ratio
20
.
100
Consider the fact that 20% of 40 is 8. We could express this as a proportion. We call the
20 the percent, the 40 the base, and the 8 the amount. Using the letter P for percent, B for base,
and A for amount, we write the proportion like this:
P A

100 B
or, in this case,
20 8

100 40
Here’s how you can tell which of the numbers to put in for P, A, and B (the 100 is always there,
to stand for cent): The number before the percent sign, or the word “percent,” is P; the number
after the word “of” is B, and the number all by itself on one side of the word “is” is A.
20 8

does not have a missing term, but there are three ways we could
100 40
alter the problem to create one. The missing term could be the percent, the amount, or the base.
The proportion
1) Missing percent: What percent of 40 is 8?
Here the “what” comes before “percent” and thus is P. The proportion becomes
P
8

100 40
2) Missing amount: What is 20% of 40?
The “what” is all by itself on one side of the “is” and is thus the amount, which gives
38
20 A

100 40
3) Missing base: 40 is 20% of what?
The “what” comes after the “of” and is the base:
20 8

100 B
Take another look at case (2), where we’re finding the amount. Hopefully you know
another way to do this type: find the product of 20% and 40. The word “of” coming after a
fraction or percent means to multiply. So you’d change 20% to a decimal (0.2) and multiply it
20 A

involves the exact same arithmetic, just in a
100 40
slightly different order. So you can do this type either way. There are shortcuts for the other two
cases as well, but it might just be easier to make them all into proportions and then follow the
algorithm for solving proportions.
by 40 to get 8. But solving the proportion
39
Worksheet for Lecture #6
If necessary, round bases and amounts to the nearest tenth and
percents to the nearest whole percent.
1) What is 32% of 451?
2) What percent of 87 is 14?
3) 230 is 160% of what?
4) What is 125% of 48?
5) 18 is 20% of what?
6) What percent of 25 is 150?
7) What is 0.3% of 691?
40
Assignment for Lecture #6: Solving Percent-Base-Amount Problems
If necessary, round bases and amounts to the nearest tenth and percents to the nearest whole
percent.
Examples:
30
a

; a  150  30 100 ; answer: 45
100 150
p
45

What percent of 150 is 45?
; p  100  45  150 ; answer: 30%
100 150
30 45

45 is 30% of what?
; b  45 100  30 ; answer: 150
100 b
What is 30% of 150?
1) What is 25% of 84?
10) What is 15% of 600?
2) What percent of 250 is 50?
11) What percent of 34 is 68?
3) 42 is 50% of what number?
12) 13 is 10% of what number?
4) What is 10% of 245.79?
13) What is 75% of 160?
5) What percent of 48 is 11?
14) What percent of 30 is 90?
6) 60 is 75% of what number?
15) 96 is 30% of what number?
7) What is 8% of 457?
16) What is 80% of 240?
8) What percent of 20 is 13?
17) What percent of 200 is 80?
9) 29 is 100% of what number?
18) 42 is 70% of what number?
41
19) What is 30% of 14?
27) 30 is 15% of what number?
20) What percent of 12 is 18?
28) What is 0.7% of 9,000?
21) 24 is 2% of what number?
22) What is 0.6% of 1,000?
23) What percent of 36 is 27?
24) 19 is 50% of what number?
25) What is 35% of 700?
26) What percent of 20 is 50?
42
Lecture #7: Unit Measure and Conversion between Units
Civilization is based on measurements, and it’s important that you understand them.
Some measurements deal with a single unit, like distance, area, volume, time, currency,
temperature, angle size, and weight. Others involve ratios of units, like speed (unit distance per
unit time), pressure (unit weight per unit area), and fuel efficiency (unit distance per unit
volume).
Units of measurement are fascinating, because they are a link to the history and cultures
of the world. We don’t have enough space or time to go into this topic in any detail. Here’s a
simple and entertaining introduction:
http://ellerbruch.nmu.edu/cs255/JoniEMi/metricsystem.html
Suffice it to say that some units are based on physically-available standards, like the foot
and the hand (for measuring the height of horses), but obviously there are problems with this
system since peoples’ size varies. During the French Revolution in the late eighteenth century,
the government decided to standardize units by keying them to a more general physical standard
– the size of the earth. The basic unit of measure was to be the meter, defined as one tenmillionth of the distance from the North Pole to the equator along the meridian running near
Dunkirk in France and Barcelona in Spain.
Of course the great advantage of the resulting metric came from how easy it is to convert
measurements, involving merely moving the decimal point a certain number of places right or
left. Our system, called the U.S. customary units of measurement, is much more cumbersome
when it comes to conversions, as no doubt you know. But both systems use the same units for
measuring time, and these units do not allow conversion by moving the decimal point, because
they are based on 60 (60 seconds per minute, etc.) instead of 10. Using 60 is convenient because
it can be broken up into so many different whole-number parts (i.e., it has many divisors: 1, 2, 3,
4, 5, 6, 10, 12, 15, 20, 30 and 60).
In some cases, many different units exist to measure the same thing. Length or distance
in particular has many units: inches, feet, yards, miles, meters, and all the ones meters form with
prefixes – milli- (one-thousandth), centi- (one-hundredth), kilo- (one thousand), and so on –
hands, furlongs, cubits, light-years (not a unit of time but of distance – how far light travels in
one year, just under six trillion miles), parsecs (again, distance, not time or angle measure), and
lots more besides.
There are two important things you need to know about units and measurement.
First, if you measure something, you have to say what unit you used. Saying something
is 5 long is absolutely meaningless. Is it 5 feet, 5 miles, 5 meters, 5 light-years, or 5 somethingelses? Of course, there are exceptions to this rule. If asked how old someone is, you can say,
“Twenty-five” without adding “years,” unless you’re referring to a toddler, about whom, if you
want to use the number 25, you’d better add “months.”
43
Second, no measurement is exact. Nothing is exactly 5 meters long. If you say it’s 5
meters long, you’re really saying that it’s between 4.5 and 5.5 meters long. You’re giving its
length to the nearest whole meter. That’s why it’s different to say something is 5.0 meters long,
even though people usually think that 5 and 5.0 are the same number. When you say it’s 5.0
meters long, you’re saying that you measured it to the nearest tenth of a meter, so you’re really
saying it’s between 4.95 and 5.05 meters. (These numbers are called boundaries, and we study
them further in Statistics.) You’ve measured it ten times more precisely than if you called its
length 5 meters. That’s a big difference in precision.
Measuring is very different than counting. If you count the people living in your
household and say that 5 people live in your household, you don’t mean that there are between
4.5 and 5.5 people living in your household. The 5 is an exact number. Counting always gives
exact numbers; measuring never does. In measuring, the numbers always need to be rounded,
because you could always measure more precisely by using a more refined measuring tool.
Equivalent Measures
Because there are different ways of measuring the same quantity, you can change the unit
used by using the appropriate arithmetic. You know some of these, like the fact that 12 inches
equal 1 foot, that there are 60 seconds in a minute, etc., and you also probably know that to
change feet to inches you multiply by 12, to change seconds to minutes you divide by 60, etc.
But sometimes the process is confusing, and we’re going to produce an algorithm for changing
units after some more explanation of equivalent measures.
I call an equation like 12 inches = 1 foot an equivalency, which is an equation which has
different units measuring the same thing on each side, and one of the sides has 1 as its numerical
1
feet, or about 0.083 feet, is another. As long as two units measure the
12
same sort of concept, like length or time or area, you can make an equivalency between the units,
with either unit being given the 1.
value. So 1 inch =
Whereas you should know a lot of these equivalencies because of being an educated
person, in fact there are thousands of them, and it’s nice to have a simple way to look them up.
Here are two URLS that will do the job for you:
https://www.google.com/search?q=unit+converter
and
http://www.unitconverters.net/
Length, Area, and Volume
People are often confused about the units used for measuring length, area, and volume,
and about the concepts themselves. The differences are important in everyday life as well as in a
math class.
44
Looking at definitions of the word “length,” I can see why people might be confused.
It’s “the distance from one end of something to the other end.” That kind of throws the meaning
back on the word “distance,” which might not be all that much help. Nevertheless I think you
probably have a pretty good idea of what it means, and you know you would use a ruler of some
sort to measure it (if it isn’t too big).
Contrasting length and area might be the best way to understand both, except that again
the definitions aren’t much help. Area is “a quantity that expresses the extent of a twodimensional surface or shape.” Okayyyy…. Think of it as being measured by covering
whatever you’re finding the area of by little square stickers. If the sticker measures 1 inch on
each side, then the sticker covers 1 square inch (also written 1 inch2). If it’s a big sticker (maybe
a tile), measuring 1 foot on each side, it covers 1 square foot, or 1 foot2.
But not all units of area have “square” attached to them. For instance, acre contains the
concept of square units; it’s equivalent to a little more than 43,560 square feet. Where do they
get these numbers? Anyway, if you talk about “square acres” you are revealing your ignorance.
Volume is the third concept. It can be defined as “the amount of 3-dimensional space an
object occupies,” again a definition of limited usefulness unless you already know the meaning
of the word. Think of dice or sugar cubes, or of pouring water into a container. If a die (singular
of “dice”) is a cube measuring 1 centimeter on each side, then its volume is 1 cubic centimeter,
or 1 cm3.
Just like with area, not all units of volume have the word “cubic” in them. Liters and
quarts are units of volume all by themselves. Even if you’re using an internet unit converter to
change one unit to another, you have to know what category a unit is in.
The concept of dimensionality is essential to understanding the use of units of length,
area, and volume. Did you notice that area is two-dimensional and volume three-dimensional?
Well, length is one-dimensional. Here’s an illustration with the dimensions numbered, but since
paper and computer screens are two-dimensional, when we represent a three-dimensional object
we are only making a projection of the object onto a two-dimensional surface. Nevertheless,
we’re so used to this that we actually perceive the object as three-dimensional.
Converting Units
If you want to convert a measurement using one unit to a measurement using a different
one, there’s a sure-fire method called dimensional analysis.
45
First, you need to know that any equivalency produces two different versions of the
1 foot
12 inches
number 1. For instance, since 1 foot = 12 inches, the fractions
and
are both
12 inches
1 foot
versions of 1 (though rather unusual versions). We called these fractions conversion factors.
Now, the key to dimensional analysis is the obvious fact that if you multiply any number by 1,
the product is the original number. That idea, coupled with treating the units as though they can
be cancelled as parts of a fraction, gives the algorithm for converting units:
1) Find the appropriate conversion factor(s) involving the original unit(s) and the unit(s) you
want to convert the measurement to.
2) Start with the number you’re converting from and go from left to right. If the first
conversion factor has the number which isn’t 1 in the numerator, multiply the number
you’re converting from by it. If the conversion factor has the number which isn’t the 1 in
the denominator, divide the number you’re converting from by it. If there is more than
one conversion factor, use them all.
Examples:
Change 5 pounds to ounces.
1) Since 1 pound = 16 ounces, the conversion factors are
1 pound
16 ounces
and
. Since
16 ounces
1 pound
5 pounds
16 ounces
the appropriate conversion factor is
, since
1
1 pound
5 pounds
16 ounces
the pounds from the
will cancel out the pounds from the
when
1
1 pound
multiplied.
5 pounds is the same as
2)
Since the 16 is in the numerator, multiply the 5 by the 16. The surviving unit is ounces.
Answer: 80 ounces.
Change 80 ounces to pounds.
1 pound
, because the ounces in 80
16 ounces
ounces, being on top, and the ounces in the conversion factor, being on the bottom, will
cancel out.
1) The appropriate conversion factor this time is
2)
Since the 16 is in the denominator, you divide the 80 by the 16, and the surviving unit is
pounds. Answer: 5 pounds.
46
When the measurement is a rate, you’ll have to use two conversion factors, one to
change the unit in the numerator, and one to change the unit in the denominator.
Example:
Change 55 miles per hour to feet per second.
55 miles
1) 55 miles per hour is
. Since 1 mile = 5,280 feet, the conversion factor we’ll
1 hour
5,280 feet
use to cancel the miles is
, and since 1 hour = 3,600 seconds, the conversion
1 mile
1 hour
factor we’ll use to cancel the hours is
. (The hour has to be on top,
3,600 sec onds
because in the original measurement it’s on the bottom.)
2)
Since the 5,280 is in the numerator and the 3,600 is in the denominator, multiply 55 by
5,280 and then divide the product by 3,600. The surviving unit is feet per second.
Answer: 80.7 feet per second, to the nearest tenth.
A Note about Length, Area, and Volume Measurements
If you want to know, for instance, how many square feet are in 5 square yards, and you
don’t have access to a handy unit converter, but you do remember that there are 3 feet in a yard,
you can use dimensional analysis to find the answer.
5 square yards, or 5 yards2, could also be called 5 yards x yards, because squaring
something means multiplying it by itself. Note that the 5 itself is not squared: you have 5 of
these things, which are square yards.
Example:
Convert 5 square yards to square feet.
1) Use the conversion factor
3 feet
, but use it twice, once for each of the yards you want
1 yard
to cancel out.
2)
Multiply 5 x 3 x 3, and remember that feet x feet is square feet.
Answer: 45 square feet, or 45 feet2.
47
Worksheet for Lecture #7
Common Conversions
1 hour = 3600 seconds
1 meter = 3.28 feet
1 meter = 1.09631 yd
1 mile = 5280 feet
1 km = 0.62 miles
1 mi = 1.6 km
1 yard = 3 feet
1 foot = 12 inches
1 inch = 2.54 cm = 25.4 mm
1. How many kilometers in 3.7 miles? Round your answer to the nearest tenth.
2. How many yards in 4.98 meters? Round your answer to the nearest hundredth.
3. How many miles per hour is a speed of 75 feet per second? Round your answer to the
nearest whole number.
4. How many square feet are in 12 sq. yds?
48
Assignment for Lecture #7: Unit Measure and Conversion between Units
1) Pick a unit from the lecture or a unit not mentioned in the lecture which you either had
never heard of or didn’t know what it was. Look up its history and equivalences, and
briefly describe both.
Set up and complete the following conversions.
2) How many miles in 5.6 kilometers?
Use the equivalence 1 mi. = 1.6 km, and round your answer to the nearest tenth.
3) How many inches in 3.27 millimeters?
Use the equivalence 1 in. = 25.40 mm, and round your answer to the nearest hundredth.
4) How many tablespoons in 2.4 quarts?
Use the equivalence 1 qt = 64 tbsp, and round your answer to the nearest tenth.
5) How many cubic feet in 50 quarts?
Use the equivalence 1 ft3 = 29.92 quarts, and round your answer to the nearest hundredth.
6) How many meters in 7.3 kilometers?
Use the equivalence 1 km = 1,000 m.
7) How many acres in 7,923 square yards?
Use the equivalence 1 acre = 4840 yd2, and round your answer to the nearest hundredth.
8) How many minutes in 3 weeks?
Use the equivalences 1 week = 7 days, 1 day = 24 hours, and 1 hour = 60 minutes.
9) How many seconds in 2 days?
Use the equivalences 1day = 24 hours, 1 hour = 60 minutes, and 1 minute = 60 seconds.
10) How many liters in 5 cubic meters?
Use the equivalence 1 m3 = 1000 L.
11) How many quarts in 18 liters?
Use the equivalence 1 qt = 0.95 L, and round to the nearest hundredth.
49
12) How many cubic inches in 0.5 gallons?
Use the equivalence 1 gal = 230 in.3.
13) How many ounces in 3 kilograms?
Use the equivalences 1 kg = 2.2 lbs and 1lb = 16 oz.
14) How many grams in 1.7 pounds?
Use the equivalences 1 kg = 2.2 lbs and 1 kg = 1000 g, and round to the nearest tenth.
15) How many feet per second is a speed of 95 miles per hour?
Use the equivalences 1 mi = 5,280 ft, 1 hour = 60 min, and 1 min = 60 sec, and round to
the nearest hundredth.
16) How many miles per hour is a speed of 40 feet per second?
Use the equivalences 1 mi = 5,280 ft, 1 hour = 60 min, and 1 min = 60 sec, and round to
the nearest tenth.
17) How many feet per hour is a speed of 83 centimeters per minute?
Use the equivalences 1 in. = 2.54 cm, 1 ft = 12 in., and 1 hr = 60 min, and round to the
nearest hundredth.
18) How many kilometers per liter is 32 miles per gallon?
Use the equivalences 1 mi = 1.6 km and 1 gal = 3.8 L, and round to the nearest tenth.
19) How many pounds per square inch is 14.3 kilograms per square centimeter?
Use the equivalences 1 kg = 2.2 lbs and 1 in2 = 6.4516 cm2, and round to the nearest
tenth.
50
Sample Test on Topics in Arithmetic
1) What are the terms of the fraction
3
?
17
2) What is the product of the numerator and the denominator in the fraction
3) If you are raising the fraction
5
?
21
5
to higher terms, and want it to have a denominator of 72,
9
what’s the building factor?
4) Raise
11
to higher terms so that its denominator is 64.
16
5) List three divisors of 60.
6) While reducing a fraction to lowest terms, you first divided by 8 and then by 3. If you
wanted to perform to reduction in one step only, what number would you have divided
by?
7) Reduce
35
to lowest terms.
49
8) List three numbers that 12 is a divisor of.
9) List two numbers which could be used as the common denominator in adding the
fractions
3
7
and
.
20
15
10) What is the difference of
5
7
and
?
8
12
11) What is the quotient of
3
9
and
?
5
10
12) What is the product of
7
8
and
?
12
21
51
13) What is the sum of
3
1
and
?
10
10
14) Algorithm for four-number mishmash
A) Let the product of the first and third numbers be the numerator of a fraction.
B) Let the sum of the second and fourth numbers be the denominator of a fraction.
C) Reduce the fraction.
Perform the four-number mishmash on these numbers:1
6
4
4
What answer do you get?
19
to a decimal using the algorithm, to what denominator
250
would you build up the fraction?
15) In changing the fraction
16) In changing the fraction
137
to a decimal, how many spaces would you put after the
10,000
decimal point?
17) In changing 0.056 to a fraction, how many 0’s would you put in the denominator after
the 1?
18) Where does the decimal point go in the number 1, 243?
19) Change
1
to a percent.
8
20) Change 0.58 to a fraction.
21) Change 2.3% to a decimal.
22) Change
4
to a decimal.
25
23) Change 1.67 to a percent.
52
24) Change 450% to a fraction (or a mixed number).
25) Which of these fractions turn into repeating decimals?
7 9
30 50
11
80
73
(Could be
120
more than one.)
26) What two whole numbers is the number
27) Change
31
between?
8
23
to a mixed number.
6
28) Locate the number 260% on this number line:
_____________________________________________________
29) Which is larger,
11
12
or
? Show enough work to support your answer.
21
23
30) In the number 265,801,453.2749, which period do the digits 801 occupy?
31) In the number 265,801,453.2749 what place is the 6 in?
32) If you’re rounding 453,908.61 to the nearest hundred, which digit is the decider?
33) If you’re told that the digit 4 is the decider, to what place are you rounding the number
684,029.256?
34) Round to the nearest thousand: 68,258.7976
35) Round to the nearest hundredth: 68,258.7976
36) What is 26% of 86? Round to the nearest tenth.
37) What is 175% of 85? Round to the nearest one (or whole).
38) 18 is what percent of 79? Round to the nearest whole percent.
39) 95 is what percent of 54? Round to the nearest whole percent.
40) 46 is 30% of what? Round to the nearest hundredth.
53
Answers
1) 3 and 17
21) 0.023
22) 0.16
23) 167%
2) 105
3) 8
1
9
or
2
2
7
73
25)
and
120
30
24) 4
44
4)
64
5) Any three of these: 1, 2, 3, 4, 5, 6,
10, 12, 15, 20, 30, 60
26) 3 and 4
6) 24
7)
5
7
27) 3
8) 12, 24, 36, 48, 60, 72, 84, 96, 108,
120, etc.
28)
11
11
 0.5238
29) 21 , because 21
and
12
 0.5217
23
30) Thousands
9) 60, 120, 180, 240, 300, 360, etc.
10)
1
24
2
3
11)
12)
2
9
31) Ten-millions
32) 0
2
5
13)
33) Ten-thousandths
34) 68,000
35) 68,258.80
2
14)
5
15) 1000 or 10,000 or 100,000, etc.
36) 22.4
16) 4
37) 149
17) 3
38) 23%
18) To the right of the 3: 1,243.
39) 176%
19) 12.5%
20)
5
6
29
50
40) 153.33
54
Lecture #8: The Arithmetic of Signed Numbers
One difference between arithmetic and algebra is that in algebra we use signed numbers,
which are numbers preceded by a sign, either a  sign (called ‘plus’), or a  sign (called ‘minus’
or ‘negative’) – except for the number 0, which doesn’t have a sign.
So a number is made up of two parts, its sign and the number part itself (which is called
the absolute value). The sign to the left of the absolute value is the sign of the number.
However, if you write a number without a sign, like 5, you are implying that its sign is
positive. The  sign is the default setting of signs. If you write 5, you cannot claim that you
meant  5 . Writing 5 by itself implies that the number is  5 .
There are many stories and analogies for signed numbers. You can think of  5 as
meaning you get $5, or gain 5 yards in football, or travel 5 miles towards a destination, or are 5
feet above sea level, and so on. With these interpretations,  5 can be thought of as meaning that
you spend $5, or lose 5 yards in football, or travel 5 miles away from a destination, or are 5 feet
below sea level.
No matter which story you tell yourself, we’re going to look at signed numbers as being
locations on a number line. In Lecture #3 we used a number line that looked like this:
Now we’ll be extending the line to the left of 0, and that’s where the negative numbers
go:
All the numbers marked on this number line are integers, which means that they have no
fractional or decimal part. Technically, the numbers 1, 2, 3, … are called the natural numbers,
the numbers 0, 1, 2, 3, … are called the whole numbers, and the numbers …,  3 ,  2 ,  1 , 0,
1, 2, 3, … are called the integers, from the Latin word for “intact,” or “untouched.”
Notice that the arrow is still on the right; we could put one on the left too, to indicate that
the negative numbers go on forever, but we’ll stick to having just the one on the right, to show
55
that this is the direction in which the numbers increase. Even though  5 has a bigger absolute
value than  3 , it is a smaller number. (If you spend $5 you have less money than if you spend
$3; if you’re 5 feet below sea level, you’re lower than if you’re 3 feet below sea level.)
To locate negative numbers on the number line, follow the same procedure as explained
in Lecture #3, splitting the segment between two numbers into the right number of equal parts,
2
and then counting over to the correct hash mark. But remember that a number like  1 is
3
between  1 and  2 , because it’s smaller than (to the left of)  1 and larger than (to the right of)
 2:
Adding Signed Numbers
Once mathematicians get ahold of a new kind of number, they immediately want to know
how it works. How do you add signed numbers? Subtract them? Multiply or divide them?
We’ll start with adding, and we’ll start with a problem we know the answer to (always
the best way): 2  3 . It’s 5. You probably first learned this by taking a pile of 2 rocks or candies
or something and then placing 3 more in the pile and counting how many you wind up with. But
we’re going to interpret addition on the number line, with the following story: You start at  2 ,
and then you go 3 to the right. (Why to the right? Because it’s the positive direction.) Finally,
you look down and see where you are. Here’s the journey:
That’s a lot of work for something you knew the answer to already! The benefit comes
when we apply the method to problems you might not have been able to answer.
But first, a note about notation. There are other ways to write the problem we just did
besides 2  3 . There’s  2  3 , putting the default + sign to the left of the 2. There’s  2   3,
putting the default + sign to the left of the 3. But when we do that we have to enclose the  3 in
parentheses, because the syntax (the correct arrangement of symbols and numerals) in algebra
doesn’t allow two signs, pluses or minuses, to be next to each without one being preceded by a
parenthesis. That’s because these signs have double meanings (triple in the case of the minus
56
sign, as you’ll see). They are operation signs, addition for plus and subtraction for minus, and
also the signs of numbers, indicating a number’s place on the number line, to the left or right of
0.
If you have a lot of parentheses lying around, you could even enclose the  2 in
parentheses, like this:  2   3, but it’s not necessary.
Let’s get on with it. Suppose you want to add  2   3 . You start at  2 , and then you
go 3 to the left (because it’s a minus 3, and left is the negative direction), and then you look
down:
The answer is  5 .
So far we’ve added two positive numbers and then two negative numbers. Now we’ll
add one of each. How about 2   3? You start at 2, go 3 to the left, and look down:
It’s  1 . And  2   3, or  2  3 ? Start at  2 , go 3 to the right, and look down:
The answer is 1, or  1 .
The Algorithm for Addition of Signed Numbers
I’m sure you’ll agree that adding signed numbers on a number line is a very cumbersome
process. You’ll be happy to hear that we won’t be using the number line again. Instead, we’ll
use an algorithm.
57
Algorithm for adding two signed numbers:
1) They have the same sign. Add the absolute values of the two numbers, and attach their
sign to the answer.
2) They have different signs.
a) If they have different absolute values, subtract the smaller absolute value from the
larger one, and attach the sign of the larger absolute value to the answer.
b) If they have the same absolute value, the answer is 0.
3) One of the numbers is 0. The sum is equal to the other number. (if they’re both 0, the
sum is 0).
Examples:
 3   5
3 5
1) 3  5  8
3   5
 3 5
1)  3   5  8 2)a) 5  3  2
Since 5 is
bigger, the
sum is  2
3   3
2)a) 5  3  2 2)b) 0
Since 5 is
bigger, the
sum is 2
30
3) 3
Numbers like 3 and  3 are called opposites, because they are on opposite sides of 0 and
the same distance away from it, one to the right and the other to the left. Their sum is always 0.
Do you see that the opposite of a negative number is its positive version?
Subtraction of Signed Numbers
As promised, no number lines will be used here, just an algorithm. This algorithm
provides for the translation of a subtraction problem into an addition problem, and from there
you can use the algorithm for addition of signed numbers.
Algorithm for Subtraction of Signed Numbers:
1) Leave the first number in the problem as it is; change the subtraction sign to an addition
sign; change the second number to its opposite. If the second number is 0, the difference
is equal to the first number.
2) Add the two numbers.
Examples:
 7   2
 2   2
70
1)  7   2
7   2
1) 7  2
1)  7  2
1)  2  2
1)  7
2)  9
2) 9
2)  5
2) 0
72
72
1) 7   2
2) 5
58
For problems like 7   2 , some people like to join the minus signs into a kind of plus sign to
emphasize that subtracting a negative is the same as adding a positive, like this:
If this helps you, feel free to do it.
Multiplication of Signed Numbers
Before giving the algorithm, let me remind you of how you first learned to multiply, by
using repeated addition. You were told that 3 4 meant that you should take three 4’s, and
add them: 4  4  4  12 . What does this mean for 3  4 ? Same idea, but this time you’re
adding three  4 ’s:  4   4   4 . Whereas the addition algorithm didn’t tell you how to add
three signed numbers, I think it’s pretty clear that the answer is  12 . If you’re not sure of this,
add the first two numbers first. This gives  8. And  8   4  12 .
Algorithm for Multiplying Signed Numbers
1) They have the same sign. Multiply the absolute values. The product is positive.
2) They have different signs. Multiply the absolute values. The product is negative.
3) At least one of the numbers is 0. The product is 0.
Examples:
3 4
 3 4
1) 12
2)  12
3  4
2)  12
 3  4
3 0
1) 12
3) 0
Of these examples, people have the most trouble with  3   4 . Can the product really be
positive? It can, and it is, and your best approach is just to accept the fact.
Division of Signed Numbers
The algorithm for division of signed numbers is very similar to that of multiplication, at
least for steps 1 and 2.
Algorithm for Dividing Signed Numbers
1)
2)
3)
4)
They have the same sign. Divide the absolute values. The quotient is positive.
They have different signs. Divide the absolute values. The quotient is negative.
The first number is 0, and the second number is not 0. The quotient is 0.
The second number is 0. The quotient is undefined.
59
Examples:
8 2
8   2
 8   2
 8   2
02
8 0
1) 4
2)  4
2)  4
1) 4
3) 0
4) Undefined
The last three examples are the ones that give people trouble. But they can be understood
if you think about how to check a division problem. Why does 8  2  4 ? Because 4  2  8 .
If  8   2 equaled  4 , then  4   2 would have to equal  8 . But we know by the
algorithm for multiplication of signed numbers that it equals 8. But  8   2  4 does check:
4   2  8 .
What about 0  2  0 ? Well, since 0  2  0 , it checks. And 8 0 ? You’d have to find a
number which multiplied times 0 gives an answer of 8, which is impossible, because 0 times any
number equals 0, not 8. So we call that quotient undefined, which means there isn’t an answer.
60
Worksheet for Lecture #8
Simplify.
9)  3   2
1)  7  9
10)  10  5
2) 3   4
11)  4  0
3) 1   4
4) 0  5
12)  14   8
5)  12   6
13)  6 8
6)  8   5
14) 4  11
7) 14   1
15) 3   11
8) 17   8
16)  5   9
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Assignment for Lecture #8: The Arithmetic of Signed Numbers
1) Which do you think is more difficult,
addition of signed numbers or
multiplication of signed numbers?
Support your choice with at least one
reason and one relevant example.
13)  18  9
2) Why do the algorithms for addition
and subtraction of signed numbers
make special reference to the cases
where one of the numbers is 0?
16) 0   2
14)  18   9
15) 18   9
17)  4  0
18)  1 0
3) Do an internet search on the question
“Why is the product of two negative
numbers positive?” State the answer
that makes the most sense to you.
19) 0   6
20)  8  0
Find the answer for each problem:
21) 0   5
4)  18   9
22)  3 0
5)  18   9
23) 0   7
6) 18   9
24)  9  0
7) 18   9
25) 0  0
8)  18  9
Pick two numbers (the first should be a
multiple of the second), and, using them in
place of the 18 and the 9, redo #4 through
#15 with the same signs and operations.
(For instance if your numbers are 24 and 8,
#4 would be  24   8 .) Number your
9)  18  9
10)  18 9
11)  18   9
problems #26 though #37.
12) 18   9
62
Lecture #9: Order of Operations and the Syntax of Algebra
What happens when two or more operations are indicated in an expression? (An
expression is basically anything that doesn’t have an equals sign.) Here’s an example: 3  2  7 .
Does it equal 35, or does it equal 17? (It’s 35 if you add before multiplying, 17 if you multiply
before adding.)
Does it matter? Well, yes, it does, because mathematics is all about precise
communication, and the last thing we want is for some people to look at 3  2  7 and think 35,
while others are thinking 17.
Everyone who uses this language has to agree on the choice, and the resulting set of rules
(called the order of operations) assures that everyone who correctly follows it will come up
with the same answer. There are reasons we might prefer one over the other, but the main thing
is that we follow the agreed-upon rules.
And the amazing thing is that we need only four different steps in an algorithm to allow
us to ensure that, no matter how complicated the expression, we’ll come up with the same
answer. I find this little short of miraculous. Of course, the steps themselves are complicated
and require quite a bit of explanation, as you’ll see, but still….there are only four of them.
I’ll begin with Steps 3 and 4, because they’re the ones that apply to our example:
3) Perform all multiplications and divisions in order from left to right.
4) Perform all additions and subtractions in order from left to right.
That settles it. I won’t ever be adding the 3 and the 2, because I have to do the
multiplication first. So
3 27 
3  14 
17
It is challenging to resist adding the 3 and the 2 first. That’s because we’re taught to go
from left to right when we learn to read, and the 3 2 just pops out and says 5. You have to
resist this temptation.
One more point: Multiplication and division have the same priority; you don’t do all the
multiplications in one sweep and then go back and pick up the divisions. The same applies to
addition and subtraction. Why is this? Well, multiplication and division are pretty much two
sides of the same coin, as are addition and subtraction, and we can’t distinguish between them.
Say you want to divide 8 by 2, in symbols 8 2 . The answer of course is 4, but you could also
1
1
 4 . The numbers 2 and are called reciprocals. You get
2
2
2
one from the other by reversing the numerator and the denominator (remember, 2  ). Their
1
get it by multiplying 8 by ½ : 8 
63
product is always 1. Each number has a reciprocal, except 0. (What is
1
1
? It’s undefined.
0
0
means the same thing as 1 0 .)
In fact, we could have made an algorithm for division, in which the first number stays the
same, the division sign is changed to multiplication, and the second number is replaced by its
reciprocal, much like our algorithm for subtraction. In other words, multiplication and division
have to have the same priority, because they’re interchangeable.
And that’s also why addition and subtraction have the same priority. Our algorithm
turned each subtraction problem into an equivalent addition problem (by equivalent, I mean
having the same answer).
A note here about notation, how we write mathematics in symbols: Just as the positive
sign is the default sign for numbers, so multiplication is the default sign for operations. You
could write 3  4 as 3  4 , and it would mean you are to multiply the numbers. When you
omit the times sign, it’s called implicit multiplication, because the multiplication is implied. If
you wrote 3  4 , nobody could claim that you meant the two numbers to be added, subtracted,
or divided. (If you left out the parentheses, it would look like a subtraction problem.) Those
operations would have to be indicated with operation signs. To give another example, you could
write 3 4 as 3 4. (You have to use parentheses here because otherwise what you wrote would
look like 34, thirty-four.)
More about notation: Division can be expressed in a couple of ways (but we don’t use
8
or as 8 2 . They all equal 4. The fraction bar
2
can always be interpreted as meaning division. (The third way is very convenient when you’re
using a keyboard.)
the long-division sign). You can write 8 2 as
Examples:
6 10 / 2
3) 6  5
4) 1
4  6 0
3) 4  0
8  2 4
3) 4 4
3  4 1
4)  1 1
10  5  3
4) 5 3
4) 4
3) 16
4) 0
4) 2
Grouping Symbols
What if you want 3  2  7 to equal 35? In that case, you have a way of overriding Steps
3 and 4, by the use of notation called grouping symbols, which include parentheses (the word
“parentheses” is the plural of “parenthesis,” which comes from the Greek word for “a putting in
beside”) as well as many other kinds. Here’s Step 1 of the order of operations algorithm:
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1) Perform operations within grouping symbols.
So if you put parentheses around 3 2 , the addition must be performed first, before the
multiplication:
3  2 7 
5 7 
35
Within the parentheses, all steps must be followed, as in this example:
5 10  2  3 
5 10  6 
5 4 
20

Remember, the fact that the 5 is right next to the parenthesis to the left of the 10 means that
multiplication is implied. If it helps you to remember, you can insert a times sign, like this:
5  10  2  3 .
This brings up some other points about notation. What if you wanted to imply the
multiplication of the 2 and the 3 instead of using a times sign? You’d have to put the 3 in
parentheses, like this: 510  2 3 . But multiple parentheses can get confusing, so you can also
use a different grouping symbol, called brackets, to make it clear which goes with which:
510  2 3. The brackets have the same meaning as parentheses, and we use them as the outer
grouping symbol. The parentheses are said to “nest” within them. If you need another set, you
can use the curly brackets, { }.
But here’s an interesting fact: The parentheses around the 3 aren’t really grouping
symbols, because there’s nothing to be done inside them. They’re visual separators, intended to
show that the 3 is a number in its own right, not part of a 23. So not all parentheses are grouping
symbols.
You’ve seen this before: 4   3. The parentheses around  3 show that it’s a number,
not something being subtracted. How about this one: 4  3 ? This shows implied multiplication
of 4 and  3 , and it equals  12 .
You may find this confusing; many people do. But if you can relax about it and just take
it in and learn it as a child learns his or her first language, you may be surprised by how secondnature it becomes. People have the capacity to master complex symbolism containing subtle
differences, which is certainly a description of the syntax of mathematics!
So the horizontal dash can have different meanings. In 4  3 it’s part of the name of a
number. In 4  3 it’s a subtraction sign. It might be helpful to have different words for the two
65
different meanings. In Lecture #8 we said that  3 could be called either “minus 3” or “negative
3.” From now on let’s stick to “negative 3” and leave the “minus” to indicate subtraction. Put it
all together and you get “4 times negative 3” for 4  3 and “4 minus 3” for 4  3 .
But wait, there’s more! That little dash has a third meaning: it’s an operator which
produces the opposite of a number. So   3 would be pronounced “the opposite of negative
3,” and I hope you can tell that it equals 3. Would it be so terrible to pronounce it “minus
negative 3?” Probably not, as long as you see it equals 3. How about “negative minus 3?” It
might not make you get the wrong answer, but it sounds bad to me, as though the answer itself is
going to be negative, which it clearly isn’t. Just like other languages, mathematics has style and
customs.
It might help you to understand that   3 is equal to 3 if you insert a 1 after the first
horizontal dash:  1 3. Can you see that this is implied multiplication, which could also be
written  1  3 , and that the answer is still 3, by the algorithm for multiplying signed
numbers? I call it the “invisible 1,” and it can be inserted whenever multiplication is implied,
because 1 is that special number of multiplication: multiplying a number by 1 does not change
the number at all. Its official name is the “identity element for multiplication.”
Let’s carry this one step farther. Consider this expression:    3 . The inside
parentheses don’t have anything inside them that must be done, so we start with the outer ones:
   3 

3

3
This might be pretty obvious to you: You take 3, change it to  3 (using the right-most dash),
then back to 3 (using the dash in the middle) and finally back to  3 (using the left-most dash).
In fact it’s probably also clear to you that an odd number of dashes means you end up with a
negative number, while with an even number of dashes the answer is positive.
One more little point here. How do you know whether  3 should be pronounced
“negative 3” or “the opposite of 3?” Here’s the great thing – it doesn’t matter a bit! It’s still the
same location on the number line, midway between  4 and  2.
Let’s move on to another grouping symbol, the fraction bar. Of course you know it
indicates division, but if either the numerator or denominator contains an operation, some
64
. Both parts of the
26
fraction contain operations. The effect of the fraction bar is to enclose each in parentheses.
5
1
Written horizontally it would be  6  4  2  6 , which is 10   4 , or , or 2 .
2
2
simplification must be done before the division. Here’s an example:
If you write the fraction bar as a slash or a division sign, you have to enclose both parts
in parentheses. To write  6  4 / 2  6 or  6  4  2  6 would indicate that 4 is to be divided by
66
2, so you’d get  6  2  6  14 instead of the correct answer. You have to write
 6  4/2  6 or  6  4  2  6 .
Another grouping symbol is the radical sign,
. You know what this is: 25 is the
positive number which, when multiplied by itself, equals 25. It’s 5, called the square root of 25.
(The word radical comes from the Latin word for root, radix, which reminds us of our word
“radish,” which is a root.) So how is the radical sign a grouping symbol? Well, if there’s an
operation under its roof, that operation must be performed before the square root is taken.
Compare 9  16 and 9  16 . In the first version there’s nothing to do under the roof, so this
expression is 3  16  19 . In the second version the 9 and the 16 must be added before the square
root is taken: 9  16  25  5 . It’s as though there were parentheses enclosing the radicand
(the expression under the roof, from the Latin for “the thing that is being rooted”):
9  16 .
The last grouping symbol we’ll consider is the absolute value operator. You know one
meaning of absolute value – the number part of a signed number, or the signed number without
its sign. But there’s a way of directing someone to change a number into its absolute value. (Of
course if a signed number is positive it won’t change anything, because positive is the default
sign.) The symbol for this is two vertical lines, one on either side of the number. For instance,
3  3 , and  3  3 . When writing the absolute value operator, be sure to make the vertical
lines long enough that the second example won’t look like you wrote l-3l, or someone would
think you meant  30 .
Another interpretation of absolute value is how far the number is from 0. In the case of 3
and  3 , they’re both three units from 0, though of course in opposite directions.
So how is the absolute value operator a grouping symbol? The same way the radical sign
is – if there’s an operation to do inside it. So 6  8   2  2 . If the problem is 8  6 , make
sure that after you get 2 you don’t leave it like that; go ahead and write what the absolute
value of 2 is, namely 2.
Exponents
I’m sure you’re wondering what happened to Step 2 of the order of operations algorithm.
Well, here it comes:
2) Apply exponents.
Remember when I described multiplication as repeated addition? Well, exponents
indicate repeated multiplication. Say you want to multiply five 2’s. You could write
2  2  2  2  2 and get 32, but there’s a more compact notation: 2 5 . The 2 is called the base,
the factor to be repeated, and the 5 is called the exponent, indicating how many copies of the
base will be multiplied. It’s pronounced “two to the fifth power,” or just “two to the fifth”. A
67
couple of exponents have nicknames: 4 2 can be pronounced “four squared,” and 4 3 “four
cubed.” These are references to area and volume.
If you typing an exponential expression (an expression which contains an exponent) on
a keyboard, indicate the exponent by using the caret, the upside-down V which is uppercase on
the 6 key, like this: 4^2.
So exponents are applied before any multiplications, divisions, additions, or subtractions
are performed. For example, 3  4 2 is 3  16 , or 19, not 7 2 , or 49. If you want it to be 49, you
2
have to enclose the 3  4 in parentheses, like this, 3  4  , because Step 1 causes the operation
within the grouping symbols to be performed first.
Likewise, 3  4 2 is 316 , or 48, not 122 , or 144. If you want it to be 144, you’d write
3 42 .
Here’s one that drives algebra students crazy:  52 versus  5 . In the first case, the 5
is the base, and after it’s squared the minus sign turns the result into its opposite:  52  25 .
You can think of the minus sign as implied multiplication by  1 ,  1 5 2 , and multiplication
doesn’t happen until Step 3. Another way is to realize that the exponent goes only as far to the
2
left as it has to. In  5 the parentheses keep the 2 from applying only to the 5. In this case,
2
2
 5 is the base, and it’s to be multiplied by itself twice:  5  5   5  25 .
So now we have all four steps of the order of operations algorithm:
1)
2)
3)
4)
Perform operations within grouping symbols.
Apply exponents.
Perform all multiplications and divisions in order from left to right.
Perform all additions and subtractions in order from left to right.
It’s deceptively simple, and only through practice and strict adherence to the steps will
you arrive at correct answers. But as stated before, it enables even the most complicated
expression to be simplified to the same answer by everyone who uses it correctly.
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Worksheet for Lecture #9
Complete the following problems by hand (step-by-step) and then verify your result with your
calculator.
1) 6  2 8  5
2)
4)
1 5
 4   6 

3) 7  4 5  32
289  81
5) 18  2  7

6) 1  3  4 2
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Assignment for Lecture #9: Order of Operations and the Syntax of Algebra
1) What would happen if Step 3 and
Step 4 of the Order of Operations
were reversed, i.e. if addition and
subtraction were done before
multiplication and division? Would
this be a problem even if everybody
agreed to the new order? Pick two of
the exercises below, and do them the
way they would be done if Steps 3
and 4 were reversed.
11) 5  3 1  2 
3

12) 5  3 1  23
13)
169  25
14)
169  25
15)
169  25
Simplify using the order of operations:
2) 7  3 5  1

3)  7  3 5  1
4)  7  3  5  1
16) 20  8  3
5)  7  3  5  1
6)
26
3   1
7)
26
3   1
17) 20  8  3
18) 20  8  3
19)  20   8  3
20) 4  6  2 5
8)
26
 3   1
21) 4  6  2 5
9)  2  6  3   1

10) 5  3 1  23
22)  4  6  2 5

70
Lecture #10: Evaluating and Simplifying Algebraic Expressions
Besides signed numbers, the other big distinction between arithmetic and algebra is that
algebra often uses letters, chiefly x, to represent numbers whose identity we don’t know. These
letters are called variables. Some people find this scary and maybe some kind of trick, but it’s a
very useful thing to do.
Let’s say you’re two years older than your friend. So when your friend was 15 you were
17; when your friend was 16 you were 18; when your friend was 17 you were 19; when your
friend was 18 you were 20…..I could go on and on, but instead I say, “Let’s call your friend’s
age x. Then your age is x  2 .” Or I could say, “Let’s call your age x. Then your friend’s age is
x  2 .”
Let’s use the first version, with your friend’s age being x. Then if I want to find out how
old you’ll be when your friend is 25, all I have to do is substitute 25 for x in the expression x  2
and get 25  2 , or 27. We call this evaluating the algebraic expression x  2 for x  25 , and
it’s something we do all the time in algebra.
Sometimes we have enough clues to find out what number or numbers the letter stands
for. This happens when we solve equations, the subject of the next lecture. But in this lecture,
we’re interested only in evaluating expressions and in writing them in as simple and compact a
form as possible.
Evaluating Algebraic Expressions
Here’s an algebraic expression: 3 x  5  4 x . It could be equal to anything, depending
on the value of the variable x. What if x  2 ? To find out is to evaluate the expression for x  2
.
(Here you need to understand that 4x means 4 times x – implied multiplication. Whereas
I could have put a multiplication sign between the 4 and the x, 4  x , or enclosed the x in
parentheses, 4 x , I don’t have to, unlike when you write 4 times a constant, for instance 4 3 or
4 3 , and you have to avoid confusion with the number forty-three, 43. The expression 4x does
not mean forty-x, with the x in the ones place.)
First, let’s put in the multiplication signs where there is implicit multiplication. Only
now, instead of using a  for multiplication, we’re going to use a raised dot . . We don’t want
to use the  because it might be mistaken for an x. Just make sure that your raised dot doesn’t
look like a decimal point!
Here it is: 3  x  5  4  x Next replace each x by a 2: 3 2  5  4  2
Now follow the order of operations to simplify the expression and find its value.
71
Work inside the first set of parentheses, which are the only grouping symbols in the expression:
3  3  4  2 . Then do the multiplications from left to right:  9  8 . Now there’s only one
operation left, addition, which gives the final answer of  1.
Putting all the steps together, we get
3  2  5  4  2  
3   3  4  2  
9  8 
1
Now let’s evaluate the same expression for x  1 :
3   1  5  4   1 
3   6  4   1 
 18   4  
 22
Here’s a somewhat more complicated expression: 2 3  x  5 x  4 . Let’s evaluate it
for x  3 :
2  3  3  5  3  4  
2  0   5  7  
0  35

 35
Now let’s evaluate 2 3  x  5 x  4 for x  5 .
2  3   5  5   5  4  
2  8  5   1 
16  5

21
(We had to put the first  5 in parentheses because we can’t have two horizontal dashes
side by side with nothing in between. The parentheses around the  5 are visual separators
rather than grouping symbols.)
Were you surprised to see 16  5 in the second-to-last line? I could have written
16   5 first, and you should if you wish, but I think you’re better off going directly from
something like 16  5  1 to 16  5 . You can recognize and execute this pattern, and you want it
to be as second-nature as language.
Here’s one last example, using the expression 4 x  x  3 . First let x  5 .
72
4  5  5  3 
45  2 
20  2 
18
I’m sure you can find various correct shortcuts in the process, and feel free to use them. For
instance, even though we simplify expressions within parentheses before multiplying outside of
them, in the third line it would work just fine to multiply 4 and 5 at the same time as subtracting
3 from 5, making the third line unnecessary. That’s because the multiplication is entirely
separate from the parentheses.
Let’s evaluate 4 x  x  3 for x  6 :
4   6   6  3 
 24   9 
 15
Simplifying Algebraic Expressions
Sometimes you don’t want to evaluate an algebraic expression for a certain value of x;
instead you want to write the expression as compactly as possible while still retaining the
variable x. This process is called simplifying, and it’s an indispensable skill in solving
equations.
Let’s look again at the first example of evaluating an expression in this lecture. This
expression was 3 x  5  4 x , and when you were asked to evaluate for x  2 you replaced the
x’s by 2’s and eventually worked out that the expression had the value  1 . Among other
operations, you were able to subtract 5 from 2 and get  3 .
But what if you didn’t know what x was, so you couldn’t work within the parentheses
first, as the order of operations rules dictate, because you don’t know what x  5 equals? You
might think you’re stopped dead in your tracks at this point, but you’d be wrong. There are
things we can do, not to get a number value for the expression, but to make the expression look
simpler and easier to handle.
To do these things, you have to understand two processes: distribution and combining
like terms.
First, distribution. Let’s say you take 4 girls and 6 boys to an arcade. You plan to give
each child $5 to spend (obviously, your stay will not be a long one). So, with 10 children, and at
$5 per child you are going to be handing over $50. What if you decide to give the girls their
money first, and then the boys? $5 per girl, 4 girls works out to $20, and it’s $30 for the boys,
for a total of . . . $50. Either way you do it, it’s the same amount of money. I’m sure this doesn’t
come as a surprise to you.
But look at what you did as expressed in algebraic notation:
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5 4  6  5  4  5  6
or
5 10   20  30
50

50
Instead of adding first and then multiplying, which would be the way the order of operations
rules would have you do the problem, you can do two separate multiplications and then add, and
you get the very same answer!
So distribution concerns the connection between multiplying and adding (or subtracting),
and it allows you to make a substantial rearrangement of the number and order of the operations
and still come up with the same answer. Obviously there’s no reason to do it if you can actually
follow the order of operations, i.e., work within parentheses first, but if you can’t it allows you to
move forward. In the case of 3 x  5  4 x , we can distribute the 3 over the x and the 5:
3  x  5  4 x 
3  x  3  5  4x 
3x  15  4 x
Some people like to draw curved arrows to remind them to multiply both the x and the 5
by 3:
Now consider the effect of distribution on the expression 4 x  x  3 . What exactly are
we distributing over x  3 ? I call it “the silent 1,” and in this case it’s actually a  1 . You can
put a factor of 1 in to remind yourself:
4x  1 x  3
You are really multiplying the x and the  3 by  1 , and, remembering that the product of two
negative numbers is positive, you get
4x  x  3
There are more complicated ways to do this, involving more steps, but you’re better off skipping
them. Just remember that the effect of distributing a negative number is to change the signs of
all terms in the parentheses. The x was really a  x , and it became  x ; the  3 became  3 .
Now for combining like terms. This is really just distribution turned around. As your
grade school teachers probably said, if you have 5 apples and then get 3 more apples, you have 8
apples. The same applies to x’s: 5x  3x  8x . By the same token, 2x  6x  4x , because
2x  6x  2x   6x , and 2   6  4 . You’re better off not thinking in terms of subtraction,
but rather seeing the dash to the left of the 6 as a negative sign.
Can you see why 5x  x  6x and  x  9x  8x ? Remember the silent 1.
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Although 5x  3x  8x , 5  3x does not equal 8x, and neither does 5x  3 . Making
equal
8x might be tempting, because we read from left to right, but remember the order of
5  3x
operations rules, in which multiplication (even implied multiplication) preceded addition. The 5
and the 3 never get a chance to be added, because the 3 is joined to the x in multiplication.
Terms without a variable (called constant terms – they’re just plain numbers and don’t vary)
can never be combined with terms with variables.
Another way to see this is to realize the only justification for simplifying algebraic
expressions: If you pick a value of the variable and evaluate the before- and after-versions of the
expression for the value, you get the exact same answer. Take 5x  3x , 8x , and 5  3x , and
evaluate all three for x  10 . The first two give the answer 80, but the third one equals 35. So
it’s perfectly correct to rewrite 5x  3x as 8x , but not to do the same to 5  3x .
What about expressions whose terms contain different variables? Consider 5 x  3 y .
Well, you have 5 of these and 3 of those, but you don’t have 8 of anything. As your gradeschool teacher would have said, “It’s apples and oranges.” It’s not that 5 x  3 y can’t be
evaluated if we’re given values for x and y; for instance, for x  2 and y  6 , 5 x  3 y is 28. But
there’s no way to compact the expression with the result that the “simplified” version has the
same value for the same values of x and y.
Remember where we left off with simplifying the expression 3x  5  4x . We had
gotten as far as 3x  15  4 x by distributing. Now we can finish the process by combining like
terms: 3x  15  4 x  7 x  15 . This could also be written  15  7 x , though it takes one more
symbol this way, because you have to write the minus sign to the left of the 15, whereas you
don’t have to write the plus sign to the left of 7x if the expression starts with 7x.
In the section in this lecture on evaluating algebraic expressions, 3x  5  4x was one
of the expressions we evaluated, for x  2 and x  1 . For x  2 we got  1 , and for x  1 we
got  22 .
The crucial thing about simplifying expressions is that the result when evaluated for a
certain value of the variable has to be equal to what it was when the original expression was
evaluated for the same variable. Check it out for 7 x  15 :
7 2  15  14  15  1 and 7  1  15  7  15  22
Algorithm for Simplifying Algebraic Expressions
1) Distribute (remove parentheses).
2) Combine like terms.
Let’s use this algorithm on the other expressions we evaluated earlier in this lecture.
First, there’s 2 3  x  5 x  4 :
75
1) 6  2x  5x  20
2)  7 x  14 , or  14  7 x
Be sure you see why there’s a minus sign to the left of the 20 in Step 1.
Now let’s make sure our answer in Step 2 is correct. When we evaluated the original
expression for x  3 we got  35 ; for x  5 we got 21. Let’s evaluate  7 x  14 for those values
of x.
 7 3  14  21  14  35 and  7  5  14  35  14  21
So our answer  7 x  14 is correct. (Remember that  35 are 21 are not the answers; they are
the checks.)
The last expression we used was 4 x  x  3 . Using the algorithm, we get
1) 4 x  x  3
2) 3x  3
When we evaluated the original expression for x  5 we got 18; for x  6 we got  15 . Now
let’s use those values of x in our answer, 3x  3 .
3  5  3  15  3  18 and 3   6  3  18  3  15
This validates the fact that our answer, 3x  3 , is in fact the correct simplification.
76
Worksheet for Lecture #10
For each expression,
a) Evaluate it for x  3
b) Evaluate it for x  5
c) Simplify it using the algorithm
d) Evaluate the simplified expression for x  3
e) Evaluate the simplified expression for x  5
Show each step of your work and clearly label parts a – e.
1) x  14  3x  8
2)
4x  5  7 x
3) 2 x  1  5 x  3
77
Assignment for Lecture #10:
Evaluating and Simplifying Algebraic Expressions
Label the parts of your answers (a, b, c, etc.) clearly.
For each expression,
a) Evaluate it for x  3
b) Evaluate it for x  5
c) Simplify it using the algorithm
d) Evaluate the simplified expression for x  3
e) Evaluate the simplified expression for x  5
Make sure that your answers for (a) and (d) are the same, and for (b) and (e).
1) 2x  x  5x
2) 3x  12  5x  15
3) 5 2 x  3  10
4) 3 x  1  4 x
5) 5 2  x  7 x
6) 3x  x  4
7) 4 x  1  5 x  2
8) 2 x  3  x  5
9) 6 3  x  3 x  4
10) 5 x  3  7 x
78
Lecture #11: Solving Linear Equations in One Variable
An equation is just two algebraic expressions connected by an equals sign (=). “One
variable” means just that – there is only one letter, often an x, though it may appear more than
once, but never an x and a y, say. “Linear” means there is no exponent attached to the x, and the
x doesn’t appear in the denominator of a fraction or under a radical sign.
To solve an equation means to find the value of the variable which, when you evaluate
the expressions on both sides of the equation for that value, you get the same number. Some are
really easy to solve. You can tell just by looking that the solution to the equation x  2  7 is
x  5 . It’s obvious. We call this solving an equation by inspection. It’s the best way, if it
works. Here’s another: 3x  12 . As long as you remember that the lack of a sign between the 3
and the x means that they’re multiplied, you’ll know the solution is x  4 .
But if the equation is more complicated, like 4 x  3  x  2  3  2 x  4 , it’s hard to
solve by inspection, and you need some tools and an algorithm to be sure of finding the answer.
The first tool is called the addition rule. Think of a scale where the two sides balance
each other. For our equation x  2  7 it would look like this:
The two sides weigh the same. So if we remove the same weight from each side, the scale will
still be balanced, and both sides will still weigh the same (though less than they did before).
Let’s remove 2 pounds, or grams, or whatever the unit of weight is from each side. The scale
will now look like this:
Hopefully it’s obvious from this simple-minded analogy that the solution is x  5 .
Here’s how we write it out without the scale:
x27
2 2
x5
We subtract 2 from (or add  2 to, since it amounts to the same thing) both sides of the equation
(remember, the equals sign separates the two sides from each other). On the right, we do a little
arithmetic: 7  2  5 . On the left, we combine like terms: x  2  2  x  0  x . The number 0 is
a special number for addition. It’s called the identity element for addition, because when you
add 0 to a number, the number retains its identity. It doesn’t change. I call this process making
79
a 0. You don’t have to write out the step x  0  x , but you have to make sure that you are in fact
making a 0.
Here are a couple of other examples:
x  4 1
4 4
and
x  7  3
7 7
x  3
x4
You might find it a little harder to solve these equations by inspection.
The second tool is the multiplication rule. I’ll skip the drawing of the scale, but I hope
it’s clear that if two things weigh the same, then dividing their weights by 3 or some other
number will result in their still weighing the same. So, if 3x  12 , then
3x 12
, and

3
3
x4
3x
There’s really a middle step,
 1x  x . I call this making a 1, because 1 is the special
3
number for multiplication, the identity element for multiplication, as stated in Lecture 9. (You
might think I should have called this rule the division rule, because I divided both sides by 3.
1
But remember that dividing by 3 is the same as multiplying by , so I could have written
3
1
1
 3x   12 . I’ll call it the multiplication rule whether you multiply or divide, just as I call
3
3
the first rule the addition rule whether I add or subtract.
Here are more examples:
 4 x  24
 4 x  24

4
4
x6
and
x
3
5
x
5  53
5
x  15
and
2
x  18
3
2
x
3  18
2
2
3
3
x  27
In the example on the left, since the coefficient of x is  4 , you have to divide both sides by  4 ,
not just 4 (if you divided by 4, you’d get  x or 1 x , which is not the same as x). In the
middle, since the equation has a 5 in the denominator, you have to multiply by 5 to make a 1, or
x
1
1
you may think of as x , and thus divide both sides by . On the right, you may have been
5
5
5
3
2
taught to multiply by the reciprocal, , instead of dividing by .
2
3
80
The above equations are called one-step equations, because you have to perform just one
operation on both sides to isolate the x -- get it alone on one side – so you can read off the
answer on the other side. But equations can sometimes require a lot of steps, like our example
4 x  3  x  2  3  2 x  4 . In that case, you need a systematic approach.
Algorithm for Solving Linear Equations in One Variable
1) Simplify each side.
2) Get variable terms on one side and constants on the other by making 0’s.
3) Isolate the variable by making a 1.
Here’s an example: 5x  7  2x  8 .
Step 1 is already done, because each side is as simple as it can get, with a variable term
and a constant term.
Step 2 can be done in four different ways. You can make 7 into a 0 by adding  7 to each
side. Be sure to write the  7 on the right side directly below the  8 to emphasize that  8 is the
term  7 will be combined with.
5x  7  2 x  8
7
7
5 x  2 x  15
Look what’s happening. The only constant term,  15 , is on the right side, so you have to get the
variable term on the left side. You have only one next move: subtract 2x from both sides to
make a 0.
5 x  2 x  15
 2x  2x
3 x  15
(A common mistake is to subtract 2 instead of 2x . But 2x  2 is most definitely not equal to 0.)
That completes Step 2. For Step 3 you must divide both sides by 3 to make a 1.
3x  15

3
3
x  5
As mentioned above, Step 2 could begin in three ways besides the one we used. One way
would be to subtract 5x from both sides. Can you finish this? What are the other two ways you
could begin Step 2, and how would they proceed?
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None of the four ways is better than any other, though I find it helpful to use one that
ensures that the coefficient of x is a positive number, which makes it less likely that I’ll forget
the negative sign when dividing by the coefficient. In future examples, I’ll just begin with
whatever step I feel like using to make 0’s.
Here’s an example that needs some work in Step 1: 3 x  2  x .
1) Simplify both sides. Only the left side needs it, so we follow the algorithm for
simplifying algebraic expressions.
1) Distribute.
3 x  2  x
3x  6  x
And that’s as far as we can go, because there are no like terms.
2) Since there’s only one constant term,  6 , and it’s on the left side, we’ll get the x terms on
the right by subtracting 3x from both sides to make a 0.
3x  6  x
 3x  3x
 6  2 x
3) Now we’ll isolate x by making a 1.
 6  2x

2 2
3 x
It doesn’t matter which side the x is on. It equals 3 either way.
Now comes the great thing about equations: they are fully checkable. Remember that a
number is the solution to an equation if, when each side is evaluated for it, you get the same
result. Always use the original equation to do this, in case you made a mistake in one of the
steps.
3 3  2  3
3 1  3
3 3
Thus we’ve shown that 3 fulfills the conditions for being the solution to the equation.
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How about that complicated equation I gave earlier in the lecture to show that solution by
inspection is not always possible? It was 4 x  3  x  2  3  2 x  4
1) Simplify both sides. They both need it. You can work on both of them at once.
4 x  3  x  2  3  2 x  4
4 x  12  x  2  3  2 x  8
3x  14  11  2 x
And that’s as simple as it gets. First we distributed, and then we combined like terms.
Be sure you do these steps correctly, because one little mistake can lead you to the wrong
solution. (Of course, you’ll find this out when you perform the check, and then you can
troubleshoot your work. Did you multiply correctly; did you remember to change the
sign when a set of parentheses is preceded by a negative sign; did you combine the like
terms correctly; etc.)
2) You have four choices to begin sorting out the constant terms and the x terms. I’ll add
2x to both sides so that the x term winds up with a positive coefficient.
3 x  14  11  2 x
 2x
 2x
5 x  14  11
At this point it’s clear that the constants belong on the right, so we’ll add 14 to both sides.
5 x  14  11
 14  14
5 x  25
3) Isolate the variable.
5 x 25

5
5
x5
And we’re done, except for the check:
4 5  3  5  2   3  2 5  4 
42  7
8  7
1
 3  2 1
 3 2
 1
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Contradictions and Identities
Here are two little curiosities. They’re not very important, but you might find them
interesting.
Some linear equations have no solutions whatsoever. No number can be found which
will make both sides evaluate out to the same answer. We call them contradictions.
Here’s one: x 1  x . It’s just asking too much of a number that it be equal to itself plus
1. But let’s say you don’t notice this, so you set about solving it using the algorithm. Step 1 is
done, and for Step 2 you decide to get the x term out of the left side, since the only constant is on
that side.
x 1  x
x
x
1 0
That last statement is definitely not true, and no value of x will make it true, because the x’s have
entirely disappeared from the equation. That’s the sign that this equation is a contradiction, and
the solution to it is written as “No solution,” which also has the delightful symbol Ø, also called
the empty set. (But don’t make your zeroes with slashes, or people won’t know if you mean
zero or the empty set.)
The other curiosity is the equation for which all numbers work. These are called
identities. You might not realize at first that the equation 2 x  3  2x  6 is an identity, but
after Step 1 of the algorithm it should become obvious.
1) Simplify both sides.
2  x  3  2 x  6
2x  6  2x  6
If not, you might go on to Step 2 and decide to subtract 2x from both sides.
2) Get the variable terms on one side by making a 0.
2x  6  2x  6
 2x   2x
66
As with the contradiction, the x’s have disappeared from both sides, but this time we’re left with
a true statement, which means that any value of x would have made the original equation true.
Try putting a number in for x. How about 5?
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2 5  3  2  5  6
2  8  10  6
16  16
No need to try any others; they’ll all work. We say that this equation is an identity and that the
solution is all real numbers (don’t worry about the word “real” here), and its symbol is R.
85
Worksheet for Lecture #11
Solve each of the following equations.
1) x 
2)
4
2

5
3
5
x  20
8
3) 5  7 x  19
4) 8x  9  3x 16
5) 33x  4  3x  28
6) 10  2x  3  x  5
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Assignment for Lecture #11: Solving Linear Equations in One Variable
11) 4  3x  x
Solve these equations. Show your work.
Give answers as integers or fractions.
Leave improper fractions as is; do not
convert them to mixed numbers.
12)  3  4x  15
13)  0.3x  0.1  0.5
1) x  8  5
14) 7 x  4  2x  11
2) x  10  23
15) 5  2x  3x  1
3) x 
16) 5 2 x  1  12
2
3

3
8
17) 4 x  3  3x  9
3
5
4) x   
7
6
18) 3 x  1  5   x
5) 6x  48
19) 2 2x  3  2x  45
3
6)  x  12
5
20) 2 20  x  3 20  x
7)
21) 2x  3 3x  5  6
x
 3
7
22) 15  3 x  1  x  2
8)  6 x  9
23) 5 x  2  4 x  1  3
x
9)   5
4
10) 
24) 6 x  3  5 x  2  8
2
2 2
x 
5
15 3
25)
87
2
x  6  1 x  10  13
3
5
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Lecture #12: Modeling Situations with Linear Equations in One Variable
You might think from the title of this lecture that we’re going to be trying clothes on and
displaying ourselves, or maybe letting equations wear clothes, but the meaning of the word
modeling here is very different than in ordinary usage. It means to create an equation which
accurately and abstractly reflects the conditions given in a word problem.
Most math students simply hate word problems. That’s why almost all textbooks now
refer to them euphemistically as “applications,” hoping to avoid the negative effect of the phrase
“word problems.” Nobody is fooled.
Here’s one: Jo and Mo have a total of 22 shirts. Jo has 6 more than Mo. How many
shirts does each of them have?
The Greeks, who didn’t have equations and x’s at their disposal as we do now, solved
problems like this using trial-and-error (what we now call “guess-and-check,” presumably
because of the negative connotation of the word “error,” though guess-and-check is really a
better name). They would make a guess: Perhaps Mo has 6 shirts. So Jo, with 6 more than Mo,
would have 12. Then their total would be 18 shirts. Doesn’t check; too few. Okay, up the guess
for Mo to 10 shirts. Then Jo would have 16, for a total of 26. Doesn’t check again; too many.
So they’d try a number for Mo’s shirts between 6 and 10, and before too long they’d pick 8 shirts
for Mo, which means 14 for Jo, and now the total number of shirts is 22, which is correct. The
answer is that Mo has 8 shirts and Jo has 14.
This method works just fine, but it can be cumbersome and long. What if together they
had 175 shirts (quite the shirt collectors!), and Jo has 17 more shirts than Mo? It might take
quite a lot of guessing-and-checking before the correct solution of 79 shirts for Mo and 96 shirts
for Jo is arrived at.
There are other ways to get the solution which don’t involve algebra, but you’d have to
know them. In the first case, subtract 6 from 22, which gives 16, and then divide 16 by 2 to get
8, which is the number of Mo’s shirts. Add 6 to that and get 14, which is the number of Jo’s
shirts. In the second case, subtract 17 from 175, which gives 158, and then divide 158 by 2 to
get 79, which is the number of shirts Mo has. Add 17 to that and get 96, the number of Jo’s
shirts.
This works, but, again, you’d have to know to do this. So in algebra we use a different
technique entirely, which depends upon our knowledge of solving linear equations. Take the
first problem, where the total number of shirts is 22. First we represent the situation like this:
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Then we have to fill in the boxes. It would be nice to call the number of shirts Jo has x, and the
number of shirts Mo has y, or, even better, to call the number of shirts Jo has J, and the number
of shirts Mo has M, but we’re dealing with linear equations in one variable here, not two. (There
is a way to use two variables, but it involves the trade-off of making two equations, plus learning
how to solve what are called systems of equations, and I don’t think it’s worth the extra work.)
So we’re stuck with using only one variable, and we have to decide which it should stand
for, the number of shirts Jo has or the number of shirts Mo has. To me, this is an easy choice.
Look for the word in the problem that denotes comparison. In this problem, it’s the word
“than.” Jo has 6 more than Mo. Always let the letter represent the quantity referred to after the
comparison word. In this case, let the variable be M, the number of shirts Mo has.
Then what do we do about representing the number of shirts Jo has? We make an
algebraic expression involving M which translates the idea that Jo has 6 more shirts than Mo.
This expression is M  6 , which has the meaning of 6 more than M. It’s true that you could
write it 6  M , because addition is commutative (reversing the terms in a sum doesn’t change
the answer), but this could get you into real trouble when you have to translate phrases like “6
less than x. This phrase translates into x  6 , which is definitely not the same as 6  x . Check it
out: Use 10 for x. Then x  6 becomes 10  6  4 , whereas 6  x becomes 6 10  4 . A big
difference! And it should be obvious to you that 4 is in fact 6 less than 10, and  4 isn’t. So you
might as well get used to writing the phrase “6 more than M” as M  6 . This is also closer to the
meaning of the phrase than 6  M , because really what we’re looking at is the fact that to find
the number of shirts Jo has, we start with the number Mo has and then add 6.
Now fill in the boxes with the expressions we’ve constructed:
(Technically, the M  6 should be in parentheses, but because there’s no negative sign to the left
of the box for the number of shirts Jo has, we can omit them.)
Now omit the boxes and the words beneath them, and write what’s left:
M  6  M  22
Presto! It’s an equation, a model of the situation in the problem, and we can solve it.
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M  6  M  22
2 M  6  22
6 6
2 M 16

2
2
M 8
Aren’t you pleased that you were able to use your recently acquired skills in solving
linear equations? But don’t get too excited about it or too complacent, because we haven’t
finished solving the problem. First, we haven’t figured out how many shirts Jo has. But that’s
easy, because all we have to do is evaluate the expression for the number of shirts Jo has, M  6 ,
for M  8 : 8  6  14 .
Even now you’re not finished. The question didn’t ask what M equals, so even though
it’s obvious that M is the number of shirts Mo has, you have to answer the question in the terms
it was asked: How many shirts does each of them have? You could write complete sentences for
this: Jo has 14 shirts, and Mo has 8 (or Mo has 8 shirts and Jo has 14). Or you could write: Mo:
8 shirts; Jo: 14 shirts. Either way is fine.
I know this is a long, drawn-out process, but it’s valuable because it’s foolproof, does not
involve guessing, and gives you practice in translation from words to symbols, which is one of
the more important skills of mathematics.
Let’s try another, much more quickly: Cam makes $50 per week less than Sam makes.
Their total weekly income is $770. How much money does each of them make per week?
The comparative word again is “than,” so let’s use the variable for Sam’s weekly
earnings, because “Sam” comes after the “than”. Then Cam earns S  50 dollars per week.
Here’s the diagram, with the boxes filled in:
Remove the boxes and the writing beneath, and you get the equation S  50  S  770 . Notice
that we don’t put dollar signs in the equation. They would just clog it up. They belong in the
answer, though, as you’ll see.
You can solve this equation, and you should get S  410 , so that S  50  360 . Since the
question is how much money each makes, you have to include the dollar signs in your answer,
or it’s not complete: Sam makes $410 per week, and Cam makes $360 per week.
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A note about checking: If you’re going to go the trouble of checking your answer (and
you certainly should, after all this work), don’t check it in the equation, because your equation
could be a faulty translation of the situation. Check it in the words of the problem themselves.
In this case, it would go like this: “Cam makes $50 per week less than Sam makes.” Yes, Cam’s
earnings of $360 are in fact $50 less than Sam’s earnings of $410. “Their total weekly income is
$770.” Yes again, because the sum of $360 and $410 is $770. You can be absolutely sure that
your answer is correct.
A Different Kind of Situation
There are a whole lot of word problems and situations which all translate into the exact
same kind of equation. I hope you’ll see the similarities as we look at various examples.
Here’s the first one: You are saving for a new laptop, which costs $365. You already
have $50 put aside for the purchase, and you plan to take $15 from your earnings each week to
add to the fund. How long will it take you to save enough money to buy the laptop?
You can find the answer without using algebra. It’s simple. You have $50 already, so
you need only $315 more. At the rate you’re saving, $15 per week, all you have to do is find
how many 15’s make 315. You divide 315 by 15 and get 21. The answer is 21 weeks.
But we want to make an equation to model this situation and use equation-solving skills
to find the answer. If you save $15 per week, after one week you will have added 15 1  15
dollars to your fund; after two weeks 15  2  30 ; after three weeks 15  3  45 dollars; and so on.
So after w weeks you will have added 15w dollars. You started with $50, and you add 15w
dollars, where w is the number of weeks you’ve worked and set aside the $15, and you want to
know when that will equal $365. Here’s the equation:
50  15w  365
You solve it in the usual way:
50  15w  365
 50
 50
15w 315

15
15
w  21
You have to attach the word “weeks” to the 21, because the question was how long it would
take. To simply say “21” doesn’t say 21 of what. And don’t put the answer as w  21 , because
the question didn’t say anything about w. That was our invention.
Try this one: When you begin a taxi ride, the meter says $2.75. This means that even if
you change your mind and don’t go anywhere in the taxi you still owe $2.75, just for getting
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picked up and getting into the taxi. But for each eighth of a mile you go in the taxi, the meter
goes up by 25¢. You have exactly $7.00 at the moment. How long can your taxi ride be?
This can also be solved simply by using common sense. After deducting the initial $2.75,
you have $4.25 left for the eighth-mile charges. How many times 25¢ is that? Now you have to
be a little careful. Finding the quotient of 4.25 and 25 is not correct, because the units are
different. You either have to write $4.25 in cents, which is 425¢, or you have to write 25¢ in
dollars, which is $.25 (or $0.25). Either way, the quotient is 17.
So what’s the answer? Saying 17 isn’t adequate, and saying 17 eighths isn’t that helpful.
1 17
1
1
 2 , so the answer is 2
8 8
8
8
You’ll have to give the answer in miles. 17 eighths is 17  
miles.
How about the equation? Note how similar it is to the laptop example. You start with
$2.75, then for x eighths of a mile traveled you add on 0.25x dollars, and you want the total to be
$7.00. You get
2.75  0.25 x  7.00
 2.75
 2.75
0.25 x 4.25

0.25 0.25
x  17
1
miles long.
8
Notice that in both cases, though the equation was about money, the variable and answer
involved very different things – time in the first problem and distance in the second.
Using the same process as before, the answer is that your taxi ride can be 2
Tax and Discounts
Here are two situations involving percents and money.
Let’s say you buy a taxable item that costs $45.00. The applicable tax rate is 6%. (What
is taxable and what the tax rate is, depend on the state, county, and sometimes even the city in
which you make the purchase.)
I’m sure you know how to figure out the total amount you owe: First find 6% of 45. It’s
2.7. So the tax is $2.70. Add that to the $45.00, and you get the answer, which is $47.50. The
arithmetic looks like this:
45.00  0.06 45.00  45.00  2.70  47.70
But what if you are told that you paid $47.70 for an item, including the 6% tax, and are
asked to find what the item cost pre-tax? Many people incorrectly think that you take 6% of
47.70, or 2.86 (to the nearest cent), and subtract that from $47.70, for an answer of $44.84.
Clearly that’s not right, since we know that the item in question cost $45.
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What’s wrong about this method? Well, since the item didn’t cost $47.70 pre-tax, taking
6% of that amount doesn’t represent the situation. You’re not being taxed on $47.70, but rather
on the $45.00 price. So what do we do? Well, consider the expression 45.00  0.06 45.00 . We
don’t know the original price, so let’s call it x. Then the expression becomes
x  0.06x
and we know that it equals 47.70. So the equation is
x  0.06x  47.70
We combine the like terms, remembering that x  1 x :
x  0.06 x  47.70
1.06 x 47.70

1.06
1.06
x  45.00
As we already knew, the answer is $45.00.
Let’s try this on a problem for which we don’t know the answer. An item which is taxed
at the rate of 7.5% winds up costing $79.55. What was its original price?
x  0.075 x  79.55
1.075 x 79.55

1.075 1.075
x  74.00
The answer is $74.00. This checks, because 7.5% of $74.00 is $5.55, which when added to
$74.00 gives $79.55.
These equations show that all you have to do to find the original price is to divide the
price which includes the tax by 1 followed by the tax rate expressed as a decimal.
Now let’s look at what happens when a store has a sale. It knocks off, or discounts, a
certain percent of an item’s former price. For instance, in a 25% off sale, an item which
formerly sold for $64 will now cost $48, because 25% of 64 is 16. Subtracting the discount of
$16 from $64 gives the sale price of $48.
64.00  0.25 64.00  64.00 16.00  48.00
What if we didn’t know the original price? We’d do what we did in the case of taxes or
mark-ups: call the price x.
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x  0.25 x  48.00
0.75 x 48.00

0.75
0.75
x  64.00
The original price was $64.00, which we already knew. (The wrong way to do this problem is to
take 25% of $48.00, which is $12.00 and add that to the $48.00, the result being $60.00. This is
wrong because the discount wasn’t 25% of the sale price; it was 25% of the original price.)
Let’s end with a situation where we don’t know the original price. An item is on sale for
30% off and ends up costing $86.80. What was the original price?
x  0.30 x  86.80
0.70 x 86.80

0.70
0.70
x  124.00
The item originally cost $124.00.
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Worksheet for Lecture #12
For each problem, clearly follow these steps:
a) state what the variable represents
b) write the equation, and
c) answer the question.
Step (c) should not include letters used as variables.
The problem numbers refer to the assignment for Lecture #12.
5) Cam makes $80 per week more than Sam makes, and their weekly total is $830. What is
Cam’s weekly income, and what is Sam’s weekly income?
10) You start with $210 and save $8 per week. You want to save $306. How long will it take
you to save this amount?
12) When 8% sales tax is included, an item costs $15.66. What did the item cost pre-tax?
17) After a 40% discount, an item cost $45.00. What was the item’s original cost?
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Assignment for Lecture #12:
Modeling Situations with Linear Equations in One Variable
For each problem,
a) state what the variable represents
b) write the equation, and
c) answer the question.
How many shirts do Jo and Mo each have?
1) Jo has 14 fewer than Mo, and together they have 40.
2) Mo has twice as many as Jo, and together they have 33.
3) Jo has 3 more than twice as many as Mo, and together they have 33.
4) Mo has one third as many as Jo, and together they have 48.
What is Cam’s weekly income, and what is Sam’s weekly income?
5) Cam makes $80 per week more than Sam makes, and their weekly total is $830.
6) Sam makes $10 less than twice as much as Cam makes per week, and their weekly total
is $410.
7) Cam makes three times as much per week as Sam makes, and their weekly total is $640.
8) Sam makes half as much per week as Cam makes, and their weekly total is $570.
How long will it take you to save the amount you want?
9) You start with $75 and save $40 per month. You want to save $275.
10) You start with $210 and save $8 per week. You want to save $306.
11) You start with $52 and save $26 per month. You want to save $286.
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What did the item cost pre-tax?
12) When 8% sales tax is included, an item costs $15.66.
13) When 7% sales tax is included, an item costs $17.12.
14) When 6.5% sales tax is included, an item costs $25.56.
What was the item’s original cost?
15) After a 20% discount, an item cost $36.00.
16) After a 35% discount, an item cost $53.30.
17) After a 40% discount, an item cost $45.00.
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Lecture #13: Linear Inequalities in One Variable
Sometimes we’re not interested in equations – making two expressions equal to each
other – but rather in making one expression be less than another or greater than another. In these
cases we have an inequality instead of an equation. But before we get to solving inequalities,
we need to examine the very nature of the relationship of numbers.
The Trichotomy Property
Here’s something very basic about numbers. You have probably known it
subconsciously since you were a small child but have never articulated it.
You may know that a dichotomy is a situation in which there are two choices or
alternatives. So a trichotomy has three choices or alternatives. If you have two numbers, a and
b, then exactly one of these three alternatives is true: either a  b , or a  b , or a  b . One of
these three statements must be true for any two numbers a and b, and the other two must be
false.
Any two of the statements can be combined in one sentence, as this table shows:
In words, if you want to express the possibility that a could equal b or that a could be less
than b (but a is not greater than b), you write a  b . This is pronounced “a is less than or equal
to b.” (Note that this is an either/or statement – you’re not saying that a equals b and a is also
less than b. Couldn’t happen.) If you want to express the possibility that a equals b or a is
greater than b (but a is not less than b),, you write a  b , which is read “a is greater than or
equal to b.” If you want to express the possibility that a is less than b or that a is greater than b
(but a is not equal to b), you write a  b , read “a is not equal to b.”
The point is that although the Trichotomy Property presents three alternative
relationships for a and b, three other relationships are implied, and these are the ones in which
two possible relationships are combined. Thus there are six symbols altogether that express the
relationship between the size (or position, if you’re thinking of a number line) of two numbers:
, , , , , and  .
Here are some examples of true inequalities:
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35
35
53
53
35
53
Note that the two statements in each row express the same idea, but from a different
perspective. If I write 3  5 and then add that 5  3 , I haven’t given any new information at all.
(If I told you that you are younger than I am, you would not be at all surprised to hear that I am
older than you.) In an equation, the expressions on either side can always change places without
affecting the equals sign (if x  2 then 2  x ), but if you switch the sides of an inequality, <
becomes >, and  becomes  , or vice versa.
Solutions on the Number Line and Interval Notation
If you have an equation like x  2 , there’s only one answer, and writing x  2 is the
solution. Of course you could show it graphically on a number line by making a solid dot on the
line where the hash mark for 2 is located:
But that seems a little much, in most cases.
How do you show the solution for an inequality? Take x  2 for example. Well, you
can simply write x  2 , but what you can’t do is list the solutions. There are simply too many.
Even if you wrote x  ..., 2.0001, 2.0002, ... , just in that little part there would be another solution
between 2.0001 and 2.0002. The number 2.00015 comes between them and would also be a
solution to x  2 . We call the numbers on the number line dense, because between any two of
them there is another, and of course if this is true then you can find an infinite number of
numbers between any two numbers, no matter how close they are.
One way to show the solution is to use a number line. This time we make the part of the
line which contains the solutions thicker (and also the arrow to the far right). But how do we
show that 2 is not a solution to the inequality (because 2 is not greater than 2)? There are two
ways. The first way is, instead of making a solid dot on the line at 2, making a circle that isn’t
filled in, called an open circle:
Anyone who knows math knows that this graph shows that any number to the right of 2, no
matter how close it is to 2, is a solution.
The second way is to make a parenthesis-shaped curve through the hash mark at 2, facing
in the direction of the thickened line:
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Again, anyone who knows math will understand that this graph shows that all numbers to the
right of 2 are solutions. You can use either the open circle or the parenthesis in your graphs.
If the original inequality had been x  2 , so that 2 is a solution, it’s pretty obvious that
using the first method, you would place a closed circle (solid dot) at 2. For the second method,
instead of using a rounded parenthesis, you would use a bracket, again facing the right:
There’s another way to represent solutions to inequalities besides graphing, and it’s a
very useful and compact way, called interval notation.
Let’s start with using interval notation to express the solution to x  2 . Think of the
number line, and scan it from left to right. When you come to the thicker line, write the number
that marked the beginning of the thick part, in this case 2. To show that 2 isn’t included, make a
parenthesis to the left of the 2 (just like the second way of graphing the solution). Follow the 2
with a comma. So far we have (2, and now comes the interesting part. If the thicker line ended
somewhere to the right, we would just follow with that number. But it doesn’t end. It goes on
forever. We have a way to symbolize that, and it’s the mathematical symbol for infinity.
You’ve seen it; it looks like a horizontal 8 and was invented as a symbol in 1657 by the English
mathematician John Wallis. Technically, its shape is called a lemniscate, which comes from a
Latin word meaning “decorated with ribbons,” and you can see why:
To show that all numbers to the right of 2 are solutions to the inequality x  2 , after (2, you write
the infinity symbol, and then end with a parenthesis: 2,  . Do you see how efficient and
compact this notation is?
How about the inequality x  2 ? On the number line, the solution looks like this:
Notice that in this case we make a thick arrow on the left end of the number line, indicating that
the solutions go on and on in that direction. Also, now that you’ve seen interval notation, I’ve
decided to use the parenthesis rather than the open circle to indicate that solutions begin just
beyond a number, not at the number. The parenthesis has to face in the direction of the thick part
of the number line.
How do you write this solution in interval notation? For starters, not like this: , 2 .
Interval notation sweeps from left to right, and  means on and on to the right. To signify on
and on to the left, we stick a negative sign in front of the infinity sign. So the correct way to
write this interval is  , 2 .
What if the inequality has a  sign or a  sign? Maybe you can guess from the way we
graphed such inequalities on the number line. Yes! – Instead of using a parenthesis, you use a
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bracket. So the solution to x  2 is 2,  , and the solution to x  2 is  , 2 . But notice that
the infinities still have parentheses, because  and   aren’t numbers in the sense of having a
place on the number line. They are concepts – on and on forever to the right, and on and on
forever to the left.
Compound Inequalities
We’ll start with a common one which looks like this: 2  x  5 . The best way to
pronounce this is “x is between 2 and 5,” which it is.
On the number line, the solution to this inequality is
In interval notation, sweeping from left to right, this becomes 2, 5 .
I’m sure you can see that the solution in interval notation to 2  x  5 is 2, 5; for
2  x  5 it’s 2, 5 , and for 2  x  5 it’s 2, 5 . Incidentally, an interval like 2, 5 is called a
closed interval, because it includes its endpoints. An interval like 2, 5 is called an open
interval, because it doesn’t, and intervals like 2, 5 and 2, 5 are called half open (or half
closed).
Another kind of compound inequality contains the word or. Here’s one:
x  3 or x  7
Here it is on the number line:
You can see that there are two parts to the solution, two separate intervals. The one on the left
would be written  , 3 , and the one on the right 7,  . To show that numbers in both
intervals are solutions, we join them with a mathematics/logic symbol pronounced “union”:  .
(Easy to remember:  is for Union.) So the solution to this compound inequality is
 , 3  7,  .
One thing you can’t do with this system is to compact it into a single compound
inequality statement. You might be tempted, because x  7 can also be written 7  x , to hook up
the two parts by writing 7  x  3 , but this is wrong for two reasons. First, because of what we
call the Transitive Property of Inequality: if one number is less than a second number, and if
that second number is less than a third number, then the first number has to be less than the third
number. But 7 is not less than 3, certainly. Second, the use of the double < signs implies that the
two parts, 7  x and x  3 , are connected by an and, and of course there aren’t any numbers
which make both parts true.
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Solving Inequalities
So far we haven’t really solved any inequalities. We’ve just looked at ways the solutions
could be written, as graphs on a number line or in interval notation. Each inequality had the x
already isolated on one side of the inequality. Now we’ll solve inequalities where the x isn’t
isolated. It’s pretty much the same process as solving equations, so the steps will be familiar to
you, and we can use the same algorithm.
Here’s one: 3x  8  12 . You do what you would if the < sign were =. Think of a scale.
If one side is lighter than the other, it will remain lighter if the same weight is added to or
subtracted from both sides, or if both sides are multiplied or divided by the same number.
3x  8  12
8 8
3x 20

3
3
20
x
3
20
2



The solution in interval notation is   ,  , or   , 6  if you prefer mixed numbers.
3 
3


How about
1
x  3  18 ?
2
1
x  3  18
2
3 3
1
x
2  15
1
1
2
2
x  30
In interval notation this is 30,  . Note the bracket, because of the great-than-or-equal-to sign.
There’s one very annoying situation that can occur in solving linear inequalities, and it
happens when the coefficient of the x is negative. Say you have to solve  4 x  12 . So you go
ahead and divide both sides by  4 and get x  3 . That seems okay, but check it out: What if
you pick 0 as a number which is greater than  3 and substitute that into the original inequality?
 4 0  12
0  12
That simply isn’t true: 0 is not greater than 12.
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The reason the procedure breaks down has to do with the relative positions of numbers on
the number line. Look at this:
Whereas 1 is to the left of 2, so that 1  2 , when you take their opposites (which can be done by
multiplying them by  1 ),  1 is to the right of  2 , so  1  2 .
The usual method for solving inequalities where the coefficient of x is negative (which
means you need to divide or multiply both sides by a negative number in order to isolate the x) is
to remember to reverse the direction of the inequality sign when you do so, replacing < by >, >
by <,  by  , and  by  . Our example would look like this:
The solution  ,  3 is correct.
 4 x  12
 4 x 12

4 4
x  3
Now let’s solve a couple of compound inequalities.
Here’s one:
 3  2x  7  5
Whereas we’ve always referred to doing the same thing to both sides of an equation or
inequality, in this case there are three sides, or parts, and we must do the same operation on each
one. Other than this difference, the procedure is the same. First we’ll add 7 to all three parts,
and then we’ll divide each part by 2:
 3  2x  7  5
7
7 7
4
2x
12


2
2
2
2  x  6
In interval notation, the solution is 2, 6 . If you take any number between 2 and 6, including 2
and 6, multiply it by 2 and then subtract 7, you’ll get a number between  3 and 5.
And finally, here’s a compound inequality with or: 5x  7  2 or
First, solve them separately:
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1
x58
4
1
x58
4
5 5
5 x  7  2
7 7
5 x 5 or

1 
4  x  3 4
5 5
4 
x 1
x  12
These most definitely cannot be written in the “between” form. If you need to, graph these on
the same number line, and you’ll see that they are the two outside parts, or tails. So for interval
notation, we’ll have to use the  :  , 1  12,  .
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Worksheet for Lecture #13
Graph these inequalities and write them in interval notation.
1. x  8
2. x  1
3. 3  x  2
4. 1  x  5
5. x  4 or x  0
Solve these inequalities and write the solutions in interval notation.
6. 4  3x  2  13
7. 5x  4  1 or 2𝑥 + 3 ≥ −1
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Assignment for Lecture #13: Linear Inequalities in One Variable
1) Do an internet search on the origin of the word “trichotomy” and report your findings.
You might want to use the word “etymology” instead of “origin.”
2) How is solving an equation like solving an inequality? How are they different?
Graph the solutions to these inequalities
on a number line and write them in
interval notation.
Solve these inequalities and write the
solution in interval notation.
3) x  4
15) 3x  5  7
4) x  4
5) x  4
6) x  4
16)
2
x  1  6
5
17)
3
x  9  3
4
7)  7  x  4
18) 7 x  2  9
8)  7  x  4
19)  3  4x  1  7
9)  7  x  4
20) 0 
10)  7  x  4
11) x  7 or x  4
21)
12) x  7 or x  4
13) x  7 or x  4
14) x  7 or x  4
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1
x46
3
2
x  6  2 or 3x 1  5
3
108
Lecture #14: Equations and Inequalities with Absolute Value
In Lecture #9 you learned about the absolute value operator, the grouping symbol that
has you simplify the expression within it and then make it positive if it wasn’t already, keep it
positive if it already was, and make it equal 0 if it equals 0. The example I used was
6  8   2  2 . But what happens when the expression inside the absolute value operator
contains a variable, so even after you simplify it you can’t tell whether it’s positive, negative or
0?
Consider this equation: x  3 . Like all equations, to solve it we find numbers which
when substituted for x make the equation true. Obviously 3 works, and I hope you can see that
 3 also works, because  3  3 . You could say that x  3 is a compact way of writing two
equations, x  3 and x  3 . So, calling 3 the constant in this equation, we could say that the
way to solve the equation is to set x (which I’ll call the contents of the absolute value) equal to
the constant and to the opposite of the constant.
If the equation were a little more complicated, like x  5  2 I would use this algorithm:
Solving Absolute Value Equations
1) Set the contents of the absolute value equal to the constant, and set the
contents of the absolute value equal to the opposite of the constant.
2) Solve both equations.
Here’s the solution:
1) x  5  2 or x  5  2
x  5  2
x5  2
2)
5 5
x7
or
5 5
x3
So 7 and 3 are both solutions to the equation, and they check. Using 7 makes the contents of the
absolute value +2, and using 3 makes it  2 .
There is something to beware of, though. What if the equation were x  3 ? Following
the algorithm, you would make two equations, x  3 and x   3 , which is the same as x  3 .
You might think that 3 and  3 are the solutions, but of course neither of them work. Absolute
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values are never negative. In fact, the correct response to this equation is “No solution,” or  .
Let’s fix up the algorithm:
Solving Absolute Value Equations
1) A) If the constant is positive, set the contents of the absolute value equal to the constant,
and set the contents of the absolute value equal to the opposite of the constant.
B) Solve both equations.
2) If the constant is negative, the answer is “No solution” or  .
3) If the constant is 0, set the contents equal to 0 and solve the equation.
The reason that absolute values are never negative is that they are actually distances.
The equation x  5  2 can be pronounced, “The distance from x to 5 is 2.” There are two ways
that could happen: x could be 2 units to the right of 5 or 2 units to the left of 5. If it’s to the
right, we add 2 to 5 and get 7. If it’s to the left, we subtract 2 from 5 and get 3. This is exactly
what we did in solving the equation using the algorithm.
Since x  0  x , x  3 could be written x  0  3 , and we could interpret it to mean that
the distance from x to 0 is 3 units, which is another way to see that the solutions are 3 and  3 ,
since there are two numbers which are 3 units away from 0.
Inequalities with Absolute Value
There are two kinds, one with < (or  ), and one with > (or  ). Let’s look at the < ones
first.
If x  3 , in other words if x  0  3 , then the distance from x to 0 is less than 3 units.
We could also say that x is within 3 units of 0. Can you see that x could stretch as far left as
(just under) 3 units and as far right as (just under) 3 units? That would allow it to be as far left as
(just before)  3 and as far right as (just before) 3? We could translate x  3 into the compound
inequality  3  x  3 . In interval notation, this would be  3, 3 .
If x  3 , on the other hand, or x  0  3 , then the distance from x to 0 is more than 3
units. There are two ways for this to happen: x could be more than 3 units to the left of 0 or
more than 3 units to the right of 0. If to the left, then x  3 ; if to the right x  3 . These are two
110
separate intervals,  ,  3 and 3,  . In interval notation we’d write the solution as
 ,  3  3,  .
Here’s an algorithm:
Solving Inequalities with Absolute Value
1) Symbol <
A) Place contents between two < signs, putting the opposite of the constant to the left
and the constant to the right.
B) Solve.
2) Symbol >
A) Set contents less than the opposite of the constant, and set contents greater than the
constant.
B) Solve both inequalities and connect with ‘or.’
Of course this doesn’t work if the constant is negative or 0, but we won’t do any of those.
Here is an example: x  3  7 . Using the algorithm (Step 1) we get
x3  7
7  x3 7
3
3 3
 10  x  4
In interval notation, the solution is  10, 4. In words we could say that all numbers between
 10 and 4 are within 7 units of  3 . (It’s  3 because we have to think of x  3 as x   3 to
get the meaning right. The distance from x to some number a is represented by x  a , not by
x  a . Think of two numbers on a number line, like 5 and 2, and you’ll see why this is so.)
Now an example with >: x  1  5 . (In words we would say that x is more than 5 units
away from 1.) Using the algorithm (Step 2) we get
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x 1  5
x  1  5 or x  1  5
1 1
1 1
x  4 or x  6
The solution in interval notation is  ,  4  6,  .
People using the algorithm for the symbol > sometimes have trouble with the <
inequality, because they see the greater-than symbol in the original inequality and can’t imagine
turning it into a less-than symbol, but of course that’s how you get numbers that are more than a
certain distance to the left of another number.
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Worksheet for Lecture #14
Show all steps for these problems. Express answers to inequalities in interval notation.
1) x  2  3
2) x  2  3
3) x  2  3
4) x  8  1
5) x  8  1
6) x  8  1
7) x  7  5
8) x  7  5
9) x  7  5
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Assignment for Lecture #14:
Equations and Inequalities with Absolute Values
Solve these equations.
1) x  1  7
9) x  4  1
2) x  4  1
10) x 
2 1

3 2
11) x 
5 2

6 3
3) x  8  3
4) x 
2 1

3 2
12) x  1  7
5) x 
5 2

6 3
13) x  7.16  9.32
6) x  7.16  9.32
14) x  4  1
Write the solutions to these inequalities in
interval notation.
15) x 
2 1

3 2
16) x 
5 2

6 3
7) x  1  7
8) x  7.16  9.32
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Lecture #15: More about Exponents
Back in Lecture #9 you were introduced to exponents, a very convenient and compact
notation that expresses repeated multiplication: multiplying a certain number (the base) as many
times as are indicated by the exponent, which is written to the right of, smaller than, and above
the level of the base. Thus 53  5  5  5  125 .
In this lecture, we’re going to extend the range of numbers which are allowed to be
exponents. Before, it was 2, 3, 4, …, though I’m sure you can see that any number raised to the
power 1 is itself. For instance 51  5 . Even though you’re not really repeating the factor 5, you
can see the logic by this pattern:
53  5  5  5  125
5 2  5  5  25
51  5
To get from a power to the next lower one, you divide by 5. Another way to see that 51  5 is to
realize that we can always put a factor of 1 in front of the 5’s, since 1 is the identity element of
multiplication. Thus 5 3  1  5  5  5  125 , and 51  1  5  5 .
But what about using 0 as an exponent? How about  2 , or any negative integer? What
about fractions and decimals? We’ll look at 0 and the negative integers as exponents in this
lecture, and the fractions and decimals later on in the course.
To understand how 0 and negative integers behave as exponents, you have to know a few
basic rules of exponential expressions. They’re very logical.
The Product Rule
If you multiply two exponential expressions which have the same base, you do it by
adding the exponents. I know it seems weird to multiply by adding, but check it out:
x 2  x3 
xxxxx 
x5
You’re multiplying 2 x’s, and then you throw in 3more factors of x, and of course you have a
total of 5. You can count them.
Two notes on this. First, the bases must be identical. If you have x 2  y 3 , you don’t have
5 of any one thing, and you can’t compact the expression. Of course you could evaluate it if
given numbers for x and y, but short of that the expression is as simple as it can get. Second, if
instead of the base being a variable like x you have a constant, it’s very tempting to multiply the
base by itself as well as add the exponents. For instance, you might think that 4 2  4 3 equals 16 5 .
But it doesn’t. 4 2  4 3 is 4  4  4  4  4 , or 4 5 not 16 16 16 16 16 .
115
The Quotient Rule
The product rule says that to multiply the same base raised to powers, you add the
exponents. The quotient rule, not surprisingly, says that to divide the same base raised to
xxxxx
x5
powers, you subtract the exponents. Take 2 . It’s short for
, and of course there’s
xx
x
always a 1 if needed in both parts of the fraction. Let’s put a 1 in the denominator,
xxxxx
,
1 x  x
. On top there are three x’s left, on the bottom just a 1. So the
x5
3
answer is x , which is what you get when you subtract the exponents: 2  x 53  x 3 .
x
and then cancel:
The Power Rule
We won’t be using this rule until later in the course, but it goes along with the other two.
If you have an exponential expression like x 3 and in turn you want to raise it to a power, as in
 
2
the expression x 3 , you can do this by multiplying the exponents. This works because
x 
3 2
 x 3  x 3  xxx xxx  x 6 . You’re taking 2 groups of 3 x’s apiece, for a total of 6 x’s.
Just make sure you don’t confuse the product rule and the power rule. In the product rule
example, x 2  y 3 , you’re multiplying 2 x’s and then 3 x’s, for a total of 5; in the power rule
you’re taking 2 groups of 3 x’s, for a total of 6.
The Number 0 as an Exponent
You can use either the product rule or the quotient rule to convince yourself that if 0 is
used as an exponent, the answer will be 1 no matter what the base is.
Take the product rule. If you want to simplify the expression x 0  x 3 , you add the
exponents: x 0  x 3  x 03  x 3 . So when x 3 is multiplied by x 0 , it doesn’t change. We have a
number which does just that – it doesn’t change a number when that number is multiplied by it.
We call it 1. So x 0  1 , and that argument will work when you make the other factor x 3 or x to
the anything.
x3
Here’s another way to look at it, using the quotient rule. What does 3 equal? Well,
x
x3
obviously 1; any number divided by itself equals 1. But the quotient rule says that 3  x 33  x 0
x
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x3
equals both 1 and x 0 , 1 and x 0 must equal each other. (We have a name for the fact
x3
that two things equal to a third thing must be equal to each other; it’s called the transitive
property of equality.)
Oh, and if you need, or want, one more explanation, remember this pattern:
. Since
53  5  5  5  125
5 2  5  5  25
51  5
If we add one more line to the pattern, we’ll get 5 0 on the left, because the exponents are
decreasing by 1 as we go down, and on the right we’ll get 1, because the right sides are formed
by dividing the one above it by 5, and 5 divided by 5 is 1.
Negative Integers as Exponents
Now we’ll look at how negative integers work as exponents. We’re a long way from 5 3
meaning you multiply 3 5’s together here. What does 5 3 mean? It certainly can’t mean that
you multiply  3 5’s together, because this doesn’t make sense.
First look at the pattern we’ve already extended:
53  5  5  5  125
5 2  5  5  25
51  5
50  1
1
. Again, we went down 1 in the exponent on the left, and we
5
1
1
1
2
divided 1 by 5 on the right. And here’s another line: 5 
, because divided by 5 is
.
25
5
25
1
2
Remember that 25  5 2 , so we could write 5  2 , and now you can see the pattern. A
5
negative exponent puts the base in the denominator of a fraction, with 1 as the numerator, and
the exponent changes from negative to positive. There’s nothing negative about the expression!
1
Let’s go one line further: 5 
x2
The quotient rule gives us another justification. What if you had 5 ? One the one hand
x
xx
we could write it as
, and in turn we could put a factor of 1 in the numerator and
xxxxx
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cancel:
. Thus we have
1
1
, or 3 . On the other hand, using the quotient
xxx
x
1
x2
3
 x 25  x 3 . So x  3 . This is the same idea we got from the pattern of the
5
x
x
powers of 5: A base to a negative exponent is equal to 1 over the base to the absolute value
(positive version) of the exponent.
rule we get
Scientific Notation
Scientific notation is a very compact way to write numbers, especially very large
numbers like 2,531,000,000,000,000 and very small numbers like 0.0000000000000000896, in
such a way that their size, or order of magnitude, becomes apparent.
To understand scientific notation you have to know the effect on numbers of multiplying
or dividing them by 10. If you multiply a whole number by 10, you simply put a 0 at the far
right of the number: 286 10  2860 . If you multiply a number with a decimal part by 10, you
move the decimal point one place to the right: 37.54 10  375.4 .
If you multiply by 10 2 , or 100, it’s like you multiplied by 10 twice, so you perform the
action described in the previous paragraph twice. Thus if you multiply by 10 5 , or 100,000, you
perform the action 5 times. If the number with a decimal part at some point in the process
becomes a whole number (because you moved the decimal point all the way to the right) you
start adding 0’s.
To summarize, to multiply a number by 10 n , where n is a positive whole number (1, 2, 3,
etc.), you move the decimal point in the number n places to the right, or add n 0’s at the end if
the number was a whole number to begin with, or do some mixture of the two.
Dividing a number by 10 has the opposite effect. It moves the decimal point one place to
the left. Dividing by 10 n moves it n places to the left. If you run out of digits, you insert 0’s as
placeholders. Thus 24.5  10 6  0.0000245 .
1
1
 10 1 . So
, and
10
10
6
6
6
dividing by 10 is the same as multiplying by 10 . We could rewrite 24.5  10 as 24.5  10 6 ,
and it too would equal 0.0000245 .
But remember that dividing by 10 is the same as multiplying by
Viewing both cases together, we can say that multiplying a number by 10 n , where n is an
integer, has the effect of moving the decimal point n places to the right if n is positive and the
absolute value of n places to the left if n is negative.
Armed with this information, we’re ready to define scientific notation. The key is to look
at the digits in the number which are not initial 0’s. (An initial 0 would be something like the
118
bold ones in 0.0012 – the one to the left of the decimal point is just something to draw attention
to the decimal point and should be omitted for the purposes of scientific notation, making the
number .0012. The 0’s in numbers like 24.0037 or 132,500 are not initial 0’s.)
Now place a decimal point just to the right of the first digit in the number which isn’t an
initial 0. Take 347. You would write 3.47. But you can’t just play fast and loose with the
decimal point, so you have to indicate where the decimal point in the original number actually is
by multiplying 3.47 by the power of 10 that will move the decimal point you placed to where it
actually belongs (by default to the right of the 7):
Hopefully you can see, by looking at the scallops under the 4 and the 7, that the decimal point we
placed after the 3 would need to move 2 places to the right to get back to where it belongs.
So 3.47  10 2 is the same number as 347, except now it’s written in scientific notation.
The number before the multiplication sign has to be less than 10 and at least as big as 1, since it
has exactly one non-zero digit to the left of the decimal point. The effect of this is that the
majority of the number’s size is contained in the power of 10, in this case 10 2 . The number 347
is bigger than 10 2 , or 100, but not as big as 10 3 , or 1,000. The exponent of 10 is called the order
of magnitude of the number and tells us the power of 10 that is less than or equal to the number
and nearest to it.
Now let’s look at a number whose scientific notation version will have a negative
exponent for the 10. Consider 0.0005603. Of course the 0 to the left of the decimal point can be
ignored, but the three directly to the right of the decimal point are initial 0’s (but not the 0
between the 6 and the 3). So we place the decimal point just to the right of the 5 and drop all 0’s
to the left: 5.603. Now we have to determine the power of 10 that will get the decimal point
back where it came from. Can you see that we have to move it four places to the left?
0.0005603 
5.603  10 4
The most common mistake is to think that the exponent of 10 should be  3 because of the three
0’s, but before you get to them you have to move the decimal point to the left of the 5, so you’re
moving the decimal point a total of four places to the left.
Let’s go back to the very large and very small numbers we began this section with. The
problem with the way they’re written (called decimal notation in contrast to scientific notation)
is that it’s difficult to get a grasp of the size of the numbers with all those 0’s.
Take 2,531,000,000,000,000. Can you see that in scientific notation it’s 2.531  1015 ?
It’s a little over 2½ quadrillion, order of magnitude 15.
119
How about 0.0000000000000000896? I can hardly count the 0’s! At least large numbers
are separated into periods, groups of three counting from the right, so it’s easy to see how far the
decimal point has to be moved. Anyway, we start with 8.96, and then it appears that we have to
move the decimal point 17 places to the left to get it back where it started. It becomes
8.96  10 17 . Its order of magnitude is  17 .
There are two advantages which expressing numbers in scientific notation has over
decimal notation. First, you can easily see how many times bigger one number is than another
by comparing their orders of magnitude. If one number’s order of magnitude is, say, 6 (i.e. it’s
something times 10 6 ), and another’s is 2 (it’s something times 10 2 ), you subtract 2 from 6, which
gives you 4, and then you take 10 to that power: 10 4  10,000 . (To find 10 to any positive
power, just write a 1 followed by as many 0’s as the exponent.) So the larger number is
approximately 10,000 times bigger than the smaller one.
The other advantage has to do with something called significant digits, which is a very
important concept in the physical sciences. Let’s say you have rounded a number, and you come
out with 12,000. You can’t tell from the way it turned out whether you rounded to the nearest 1,
the nearest 10, the nearest 100, or the nearest 1000. It could have been any one of the four. (For
instance, to the nearest one, 11,999.6 rounds to 12,000; to the nearest ten, 12,004 rounds to
12,000; to the nearest hundred, 11,967 rounds to 12,000; and to the nearest thousand, 12,499
rounds to 12,000.) But writing the rounded number in scientific notation would reveal how you
rounded. If you rounded to the nearest one, you would write the answer as 1.2000 10 4 , if to the
nearest ten as 1.200 10 4 , if to the nearest hundred as 1.20 10 4 , and if to the nearest thousand as
1.2 10 4 .
This might not seem like such a big deal to you. What does it matter to what place the
number was rounded? But the issue is important in the physical sciences because care has to be
taken not to exaggerate the accuracy of measurements.
Here’s a joke that brings this point home: A museum contains the skeleton of a dinosaur.
A visitor asks the museum guard, “How old is that skeleton?” The guard answers, “Seventy
million and eight years.” The visitor is surprised and asks how the guard knows so precisely.
The guard replies, “Well, I’ve been working here for eight years, and when I started I was told
that the skeleton was seventy million years old.”
120
Worksheet for Lecture #15
Simplify as much as possible.
Write answers with positive exponents.
1) x 6  x 3
Write the answer as a fraction.
3
8) 5
2) x 4  x
3) x 5  x 5
4)
Write in scientific notation.
9) 43,592,000
x9
x5
10) 0.00217
x2
5) 5
x
6)
x 
7)
x 
Write in decimal notation.
11) 4.18 108
8 3
12) 1.7 10 4
8 3
121
122
Assignment for Lecture #15: More about Exponents
Simplify as much as possible. Write
answers with positive exponents.
12)
x 2
x7
13)
x 2
x 7
1) x 7  x 2
2) x 7  x 2
3) x 7  x 2
x7
14) 7
x
4) x 7  x 2
5) x 7  x 7
6)
15)
x 2
x 2
 
x7
x2
16) x 7
 
x7
7) 2
x
17) x 7
2
 
2
18) x 7
x 7
8)
x2
9)
x 7
x 2
10)
x2
x7
2
 
19) x 7
2
Pick two natural numbers and, using
them in place of the 7 and the 2, redo #1
through #19 with the same signs and
operations. (For instance if your
numbers are 6 and 4, #1 would be x 6  x 4 .)
Number your problems #20 though #38.
x2
11) 7
x
123
Simplify. Write answers as fractions if
necessary, not as decimals.
39) 4 3
1
40)  
4
3
41) 4 3
1
42)  
4
3
Write in decimal notation.
49) 8.302  10 3
43) 6 2
1
44)  
6
50) 1.794  10 9
2
51) 2.5  10 5
52) 6.14  10 8
Write in scientific notation.
45) 5,869,000
46) 872
47) 0.00000534
48) 0.029
124
Lecture #16: Roots, Radicals, and Rational Exponents
When you do something to a number, there’s always a way to undo it, to get back to the
original number. If you add 3 to a number, you can get back to the original number by
subtracting 3 from your result. If you multiply a number by 5, you can get back to the original
number by dividing your result by 5. And so on.
Operations that undo, or reverse each other, are called inverses. Subtracting 3 is the
inverse of adding 3 (and vice versa). Dividing by 5 is the inverse of multiplying by 5 (and vice
versa).
What’s the inverse operation for squaring a number (multiplying it by itself)? We call it
taking the square root, as discussed in Lecture #9. If you square 5 (raise it to the power 2), you
get 25, and when we say “the square root of 25” we mean the number which when squared
becomes 25. It’s 5. (Of course there’s another number which squares out to be 25,  5 , but let’s
consider only positive numbers here.) As you saw, the way to write this is 25 .
Of course you don’t have to square a number first to be able to take its square root. You
can ask what the square root of 100 or the square root of 6.25 is. We would write 100  10 and
6.25  2.5 , because 10 2  100 and 2.5 2  6.25 . Taking the square root and squaring a number
are inverse operations.
Some numbers, like 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100, have square roots which
are whole numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. These numbers are called perfect squares. A
perfect square is a number which can be written as n 2 , where n is a whole number. So 441 is a
perfect square, because 441  212 .
Other numbers have square roots which are fractions whose decimal equivalents are
either terminating decimals or repeating decimals. I call these pretty good squares (but not
perfect). For example, 6.25  2.5 , because 2.5 2  6.25 , and
4 2
 , because
9 3
2
2 2 4
2
     .
3 3 9
3
But some numbers have square roots which aren’t fractions at all, and the decimal
equivalents of these square roots neither terminate nor repeat – they go on and on forever without
ending or forming a pattern. These square roots are called irrational numbers, not because they
don’t use reason, but because they can’t be written as ratios, or fractions. Take 5 . The
calculator tells us that it equals 2.236067977, and you might think it does, but in fact it goes on
and on forever after that last 7, without forming a pattern or ending. So if you’re going to write
it in decimal form, you have to round it at some point, which is what the calculator did when it
made that last digit a 7 rather than an 8 (because in fact the digit after that 7 is a 4). You could
round it to the nearest hundredth and get 2.24, or to the nearest thousandth and get 2.236, or to
the nearest ten-thousandth and get 2.2361, and so on, but you have to round it somewhere.
125
The discovery of irrational numbers caused a huge crisis in ancient Greek mathematics.
Greek mathematicians just couldn’t get their heads around the idea that there were numbers that
couldn’t be completely written out. There’s a lot more to this story.
Higher Roots
What’s the inverse of cubing a number (raising a number to the third power)? We call
this inverse the cube root. If you cube 2 you get 8 ( 2 3  8 ), so the cube root of 8 is 2. We write
it like this: 3 8  2 . The little 3 inside the crook of the radical sign is called the index. It tells us
what root we’re taking. So the square root of 5 could also be written as 2 5 , but we don’t have
to write the index 2; it is the default index.
Some numbers are perfect cubes: 0, 1, 8, 27, 64, 125, and so on, because they are the
3
2
  , and 15.625,
3
3
because it is 2.5 , and some numbers have irrational cube roots which can’t be written as
fractions or decimals which terminate or repeat.
cubes of whole numbers. Others are pretty good cubes, like
8
, because it is
27
Higher indices (the plural of index) indicate other roots. Since 3  81 , the fourth root
4
of 81,
4
81 , is 3. Since 10  100,000 , the fifth root of 100,000,
5
5
100,000 is 10. And so on.
The number 1 is special. Since 1 to any power is 1, n 1 is 1 no matter what n is.
Unit Fractions as Exponents
Now we can expand our notion of exponents to include unit fractions. A unit fraction is
simply a fraction whose numerator is 1.
First notice what happens when you multiply the square root of a number by itself. For
example, 25  25  5  5  25 . The square root of a number times itself gives you the original,
the radicand. So what would 25
1
1
25 2  25 2  25
1 1
2 2
1
2
be? Using the product rule from Lecture #15, we can see that
1
 25  25 . But 25  25 also equals 25. So 25 2 must equal
1
Taking a number to the power is just another way to take its square root.
2
1
25 .
Another way to see that the one-half power is the square root is to look at the power rule
from Lecture #15. It said that to raise the power of a number to another power, you multiply the
2
1
1
2
1
exponents. So  25 2   25 2  251  25 . So 25 2 is the number which, when squared, gives


25. But that’s exactly what the square root of 25 is!
126
Now you can see the meaning of other unit fractions used as exponents. The one-third
power is the cube root; the one-fourth power is the fourth root; and so on. Here are some
examples:
8
1
3
3 8 2
10,000
729
1
1
37
1
6
1
23  8
 4 10,000  10
4
10 4  10,000
because
36  729
 6 729  3
137  1
 37 1  1
Other Fractions as Exponents
Okay, we’ve managed to figure out the only possible meaning of unit fractions as
exponents, but what about those fractions whose numerators aren’t 1? Piece of cake! We use
the power rule again, plus the fact that any fraction can be written as the product of a unit
fraction and a whole number. Take
1
2
3
2
1
. It’s equal to  2 . So by the power rule,
3
3
 
2
  8 3   3 8  2 2  4 . The denominator of the fractional exponents tells you the
 
root you’re taking (the cube root in this case), and the numerator tells you what power to take
that root to (the second power in this case). Of course you could reverse the order, namely
8
2
3
8
1
2
square first and then take the cube root, because multiplication is commutative and
1
 2 is the
3
1
3
same as 2  . But then you’d be making the number bigger before making it smaller, and it
might be harder to figure out: 8
2
8
3
2
 
1
3
 82
1
3
 64
1
3
 3 64  4 .
Let’s try some. You can write out as many of these steps as are necessary to understand
and arrive at the correct answer, but if you can do any of them in your head, or skip them
entirely, that’s perfectly all right.
1
3
4
3
 
 16 4   4 16  2 3  8 . In words, you take the


fourth root of 16, which is 2, and cube it, which gives you 8.
3
4
What is 16 ? Well, 16
3
4
 16
1
3
5
How about 8 3 ? It’s the cube root of 8, which is 2, raised to the fifth power, which is 32.
Written out the long way we get 8
5
1
3
5
5
1
 8 3   8 3  
 
 8  2
3
5
5
 32 .
And don’t forget negative exponents; they can be fractions too. Take care of the negative
first by putting the base to the positive version of the exponent in the denominator of a fraction,
127
under 1. Consider 1000
2
3
1
. It equals
1000
10, squared, which is 100. So 1000
2
3

2
. And 1000
2
3
is the cube root of 1000, which is
3
1
.
100
These fractional exponents may seem very strange to you, but if we want to follow the
rules of exponents (and believe me, we do!) we’ve defined them in the only way possible.
128
Worksheet for Lecture #16
Write each of these radical expressions as an exponential expression with a rational
exponent. Evaluate the expression. If the answer is rational, write it as a fraction if the
radicand is a fraction and as a decimal if the radicand is a decimal. If the answer is
irrational, round it to the nearest hundredth.
49

4
4.
5
0.000064 
5.
 2
57 
6.
 10 
1.
√√
2.
3.
3
4
4
5
3
Write each of these exponential expressions as a radical expression. Evaluate the
expression. If the answer is irrational, round it to the nearest thousandth.
7. 125
1
8. 489
9. 16
5
4
3
1
1

10. 7

11. 36
4

2
3
2
12. 1000000
129
1
6
130
Assignment for Lecture #16: Roots, Radicals, and Rational Exponents
Write each of these radical expressions as
an exponential expression with a rational
exponent. Evaluate the expression. If the
answer is rational, write it as a fraction if
the radicand is a fraction and as a
decimal if the radicand is a decimal. If
the answer is irrational, round it to the
nearest hundredth.
11) 4 50,625
12) 4 57
13)
5
32
3125
1)
81
14) 5 2.48832
2)
49
4
15) 5 7776
16) 5 10
3)
0.0064
17) 6 179
4)
409
18) 8 98
3
1000
6)
3
343
7)
3
0.000064
8)
3
28
9)
4
8.3521
10)
4
1
16
5)
19) 8
64
1
256
20) 10 4000
Continued on p. 132
131
Write each of these exponential
expressions as a radical expression.
Evaluate the expression. If the answer is
irrational, round it to the nearest
thousandth.
21) 45
1
5
33) 32
34) 32
1
1
24) 1.35
35) 32
36) 64
26) 489
1
37) 64
1
27) 1024
1
38) 64
30) 16
31) 16
1
3
6
5
1
2
1
3
1
6
3
2
3
4
1
5
2
3
5
3
40) 64
29) 16
3
3
39) 64
1
1
5
2
27
25)  
 8 
28) 16
4
2
22) 529
23) 125
32) 16
2
4
4
132
5
6
10) Simplify the expression 2 x  5  6 x
.
Sample Test on Assorted Topics in
Algebra
Covers Lectures # 8-16
11) Evaluate the expression
3x  2  4 x  1 for x  5 .
1) In adding 3 and  8 , what is the first
step according to the algorithm for
adding signed numbers?
12) Evaluate the expression 2 x  5  6 x
for x  4 .
2) In performing the operation
 2   5 , what is the first step
according to the algorithm for
subtracting signed numbers?
13) Solve the equation x  7  18 .
11 2
 . Give
12 3
your answer as a fraction reduced to
lowest terms.
2
15) Solve the equation x  18 .
3
14) Solve the equation x 
3) If you’re multiplying six numbers,
three of which are negative and three
of which are positive, what sign will
the product have?
4) Simplify 4  3 2  5 using the
order of operations.
16) Solve the equation 4x  7  11 .
17) Solve the equation 2x  9  5x  6 .
5) Simplify 9  2  1  3 using the
order of operations.
18) If C stands for Cam’s weekly income
in dollars, and if Sam makes $40 per
week less than Cam, and their total
weekly income is $820, write an
equation which models the situation.
1 8
using the order
13   1
of operations and express your
answer as a fraction reduced to
lowest terms.
6) Simplify

19) You have $54 and plan to save $24
per week until you have a total of
$318. If w stands for the number of
weeks this will take, write an
equation that models the situation.

7) Simplify 5  2 1  32 using the order
of operations.
8) Simplify 25  9 using the order of
operations.
20) If P stands for the pre-tax price of an
item, the tax rate is 8%, and the cost
including tax is $17.28, write an
equation which models the situation.
9) In simplifying the expression
3x  2  4 x  1 , what would the
expression be after performing the
first step of the algorithm for
simplifying algebraic expressions?
21) If P stands for the original price of
an item, and after a 40% discount the
item costs $15.60, write an equation
which models the situation.
133
22) Write the solution to this inequality
in interval notation: x  1.2 .
35) Write in scientific notation:
23,400,000,000.
23) Write the solution to this inequality
in interval notation: x  3 .
36) Write in decimal notation:
3.15  10 3 .
24) Write the solution to this inequality
in interval notation:  2  x  1.
37) 4 21 is the same as taking 21 to what
power?
25) Solve this inequality and write the
solution in interval notation:
4x  5  3 .
38) Find 6 19 to the nearest thousandth.
1
32  5
39) Find 
 and express the
 243 
answer as a fraction.
26) Solve this equation: x  1  3
27) Solve this inequality and write the
solution in interval notation:
x 4  9.
40) Find 185
28) Solve this inequality and write the
solution in interval notation:
x3  7.
29) Simplify as much as possible:
x3  x4 .
30) Simplify as much as possible:
x 8  x 8 .
31) Simplify as much as possible:
x6
.
x8
32) Simplify as much as possible:
x3
.
x 5
 
33) Simplify as much as possible: x 2
.
5
34) Rewrite x 7 using a positive
exponent.
134
2
3
to the nearest hundredth.
21) P  0.40P  15.60
Answers
1) Subtract 3 from 8, or 8  3 .
22) 1.2, 
2) Change the subtraction sign to
addition and change the  5 to  5
or  2  5 .
23)  ,  3
24)  2,  1
3) Negative
1

25)   , 
2

4) 13
5) 17
6) 
26) x  2,  4
3
4
27)  5, 13
 , 10  4, 
7)  15
28)
8)  4
29) x 7
9) 3x  6  4x  4
30) 1
10)  4x  10 or 10  4 x
31) x 2
11)  15
32) x 8
12) 26
33) x 10
13) x  11
34)
14) x  
1
4
1
x7
35) 2.34  1010
15) x  27
36) 0.00315
16) x  1
37)
17) x  5
18) C  40  C  820
1
4
38) 1.634
19) 54  24w  318
2
3
40) 32.47
39)
20) P  0.08P  17.28
135
Lecture #17: Linear Equations in Two Variables
Now we are going to begin the study of a different kind of linear equation, one which has
two variables instead of just one. They’re usually called x and y.
You’ll notice right away that the solutions of this kind of equation are entirely different
from the solutions of linear equations with only one variable. Let’s examine the equation
y  3x  2 .
First, unlike an equation that has only one variable, say an x, it’s not enough to give the
solution as the number that x must equal to make both sides of the equation name the same
number when you evaluate them for the solution. Saying x  2 doesn’t do you any good,
because even though evaluating 3x  2 for x  2 gives 3  2  2  6  2  4 , you still have y on
the left side, not 4. But this points the way: if we say that x  2 and y  4 , then when we
evaluate both sides of the equations for these values of x and y, we get 4  4 .
So a solution to a linear equation with two variables is a pair of numbers, one for x and
one for y. For reasons which will become clear in the next lecture, we write this solution in the
compact form 2, 4 . It’s called an ordered pair, because the order tells you that the first
number is meant to be the x and the second number the y. Reversing the two doesn’t work. If
x  4 and y  2 we don’t have a solution to the equation, because the left side would equal 2
and the right side would equal 3  4  2  12  2  10 , and 2 does not equal 10.
The second difference between linear equations with one variable and those with two is
that whereas equations with one variable generally have only one solution, one x-value that
makes both sides equal the same number, linear equations with two variables have an infinite
number of solutions, each one a pair of values for x and y. There are an infinite number of
ordered pairs which make the equation true.
What if x  10 ? Then y  3  10  2  30  2  28 . So the ordered pair 10, 28 is another
solution to y  3x  2 . What if x  0 ? Then y  3  0  2  0  2  2 . So the ordered pair
0,  2 is also a solution to y  3x  2 . How about x  1? Then y  3  1  2  3  2  5 .
This gives us the solution 1,  5 . What if x  1.5 ? Then y  3 1.5  2  4.5  2  2.5 , which
gives us the order pair 1.5, 2.5 . In fact, for any value of x you pick, you’ll get a value for y,
and thus another ordered pair. This may seem like overkill, and maybe not very useful – I mean,
you could hardly be expected to list all the solutions to y  3x  2 , since there are infinitely
many, so how useful can they be?
But in fact equations of this sort are very useful and come up in all sorts of contexts.
Even this very ordinary equation we’ve been investigating could be useful. Let’s say that you
are paid $3 for each smoothie you make, but you have to pay $2 for the use of the smoothie
machine. If x represents the number of smoothies you make, then y will be how much money
you make (after paying the $2 rental fee) with your smoothie business. If you make 2 smoothies,
you will have earned $4. If you make 10 smoothies, you will have earned $28. If you don’t
136
make any smoothies (i.e. you make 0 smoothies), you still owe the $2 for the use of the machine,
so you are $2 in the hole ( y  2 ). If you make  1 smoothies…oh, well, that’s impossible. As
is 1.5 smoothies. We have to restrict x to the non-negative integers, those that aren’t negative,
in other words the whole numbers. But even so you will now be able to find your earnings for
any reasonable number of smoothies, as long as you keep making $3 per smoothie and have to
pay $2 to use the machine.
And part of the beauty is that the same equation can be interpreted in many different
ways. Say you’re on a number line. At the moment you’re 2 units to the left of 0, at  2 . But
you’ve been at this for many minutes, and each minute you move 3 units to the right. Then
y  3x  2 tells you where you are x minutes after the present moment. Two minutes from now
you’ll be at 4. Ten minutes from now you’ll be at 28. Right now ( x  0 ) you’re at  2 . One
minute ago ( x  1 ) you were at  5 . And in 1.5 minutes, you’ll be at 2.5. All numbers work
for x in this interpretation, as long as we look at negative numbers as representing times before
the present moment. Where will you 56 minutes from now? Since y  3 56  2  168  2  166 ,
you’ll be at 166.
Two-Variable Linear Equations in Standard Form
It’s really simple to find solutions to equations like y  3x  2 because all you have to do
is pick an x, any x, evaluate 3x  2 for your choice, and assign the answer to y. That’s because
in the equation y  3x  2 is solved for y. This means that y is isolated on one side of the
equation (i.e. all by itself).
This is how write the general form of equations with two variables which are solved for
y: y  mx  b . The m is the coefficient of x. A coefficient is a constant (a definite number as
opposed to a variable) which multiplies the variable. The b is a constant. In the case of
y  3x  2 , m is 3 and b is  2 . We put the negative sign with the 2 because to match the form
y  mx  b we would have to write y  3x  2 as y  3x   2 .
But linear equations don’t always come in such a convenient form, with the y isolated.
Sometimes they’re in what’s called standard form.
Here’s an example: 3 x  2 y  6 . Notice that one side of the equation has an x term and a
y term, each with a coefficient, and the other side consists solely of a constant. In this case, the
coefficient of x is 3, and the coefficient of y is  2 . We put the negative sign with the 2 because
the structure of the standard form is written Ax  By  C , with a plus sign in front of the
coefficient of y, so you have to think of the equation as 3x   2y  6 . So A  3 , B  2 , and
C  6.
To find solutions to this equation, you can still choose an x and then find out what y
needs to be, but you have to do some algebra in the process. Let’s pick 5 for x. We replace the x
by 5 and get 3  5  2 y  6 . Now we need to solve this equation for y:
137
35  2y  6
15  2 y  6
 15
 15
 2y  9

2 2
9
y
2
1
 9

As an ordered pair, this solution is  5,  , or  5, 4  , or 5, 4.5 .
2
 2

You could also start by picking a number for y, but you then would have to solve for x so
it would still involve doing some algebra.
Notice that the coefficients of x and y, 3 and  2 , are both divisors of 6. When this
happens, it’s very easy to get two different solutions of the equation. First, let x be 0. It’s pretty
easy to see, by covering up the x term (because the product of any number and 0 is 0), that y is
 3 . Be sure to be careful with the signs: the quotient of a positive number and a negative
number is negative. Then let y be 0. Covering up the y term results in x being equal to 2. So we
have two easy solutions, 0,  3 and 2, 0 , and, as you’ll see in the next lecture, two solutions
are enough for our purposes.
Looking at the standard form Ax  By  C , you can see that these easy solutions will be
obtainable whenever A and B are factors of C. And even if they’re not factors of C, putting 0 in
for x and then for y simplifies finding the value of the other variables. Solutions which have 0 as
their x- or y-coordinate are very important, as you’ll see in the next lecture.
Another way to handle equations in standard form is to rearrange them to solve for y.
Solving for y means getting y all by itself on one side of the equation, the way it was in the
equation we first looked at in the lecture, y  3x  2 . The advantage is that then you can just
pick a value for x and plug it in to find y, as we did before.
The skill of solving for y or some other variable is an important one when we discuss
formulas in a future lecture. So let’s practice on 3 x  2 y  6 . It’s similar to what we did when
we let x be 5.
3x  2 y  6
 3x
 3x
 2 y  3x  6

2
2
3
6
y
x
2
2
3
y  x 3
2
138
A couple of things to note here. First, we wrote  3x  6 rather than 6  3x in the third line. We
led with the x term. It’s just a matter of style, as you’ll see in the next lecture. Second, in
dividing the right side by  2 , we divided both terms, the  3x and the  6 by  2 . You don’t
have to write out as many steps as I did, but you must get the signs right.
3
x  3 , an equivalent form of the equation 3 x  2 y  6 (one that
2
has the same solutions), it will be much easier to find solutions. We can just pick numbers for x.
But we might as well go easy on ourselves and choose only numbers that don’t result in
fractions. Which numbers would those be? Well, anything that will wipe out that 2 in the
3
denominator – in other words, multiples of 2. Let’s use 8:  8  3  12  3  9 . This gives us
2
the solution 8, 9 . Notice that this solution checks in the original equation, 3 x  2 y  6 , because
3  8  2  9  24 18  6 . Let’s end with choosing  2 for x. Then y has to be
3
 2  3  3  3  6 . This solution is  2,  6 . This solution also checks:
2
3  2  2  6  6   12  6  12  6 .
Now that we have y 
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Worksheet for Lecture #17
For these equations, find the ordered pairs which are solutions, using
a) x  3
b) x  0
c) x  5
Write the solutions as integers or fractions.
1. y  4 x  1
2. y  3x  5
3. y 
2
x 1
3
In (a) and (b), for these equations, find the ordered pairs which are solutions, using
a) x  0
b) y  0
c) Solve the equation for y.
4. 7 x  3 y  21
5. 3x  2 y  12
140
Assignment for Lecture #17: Linear Equations in Two Variables
For these equations, find the ordered
pairs which are solutions, using
In (a) and (b), for these equations, find
the ordered pairs which are solutions,
using
a) x  0
b) y  0
c) Solve the equation for y.
a) x  5
b) x  0
c) x  3
11) 2 x  4 y  12
Write solutions as integers or fractions.
12)  x  y  7
1) y  x  2
13) 5 x  3 y  15
2) y  5 x  4
14)  3x  y  6
3) y  4 x  1
15) 6 x  4 y  24
4) y   x  2
5) y  5 x  3
16) To solve the equation Ax  By  C
for y, you can do the following steps:
6) y  3x  5
7) y 
3
x
4
Ax  By  C
By   Ax  C
1
8) y   x
5
2
9) y  x  1
3
1
10) y   x  6
2
y
 Ax  C
B
y
A
C
x
B
B
Explain what was done to each step
to get to the next.
141
142
Lecture #18: The Rectangular Coordinate System
In this lecture you will encounter what is probably one of the two most important
concepts in all mathematics. If you have ever played a video game (and who hasn’t, really?) you
have used the results of this concept without even knowing it. In fact, every aspect of our
modern life would shut down if the consequences of this concept were removed and not allowed
to function.
First, a story. This may or may not be true, but I don’t really care because it makes such
a great point. René Descartes (1596-1650), the French philosopher, mathematician and writer
probably best known to you as the person who said, “I think, therefore I am,” (though he said it
in Latin, the scholarly language of the day, in which it came out “Cogito ergo sum”), was sick
one day, and he was lying in bed, and fortunately for humankind he was lying on his back. He
looked up at the ceiling and saw a fly. He realized that he could completely specify the position
of the fly by stating its distance from one edge of the ceiling and its distance from an adjacent
edge.
Of course it was vital to be clear on which edge which distance was measured from.
Being 2 feet from one edge and 3 feet from an adjacent edge is an altogether different position
than if we switched the 2 and the 3.
From this came the concept of the rectangular coordinate system. You take two
number lines, make them perpendicular to each other, and have them cross each other at their
0’s:
Now name the horizontal number line the x-axis, and the vertical one the y-axis. Show this by
writing an x to the right of its axis and a y above its axis:
143
Note that I omitted the numbers on the x- and y-axes. If they’re not written in, we’ll assume that
the hash marks represent 1, 2, 3, etc. If we need to use another scale, because we’re dealing with
large numbers, we’ll write in the first of the numbers on each axis to indicate this.
Now we’ll use Descartes’ insight to give an “address” to each point on the plane. Start at
the origin, the point where the two axes cross, and move horizontally until you’re either above,
below, or at the point whose address you’re finding. How far you moved gives you the first
number in the ordered pair, called the x-coordinate. If you moved left, the x-coordinate will be
negative; if right, positive; and if you didn’t have to move, it will be 0. From there, move
vertically until you’ve reached the point in question, either up or down, or not at all. How far
you moved gives you the second number in the ordered pair, called the y-coordinate. It will be
negative if you moved down, positive if you moved up, and 0 if you didn’t move vertically at all.
This graph shows the points whose addresses – called coordinates – are 2, 5 and 5, 2 .
Make sure you understand which coordinates belong to which point and why. The horizontal
movement always comes first.
Now I’ll add some other points:
144
Notice that the two axes divide the plane of the page into four parts, and each part is
called a quadrant. The upper right quadrant is called Quadrant I, pronounced “Quadrant One.”
We continue numbering the quadrants counter-clockwise, like this:
In each quadrant, all points have coordinates with matching signs. In Quadrant I, it’s
,  , meaning that both the x- and y-coordinates are positive, because you’re going right and
up from the origin. In Quadrant II, it’s ,   , left and up. In Quadrant III it’s ,  , left and
down. And in Quadrant IV, it’s ,  , right and down.
Of course, some points aren’t in any quadrant; they’re on the axes themselves. They’re
particularly easy to label incorrectly. If you don’t go left or right, they’re on the y-axis, and their
x-coordinate is 0; if you don’t go up or down, they’re on the y-axis, and their y-coordinate is 0.
Here are some examples:
145
This arrangement of the axes at right angles to each other, along with the method of
labeling the points with coordinates, is called the rectangular coordinate system. It is
sometimes referred to as the xy-plane, for obvious reasons, and also the Cartesian plane,
because of its inventor, Descartes.
Graphing Linear Equations with Two Variables
Now we get to the best part. Remember in Lecture #17 how there were infinitely many
pairs of x’s and y’s that make a linear equation with two variables true? You might have
wondered, “When can I stop listing them?” Well, as the saying goes, a picture is worth a
thousand words, and you’ll see what I mean when we consider the ordered-pair solutions to our
equation y  3x  2 as locations or coordinates of points. We got 2, 4 , 0,  2 , and 1,  5 ,
among others. Look at them on the rectangular coordinate plane:
The three points appear to lie on a straight line! And in fact they do, which is why we call
equations like y  3x  2 linear equations. Any equation of the form y  mx  b or
Ax  By  C has solutions which form a line. By drawing that line, we have the most compact
method of displaying all the infinite number of solutions to a linear equation. It’s called
graphing the equation.
One way of listing enough of these solutions to form the line is called a T-table (as
opposed to a coffee table??). We make a long lower-case t, like this:
and label the left-hand column x and the right-hand column y:
146
x
y
We fill in values for x in the x column and the resulting values for y in the y column:
x y
2 4
0 -2
-1 -5
It’s just a convenient way to organize the work. We make each pair of x- and y-values into an
ordered pair and locate it on the plane.
If you need more proof that the points lie on a line, see what y is when x is 1:
3 1  2  3  2  1. So the point 1, 1 is also on the graph. I’ll add it and draw the straight line
that goes through all four points:
This picture is called the graph of the equation y  3x  2 and is the best way to show its
solutions. It doesn’t matter what number you choose for x. Together with the y-value that
results, the ordered pair will lie on this line. And if you find the coordinates of any point not on
the line, its coordinates will not satisfy the equation y  3x  2 .
This was the genius of Descartes. There are many other shapes of graphs besides straight
lines, and they are each characteristic of a different kind of equation with two variables. And
when the people, etc., that you see on your video game move, it’s because the software contains
147
equations whose graphs determine which pixels will be displayed at which time in which
position.
2
Let’s graph a new equation, y   x  1 . Here’s a T-table. Make sure you can get the
3
correct y’s for the x’s listed:
x y
6 -3
3 -1
0 1
-3 3
-6 5
2
Notice that I used only those numbers for x that make  x turn out to be an integer. With all
3
numbers to choose from, why not pick those that don’t result in fractional coordinates, which are
harder to locate precisely? Here’s the graph:
Notice also that you could actually get away with finding only two points and drawing
the line through them (as long as they’re correct!). As the third-century BCE Greek geometer
Euclid wrote, two points determine a line. This means that there is only one straight line
connecting two points, which seems obvious enough.
The Y-intercept and the Slope
Now let’s look at the m and the b in the form y  mx  b and see how they affect the line.
First the b. Notice that the graph of y  3x  2 crosses the y-axis at  2 . We call  2 the
y-intercept of the line, because it’s where the line intersects, or intercepts, the y-axis. For
2
y   x  1 the line crosses the y-axis at 1, or  1 . After all, the x-coordinate of every point on
3
the y-axis is 0, and if you put 0 in for x in mx  b you get m  0  b  0  b  b . So b is the y-
148
intercept of the equation of a line in the form y  mx  b . But watch the signs: You have to
rewrite (or at least think of) y  3x  2 as y  3x   2 to get its y-intercept to be  2 .
The meaning of the m is a little harder to see. In the graphs of y  3x  2 and
2
y   x  1 I’ve drawn little stepping-stair patterns and labeled them:
3
On the left, you can see that I started at the point 1,  5 and went 3 units up and then 1
unit to the right. We call 3 the rise and 1 the run. In other words, we travel from one point on
the line to another by first moving vertically until we’re on the same level as the destination
point and then horizontally until we’re at the point.
We then make a ratio of these two numbers, which we call the slope of the line. It is a
3
fraction: the rise over the run. In this case it’s , or 3. And it doesn’t matter which two points
1
on the line we choose to make this ratio, or which point we start at. I’ve shown two other slope
“journeys”, from 0,  2 to 1, 1 , and from 1, 1 to 2, 4 . In each case, the rise is 3 and the
run is 1, so the slope is 3. Even if I’d gone from 1,  5 to 2, 4 , the slope would work out to
9
be the same: The rise would be 9 and the run would be 3, and
equals 3. (But always
3
remember to make the rise the numerator of the slope and the run the denominator.)
You don’t even have to count using the graph to get the rise and the run. You can use
what’s called the slope fraction. Simply make a fraction whose numerator is the difference of
the y-coordinates and whose denominator is the difference of the x-coordinates, as long as you
make both differences start with the coordinates of the same point. Using 1, 1 and 2, 4 gives
4   5 9
4 1 3
1 4  3
us
  3 . The
  3 (or

 3 ). Using 1,  5 and 2, 4 , we get
2 1 1
1  2 1
2   1 3
slope is always going to be 3 for this line. Just make sure that you use the negative signs from
the  5 and the  1 as well as the minus signs that are part of the slope fraction formula.
149
2
Now look at the graph of y   x  1 with the rises and runs drawn in. Here I’ve shown
3
three different possibilities. First look at the path from  6, 5 to  3, 3 . Here the “rise” is 2
2
2
units down, which we call  2 . The run is 3 units to the right, or 3. The slope is
, or  .
3
3
35
2
2
53
2
2
Using the slope fraction, we get

  (or

  ). I also show
 3   6 3
3
 6   3  3
3
the path from 6,  3 to 3, 1 . The rise is 2, but the run, since it’s to the left, is a negative
2
2
number,  3 . So the rise over run is
, and the slope is still  . The final path indicated,
3
3
4
2
from 3, 1 to  3, 3, has a rise of 4 and a run of  6 , and
 .
6
3
Notice the effect on the sign of the slope caused by whether the line rises or falls as you
go from left to right. (We always go from left to right in describing the line as rising or falling.)
2
2
For y  3x  2 , which has slope 3, the line rises. For y   x  1 , which has slope  , the line
3
3
falls. Lines with positive slopes rise as they go from left to right; lines with negative slopes fall.
Also notice that the slope itself is contained in the equation. In fact, it is the coefficient
of x. How convenient is that?! In equations in the form y  mx  b , not only does the b tell us
the y-intercept, but the m tells us the slope. In fact, we use the letter m for the slope because of
the French word monter, meaning “to climb.” And the form y  mx  b is called the slopeintercept form of a linear equation, for reasons which should now be clear to you.
Graphing Using the Slope and Y-intercept
You can actually graph linear equation in slope-intercept form just by using your
knowledge of the form.
3
x  2 . First make a dot at 0,  2 , because the
4
y-intercept, b, is  2 , so the line crosses the y-axis at  2 . Starting from 0,  2 , use the slope,
3
, to find another point on the line. Rise 3 and run 4, and make a dot at your destination, which
4
will be 4, 1 . Then draw a line connecting the two points:
Say you want to graph the equation y 
150
Now let’s try y  4 x  3 . The y-intercept is 3, so make a dot at 0, 3 . With the slope,  4 , you
have some choices to make. First, you have to write it as a fraction, and then you have to decide
which part of the fraction gets the minus sign. (Either the numerator or the denominator is fine,
but don’t give it to both, because the quotient of two negative numbers is positive!) Using the
4
4
simplest possible ratio, we could call it either
or
. This graph shows both ways:
1
1
Graphing Equations in Standard Form
As you saw in the last lecture, not all linear equations are written in slope-intercept form,
y  mx  b . Some are written in standard form, Ax  By  C like our example 3 x  2 y  6 . We
saw that there are two particularly easy solutions to this equation, which are contained in this Ttable:
x y
2 0
0 -3
Since all you need to graph a linear equation are two points, these points are sufficient. The
point 0,  3 is, of course, the y-intercept. It’s on the y-axis. It’s not surprising that the point
2, 0 is called the x-intercept. It’s where the line crosses the x-axis. Here’s the graph:
151
For linear equations in standard form, even if the A and the B are not divisors or C, you
could still get the two intercepts and use those points to graph the line, at least
approximately…unless C is 0. Look at this equation: 3 x  4 y  0 . If you let y be 0, x is 0. And
vice versa. So the x-intercept and the y-intercept are at the same point. The line goes through
the origin. You’ll need to find another point, which you can do by trial-and-error, or by solving
the equation for y:
3x  4 y  0
 3x
 3x
4 y  3x

4
4
3
y x
4
So you’ll use a multiple of 4 for x to get another point, for instance 4, which gives a y-value of
 3 and the ordered pair 4,  3 . Here’s the graph:
3
3
The equation y   x is in slope-intercept form. It could be written as y   x  0 ,
4
4
which reminds us that its y-intercept is in fact 0.
Horizontal Lines
Here’s an equation that looks odd in a lecture about linear equations with two variables:
y  2 . It has only one variable, but we can still graph it on the Cartesian plane. We just have to
understand that x has no restrictions on it (it can be any number at all), but y has to be 2. Here’s
a T-table:
x
5
2
0
-1
-4
y
2
2
2
2
2
When we graph these points, we get this:
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It’s a horizontal line at the level of 2. In fact, any equation of the form y  k , where k is some
constant, has a graph which is a horizontal line. It may seem odd to you that an equation which
mentions only y, which we associate with the vertical direction, has a graph which is a horizontal
line, but that’s the way it works out.
What is the slope of y  2 ? Using the rise and the run, we can see that going from one
point to another on the line involves a rise of 0 (you don’t go up or down), so whatever number
the run is, 0 divided by that number is 0. So the slope is 0. It’s a flat line; it neither rises nor
falls from left to right. Another way to see that the slope is 0 is to flesh the equation out so that it
looks like y  mx  b . It’s our trick with 0: y  2  0  2  0  x  2 . The slope, m, is 0.
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Worksheet for Lecture #18
1) Give the coordinates of the points A
through D in the graph below, and for each
point state the quadrant it is in, if any.
1
4) y  x  2
3
A:
B:
C:
D:
2) Draw the line containing the points 1, 1
and 2,  3 . What is its slope?
5) x  2 y  6
Graph each equation:
6) y  3
3) y  x  2
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Assignment for Lecture #18: The Rectangular Coordinate System
8) y   x  2
1) Give the coordinates of the points A
through G in the graph below, and for each
point state the quadrant it is in, if any.
9) y  2 x  1
10) y 
1
x
2
11) y 
2
x
3
1
12) y   x  4
2
13) x  y  5
14) 2 x  y  6
2) On graph paper, draw a set of x- and yaxes. Locate and label these points:
A
B
C
D
E
F
G
15) x  y  3
 2,  1
0,  5
3,  4
 2, 0
 3, 4
0, 1
5, 0
16) x  2 y  4
17) y  2
18) y  5
19) Draw the line containing the points
5, 0 and 0, 3 . What is its slope?
Graph these lines. Show them going into
every quadrant they pass through.
Graph each line on a separate set of axes.
20) Draw the line containing the points
1,  2 and 1, 4 . What is its slope?
3) y  x
21) Draw the line containing the points
1,  3 and  2, 3 . What is its slope?
4) y  x  3
22) Draw the line containing the points
1, 3 and 5, 3 . What is its slope?
5) y  x  1
6) y  4 x  3
7) y   x
155
156
Lecture #19: Functions and Function Notation
Besides the rectangular coordinate system, which led to graphing equations, the other
most important mathematical concept to come from algebra is what we call functions, and
function notation. These don’t figure as obviously in video games and other places, but they
are there behind the scenes.
Let me explain it like this: You have two equations with two variables, say y  3x  2
and y  x 2 . I ask you, “What is y when x is 5?” The proper answer would be for you to ask me,
“Which y are you talking about?” I would then have to specify which equation I meant.
There’s a better way. I give each equation a separate name. Instead of starting each one
with y, I call one of them f x  and the other one g x  . These are pronounced “f of x” and “g of
x.” I write f x  3x  2 and g x   x 2 . The x is called the argument; f and g are called
functions, which is why we use the letter f so often. Almost anything can be a function; the only
restriction is that the function has to produce one definite answer (i.e., the y) for each x (though
several x’s can have the same y for an answer).
Now instead of asking what y is when x is 5, I can ask you to find f 5 or g 5 . Now 5
is the argument. f 5 means that we evaluate the f-equation for x  5 :
f 5  3 5  2  15  2  17 . g 5 means that we evaluate the g-equation for x  5 :
2
g 5  5  25 . So f 5 is 17, and g 5 is 25, and we didn’t have to say which y we were
talking about.
Notice that f x  and g x  both satisfy the condition that they produce one definite
answer for each x. There’s no way that f 5 could equal anything except 17, or that g 5 could
equal anything other than 25.
How about f  1 and g  1 ? Here the argument is  1 .
f  1  3  1  2  3  2  1 , and g  1   1  1 . Remember that in g x  you’re
2
squaring  1 , and you have to put the  1 in parentheses. If you wrote  12 you (or your
calculator) might think you were taking the opposite of 1 squared, and you’d wind up with the
answer  1 .
In function notation, f x  and g x  are simply synonyms for y. But understand that
f x  isn’t always equal to 3x  2 . Each time f x  is mentioned, you have to be told the
equation for it. We don’t have enough letters to have them permanently attached to any one
formula.
In case you’re wondering how anything could possibly violate the requirement that
there’s only one value for f x  for any value of x and hence not be a function, let me give a
non-mathematical explanation. Functions don’t have to be mathematical in nature. Their
arguments and results could be almost anything. Let’s call C x  the capital of state x. (You see
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here that the letter naming the function doesn’t have to be f or g. It could be any letter, and
sometimes it’s helpful to use a letter that reminds you what the function is about – C is for
capital. So C California   Sacramento . The argument is a state, and the answer is a city. Each
state has only one capital. So C x  is a function.
But how about this: S x  is the name of Person x’s son. Jo has two sons, Jim and Tim.
So is S Jo  equal to Jim, or Tim? We haven’t been given any basis to choose, so there are two
possible answers for the argument Jo. S x  isn’t a function.
Domain and Range
There are two very important sets of numbers associated with a function. One tells what
numbers can be used as x’s. This set is called the domain. The other tells what numbers can be
used as y’s (or f x  ’s). This set is called the range.
Let’s find the domain and range of the two mathematical functions we’ve discussed in
this lecture. First, y  3 x  2 . Well, you can certainly multiply any number by 3 and then add 2
to the product, so there’s absolutely no restriction on what we can use for x. In this case, we say
that the domain consists of all real numbers, and in Lecture #11 we used the symbol R to
represent this set. We could also use interval notation here. You start all the way to the left on
the number line and go all the way to the right:  ,  .
How about the range? Well, any number can be the result of taking some number,
multiplying it by 3, and then adding 2. In fact, if we know what y is, we can find the x that will
produce that y when it’s multiplied by 3 and 2 is added to the product. We do it by solving
y  3 x  2 for x:
y  3x  2
2
2
y  2 3x

3
3
y2
x
3
In other words, if we want to be sure that a certain y can be the result of 3x  2 for some x, just
subtract 2 from the y and divide the difference by 3. Do you want to be sure that 29 can be the
result of 3x  2 ? Subtract 2 from 29, which gives 27, and then divide 27 by 3, which is 9. So
f 9  29 , and 29 is an obtainable value for y. So the range of this function is also R.
You might think that all functions can have any number at all for x and for y, so their
domains and ranges are all R, but looking at y  g x   x 2 will show us otherwise. It’s true that
the domain is R, because of course you can take any number and multiply it by itself, thus
squaring it. But what about the range? Are all numbers the result of squaring another number?
No way. You can’t find any real numbers that square out to a negative number, because the
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product of two identical numbers is positive, except if the number is 0, and 0 2  0 . And any
positive number can be a y in this function. To find the x just take the square root of y, in other
2
1
1
2
words take y 2 For instance, 5 can be y because 5  5   5 2  . So the range consists of all


numbers greater than or equal to 0 (also called nonnegative numbers, because they’re not
negative). In interval notation, this is 0,  . Notice the bracket on the left, which shows that 0
is part of the range.
Inverse Functions
In Lecture #16 we discussed inverse operations, processes that undo each other, like
adding 3 and subtracting 3. Now we discuss inverse functions, a very similar idea using
function notation.
Here’s the function that adds 3 to a number: f x  x  3 . Here’s one that subtracts 3:
g x   x  3 . Let’s say we start with an argument, or x value, of 2: f 2  2  3  5 . Now take
that 5 and use that argument in the g function: g 5  5  3  2 . Not surprisingly, we’re back to
the original argument. We have a special notation for an inverse function like g x  . We write it
f 1 x  , pronounced “f inverse of x.” The  1 isn’t an exponent, putting a base in the
denominator of a fraction. It’s just how we express inverse functions. Do you see that we could
also have called f x  the inverse of the g function? It could have been written g 1 x  . Inverse
functions come in pairs; they are inverses of each other.
In discussing the function f x  3x  2 early in this lecture, we actually came up with
its inverse when we demonstrated that we could find an argument for x that produced whatever y
y2
we wanted. We wrote it as x 
because we were solving for x, but now we want to write it
3
x2 1
2
as a function of x: f 1  x  
 x  . Let’s use 7 as an argument in f x  . Then
3
3
3
f 7  3 7  2  21  2  23 . Now use 23 as the argument in f 1 x  . We get
23  2 21
f 1 23 

 7 . We’re back where we started. We could go the other way too. For
3
3
10  2 8
8
example, use 10 as an argument in f 1 x  . Then we get f 1 10  
as
 . Now using
3
3
3
8
8
the argument in f x  , we get f    3    2  8  2  10 . Again we’re back where we
3
3
started. The two functions really do reverse or undo each other.
159
Graphs of Inverse Functions
An interesting phenomenon occurs if you graph a function and its inverse on the same set
1
2
of axes. Here I’ve graphed y  3 x  2 and y  x  , plus the line y  x :
3
3
Can you see that the y  x line acts as a kind of mirror for the other two lines? If you folded the
graph along that line, the other two lines would wind up right on top of each other. We say that
1
2
the lines for y  3 x  2 and y  x  are reflections of each other across the line y  x . And
3
3
this property is true of all pairs of inverse functions, because in a sense what you’re doing in
finding an inverse function is switching the x and the y. Every point on one line has a
corresponding point on the other line in which the x- and y-coordinates are reversed. For
1
2
instance, y  3 x  2 has the point 0, 2 on its line, while y  x  has the point 2, 0 .
3
3
160
Worksheet for Lecture #19
For each function, find
a) f 8
b) f 0.1
c) f 0
d) f  0.2
e) f  5
Give all non-integer answers as decimals. If necessary, round to the nearest hundredth. If
f(x) isn’t defined for an argument, say so.
1. f x  2x  5
4. f x   x  3
5. f x   6
2. f x   x  4
3. f x   x  3
161
Assignment for Lecture #19: Functions and Function Notation
For each function, find
a) f 8
b) f 0.1
For the functions in #8 and #9, find the
inverse of the function, and graph both
functions on the same xy-plane.
c) f 0
d) f  0.2
8)
f x   2 x
9)
f x   x  4
e) f  5
Give all non-integer answers as decimals.
If necessary, round to the nearest
hundredth. If f(x) isn’t defined for an
argument, say so.
1)
f x   5 x  7
2)
f x   x 2  3
3)
f x   x
4)
f x  
5)
f x   x 2
6)
f x   x
7)
f x   3
x
1
3
State the domain and the range of the
functions in #1 through #4. If it’s all real
numbers, write R. If it’s an interval,
write the answer in interval notation.
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Lecture #20: Types of Functions
So far we’ve looked only at linear functions, like y  f x  3x  2 , of the general form
y  f x  mx  b , whose graphs are straight lines. But there are many other kinds of functions,
which we’ll consider in this lecture.
First, here’s another linear function, one that we discussed in Lecture #18. The example I
used was y  2 , and the general form was y  k . It’s called a constant function for obvious
reasons. Its graph was a horizontal line going through the point 0, 2 , or in general 0, k  .
What is the domain of y  f x  2 ? Well, remember that you could write the function as
y  f x  0  x  2 , and from this we can see that any number can be used for x, because you can
multiply any number by 0, for a product of 0, and then add 2 to that. So the domain is R, or
 ,  . But the range – well, there’s only one number that shows up as a y-value: 2. So the
range of the function is the set consisting solely of the number 2. We show the range as a set by
putting 2 between set brackets, like this: {2} .
By the way, y  2 does not have an inverse function, because 2 is the y for any x-value.
We have no way of getting back to the argument we used for x.
Absolute-Value Functions
Absolute-value functions, obviously, are functions whose equations include the absolutevalue sign. The most basic one is f x   x . Look at this T-table for f x   x :
x
3
2
1
0
-1
-2
-3
y
3
2
1
0
1
2
3
Graphing these points produces the characteristic V shape of graphs involving absolute value:
163
It’s made up of parts of two straight lines. To the right of the y-axis, it’s the same as the graph of
y  x , because positive numbers are equal to their absolute values. To the left, it’s the same as
the graph of y   x , because a the opposite of a negative number is a positive number. And of
course 0 is its own absolute value.
Domain? Well, you can take the absolute value of any number, so the domain is R. The
range should remind you of the range of the function y  x 2 . A negative number can’t be the
absolute value of any number. The smallest absolute value is 0, and there’s no limit to the
largest. So the range is 0,  .
You can alter the function f x   x to move the V left or right, or up or down, and to
change how sharp a V it is. To move it up, add a number after the absolute value. Here’s
y  x  2 , first the T-table and then the graph:
x
3
2
1
0
-1
-2
-3
y
5
4
3
2
3
4
5
Its domain is still R, but since the lowest it goes is 2, its range is 2,  . If we wanted to move
the V down, we would subtract a number after the absolute value instead of adding a number.
Moving the V up or down is called a vertical translation. We don’t change the shape of the
graph (we could use a rubber stamp to make the new graph), but we move it vertically. We
could call this new function y  x  k . If k is positive the graph moves up k units; if it’s
negative the graph moves down k units.
To move the graph left or right, called a horizontal translation, add or subtract a number
inside the absolute-value sign. Oddly, adding a number moves the graph to the left, while
subtracting a number moves it to the right. Compare these functions and their graphs:
164
y  x2:
x
3
2
1
0
-1
-2
-3
y
5
4
3
2
1
0
1
y  x2:
x
3
2
1
0
-1
-2
-3
y
1
0
1
2
3
4
5
Both vertical and horizontal translations are part of a general category of ways that
graphs can be changed, or transformed, called transformations. By changing the equation of a
function we can move its graph up or down, left or right. We can also turn it upside down, by
putting a negative sign outside and to the left of the absolute-value sign.
y  x :
x
3
2
1
0
-1
-2
-3
y
-3
-2
-1
0
-1
-2
-3
This isn’t a translation; we’re not merely sliding the graph this time. We’re flipping it. The xaxis, the line y  0 , acts as a mirror reflecting y  x onto y   x . We call this kind of
transformation a reflection, naturally enough, and since the x-axis serves as the line across which
the graph is flipped, we call it a reflection across the x-axis.
Horizontal and vertical translations, plus reflections, are all called rigid
transformations, because they don’t alter the shape of the graph. Other transformations do alter
the shape, in the case of absolute-value functions by making the V sharper or blunter. These are
called stretching and shrinking. (We’ll use those words interchangeably even though there’s a
mathematical distinction between them.) We accomplish these by putting a coefficient outside
the absolute value sign. Here are two examples:
165
y  2x :
x
3
2
1
0
-1
-2
-3
y
y
6
4
2
0
2
4
6
1
x:
2
x
6
4
2
0
-2
-4
-6
y
3
2
1
0
1
2
3
Now let’s do one that puts several of these transformations together in one function. We’ll try
y  2 x  1  5 :
x
4
2
0
-1
-3
-4
-6
y
-5
-1
3
5
1
-1
-5
This graph took y  x and translated it horizontally and vertically, reflected it, and stretched it.
But in what order? There are several paths, and here’s one:
y x:
y 2x :
y  2 x 1 :
y  2 x  1 :
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y  2 x  1  5
Quadratic Functions
Some functions are not linear; their graphs aren’t straight lines. We saw one of them in
Lecture #19, y  x 2 . It’s called “quadratic” because it involves squaring. Here’s a T-table for
this function:
x
2
1
1/2
0
- 1/2
-1
-2
y
4
1
1/4
0
1/4
1
4
Here’s the graph. Note how rounded it is at the bottom (which is why I included 1 and  1
2
2
as arguments).
This characteristic shape is called a parabola, from the Greek words for throwing beside or
beyond. You see this shape or something very similar to it – in its upside-down, or openingdownward, orientation – every time you throw or hit a ball into the air or water your garden with
a hose. It is the physical expression of gravity. Something is propelled upward, but it goes more
and more slowly as the force of gravity overcomes it; it reaches its peak of climbing and starts
going downward. We already discussed the domain and range of y  x 2 , R and 0,  .
To transform the basic parabola, in addition to putting a coefficient in front of the x 2 , we
1
can add in a linear function. Here’s one: y   x 2  x  4 . I’ll let the calculator graph it.
2
167
It’s the basic parabola translated left and up, reflected, and shrunk (flattened, though I admit it
doesn’t look as if it’s been flattened).
The main thing about functions like this, which we generally write as y  ax 2  bx  c is
that they have one reversal of direction going from left to right. If a is positive, they start high
on the left, go down to a lowest point, and then rise again. If a is negative, they start low, go up
to a highest point, and then go down again. Their domains are R. If a is negative, their range is
 , k  for some constant k (the y-value of the highest point). If a is positive, their range is
k ,  for some constant k (the y-value of the lowest point). The point where the parabola
reverses direction is called its vertex, from the Latin word meaning “to turn”.
Polynomial Functions
Linear functions (including constant functions) and quadratic functions are
representatives of a larger group of functions, called polynomial functions.
A polynomial is made up of one or more terms, each containing a constant coefficient,
an x, and an exponent to which the x is raised. They’re usually written so that the exponents get
smaller as you go from left to right, which is called descending order. That’s because the two
most important aspects of a polynomial are its degree, which is the highest exponents it contains,
and its leading coefficient, which is the coefficient of the term with the highest exponent. If a
polynomial has one term, it’s called a monomial, like this one: 4x 2 . If it’s got two terms, it’s a
binomial: 4 x 2  3 x . If it has three terms it’s a trinomial: 4 x 2  3x  1 . Otherwise it’s just
called a polynomial with n terms. The polynomial 7 x  2 has degree 1, because 7 x  2 can be
written 7 x1  2 . The polynomial 7 has degree 0, because 7 can be written 7 x 0 .
The degree of the polynomial determines the shape of its graph. If the degree is 0 or 1, as
you’ve seen, the graph is a straight line, with no turns. If the degree is two, the graph is a
parabola, with one turn. If the degree is 3, as in y  0.1x 3  x 2  0.8x  6.4 , whose graph is seen
here,
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there can be 2 turns, but there can also be 0 turns, as in y  0.01x 3 :
In fact, a polynomial of any odd degree can have one fewer turns than the degree, or any other
even number of turns smaller than the degree, depending on its equation. Polynomials of even
degrees can have one fewer turn than the degree, or any other odd number of turns smaller than
the degree.
The leading coefficient determines what direction the graph goes at its left and right ends.
If the degree of the polynomial is even, and its leading coefficient is positive, , the graph will
start and end high, like y  x 2 . If the degree is even and the leading coefficient is negative, the
1
graph will start and end low, like y   x 2  x  4 . If the degree of the polynomial is odd, and
2
its leading coefficient is positive, the graph will start low and end high, like
y  0.1x 3  x 2  0.8x  6.4 and y  0.01x 3 . If the degree is odd and the leading coefficient is
negative, the graph will start high and end low, like y  0.01x 3  0.03x 2  0.03x  0.03 :
That’s enough about polynomials functions and their graphs; I just wanted to give you an
idea of how their characteristics, like degree and leading coefficient, control the shape of their
graphs.
Exponential Functions
Polynomial functions have x’s as the base of exponents. In exponential functions the x’s
are the exponents. Here’s a typical one: y  2 x . The base is 2, and the exponent is x. Here is a
T-table:
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x
3
2
1
0
-1
-2
-3
y
8
4
2
1
1/2
1/4
1/8
The graph of this function shows the characteristics of exponential functions:
Going to the left of the y-axis, the graph quickly approaches 0, because 2 raised to a negative
exponent results in a fraction. The y-intercept is 1, because 2 0 equals 1. From then on, the
graph rises rapidly. This pattern is called exponential growth, for obvious reasons. It shows the
growth of plants, of investments, and other things. It occurs whenever the base is greater than 1.
(What happens when the base is 1?)
When the base is between 0 and 1, we have exponential decay. Here’s a T-table for
x
1
y   :
2
x
3
2
1
0
-1
-2
-3
y
1/8
1/4
1/2
1
2
4
8
And here it is, graphed along with y  2 x :
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This time, the graph starts high on the left, crosses the y-axis at 1, and then gets closer and closer
to the x-axis. You can see why it’s called decay. It shows how radioactivity works.
Notice that the two graphs are mirror images, reflections across the y-axis. This
 
x
x
1
happens because we could rewrite y    as y  2 1 or , using the power rule of exponents,
2
x
as y  2 . When a negative sign is put in front of the x, it causes the graph to be reflected
across the y-axis.
What about using 0 or a negative number as the base? Neither can be done. You can’t
have a function y  0 x , because when a negative argument is used, you would be raising 0 to a
negative power, which would involve dividing by 0. You can’t use a negative number as a base,
1
because when
or some other fraction with an even denominator is the argument, you’d be
2
taking the square root or some other even root of a negative number, which is not a real number.
1
(Remember, the power is the square root.)
2
What is the domain of an exponential function? Well, looking at y  2 x , you can see that
2 can be raised to any exponent at all, so the domain is R. As for the range, though, notice that
the graph never gets as low as the x-axis, so y must always be a positive number. No power of
any number (except 0) equals 0. There’s no power of a positive base which is negative.
(Negative exponents produce positive fractions.) So the range of any exponential function is
0,  .
Logarithmic Functions
Do exponential functions have inverses? Yes, they do, and they’re called logarithmic
functions. They require a certain amount of explanation. We’d have to be able to solve y  2 x
for x, and we don’t have any way of doing this in the methods of solving equations for a variable
that we’ve encountered so far. So basically a new notation was invented to accomplish this task.
To turn y  2 x into an equation of the form x  , we write x  log 2 y , which is pronounced “log
base 2 of y.” Informally you should think of x as the power you have to raise 2 to, in order to
turn it into y. For instance, 8  2 3 , so 3 is the power you have to raise 2 to, in order to turn it
into 8, or 3  log 2 8 , pronounced “3 equals the log base 2 of 8.”
Try these examples: What is log 5 5 ? The answer is 1, because 51  5 . How about
1
1
log 6 ? It’s  1 , because 6 1  . And log 4 1 ? It’s 0, because 40  1 .
6
6
We rearrange x  log 2 y to make it a function of x: y  f x  log 2 x . The domain of
the function is 0,  , because no exponent (logs are exponents) can turn 2 into 0 or a negative
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number. The range, though, is R, because any number can be an exponent applied to 2.
(Remember, y  log 2 x means exactly the same thing as 2 y  x .)
To illustrate that the functions y  log 2 x and y  2 x are inverses of each other, here are
their T-tables side by side:
y  2x
y  log 2 x
x
8
4
2
1
1/2
1/4
1/8
y
3
2
1
0
-1
-2
-3
x
3
2
1
0
-1
-2
-3
y
8
4
2
1
1/2
1/4
1/8
Notice that the x-column and y-column are interchanged here, and that is the essence of inverse
functions.
Here they are, graphed on the same Cartesian plane, along with the line y  x . As you
can see, they are reflections across the line y  x , which is always true of inverse functions:
Functions of Several Variables
So far, all the functions we’ve looked at are of the form y  f x , which means they
have only one argument. If you know x, you can calculate y. But there are also very useful
functions which have two or more arguments. In general they are called functions of several
variables. You indicate this by writing the variables inside the parentheses, separated by
commas.
Here’s a function of two variables: f x, y   x  y . We could give it a single-letter
name if we wanted, but obviously we couldn’t call it y. How about z: z  f x, y   x  y . To
get a value for z we would have to know the values for both arguments. For example,
f  5,  3   5   3   2  2 , so z  2 when x  5 and y  3 .
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It would also be possible to provide a specific number as one of the arguments but leave
the other as a variable. So f x, 4 would equal x  4 , and that’s as far as we could go without
knowing what x equals.
Do you wonder about graphing these functions? You’d need three axes, including a zaxis, which is usually shown as vertical. It’s hard to convey such a graph in two dimensions, but
here’s a crude attempt to do so:
A point would have three coordinates and be of the form x, y, z  . Thus the point  5,  3, 2
would be on the graph. There are sophisticated computer programs which provide a twodimensional view of these three-dimensional objects. Here’s one for our function
f x, y   x  y :
These three-dimensional graphs can be amazing shapes – this one is pretty rudimentary.
If a function has more than two variables, it could still be very useful, but it wouldn’t be
graph-able, because we’ve run out of dimensions. If the function were f x, y, z   x  y  z (a
function we’ll encounter in the next lecture), we could represent the arguments as a point in
space with three coordinates, x, y, z  , but where would we put f x, y, z  ?
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Worksheet for Lecture #20
Identify each of the following functions
relative to the function y  1.3 x , using the
words horizontal translation, vertical
translation, reflection across the x-axis,
reflection across the y-axis,
shrinking/stretching:
8) log 17 17
9) log 4 1
64
10) log 9 1
1) y  1.3 x  5
2) y  1.32 x
Evaluate the functions for the given
arguments:
3) y  1.3 x
4) y  1.3 x4
11) Given f x, y   x  2 y , find:
f 3, 4
5) y  1.3 x
f  1,  2
Which of these are rigid transformations?
f  5, 7
Guide to logarithms:
𝑙𝑜𝑔𝑏 𝑏 𝑛 = 𝑛
𝑙𝑜𝑔𝑏 𝑏 = 1
𝑙𝑜𝑔𝑏 1 = 0
𝑛
𝑙𝑜𝑔𝑏 √𝑏 = 1⁄𝑛
𝑙𝑜𝑔𝑏 1⁄𝑏 𝑛 = −𝑛
12) Given f x, y, z   x  y  z , find:
f 3, 1, 5
Evaluate these logarithms:
6) log 10 10,000,000
f  5, 8, 2
7) log 9 3
f 0,  3,  2
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Assignment for Lecture #20: Types of Functions
For #1-7, pick the best description of the
function from this list:
For #8-#15, graph both functions on your
calculator and then list all the words that
apply to the second function relative to
the first.
Linear function
Constant function
Absolute-value function
Quadratic function
Polynomial function
Exponential function
Logarithmic function
Vertical translation
Horizontal translation
Rigid transformation
Reflection across the x-axis
Reflection across the y-axis
Shrinking/Stretching
1) y  x  4
8) y  x , y   x
2) y  8 x
9) y  x , y 
3) y   x 2  3x
1
x
2
10) y  x 3 , y   x 3
4) y  6
11) y  x 2 , y  x  52
5) y  log 3 x
12) y  x 2 , y  3x 2
13) y  2 x , y  2 x  5
6) y  2 x 3  5 x 2  x  7
14) y  3 x , y  3 x
7) y 
2
x2
3
15) y  4 x , y  4 x
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16) Rewrite this function in descending
order: y  3  2 x 3  4 x  5x 2
27) If f x, y   x  y , find f  8,5 .
28) If f x, y, z   x  y  z , find
f 7,4,10 .
17) Graph the function
y  0.2 x 3  4 x 2  5x  15 on your
calculator. It has two turns (changes
of direction). Find a window that
allows you to see both turns. List
Xmin, Xmax, Ymin, and Ymax.
18) Is this function y  10 x an example
of exponential growth or exponential
decay?
For #19-26, find these logarithms:
19) log 10 10,000
20) log 8 1
21) log 3 729
22) log 10 0.001
23) log 49 7
24) log 2 256
25) log 5 5
26) log 6
1
6
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Lecture #21: Formulas
A formula is just a function in which the variables have special meanings. A formula
may be a function of one, two, three, or even more variables. We’ll look at a collection of
geometric formulas in this lecture.
Squares
Let’s start with squares. As you know, a square is a four-sided figure (called a
quadrilateral – combining the Latin words for “four” and “side”) in which all four sides are
equally long, and all four angles are right (90o) angles. Here’s one:
We have to label only one side’s length, which we’ll call s, because all four sides have the same
length.
In general any quadrilateral with four equal sides is called a rhombus (from the Greek
word meaning “something that spins”). So a square is a rhombus (but not necessarily vice
versa). Here are two:
The one on the left looks like someone leaned on a square and partially collapsed it. The one on
the right is commonly called a diamond, and I can imagine it spinning.
There are two things we want to know about squares, or really any closed (by which we
mean that it separates the paper into an inside part and an outside part) geometric figure: how far
is it to go around the whole thing, and how much area is contained within it?
The distance from one point on the square, going around it until you get back where you
started, is called the perimeter (from the Latin words for “measure” and “around”). You can
easily see that you would travel four times the length of the side. So the formula for the
perimeter of a square is Ps   4s . This could be read as “The perimeter of a square whose side
measures s units is 4s units.” (Usually the formula is written simply P  4s , but I’m writing it
as a function to emphasize that if you know the length of the square’s side, you know its
perimeter.) A square with side 5 inches has perimeter P5  4  5  20 inches. Whatever the unit
of the side’s length – inches, centimeters, miles, or whatever – the perimeter has the same unit.
We’re talking about what’s called linear measurement here, something that can be measured
with a ruler, even if it has to be a pretty long ruler.
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The other thing we’re interested in, the square’s area, is a measure of how much surface
is contained within the square. This square, whose side measures 5 inches (not really!), has been
divided into smaller squares, each of which measures 1 inch on a side:
The area within each of those little squares is 1 square inch, or 1 sq. in., or 1 in2. You can see
that there are 25 such squares (five rows each containing five squares). So the area of this square
is 25 square inches, and the formula for the area of a square is As   s 2 . Now you can see why
raising a number to the exponent 2 is also called “squaring.”
Whatever linear measurement is used for the sides of the square, the area’s unit is that
measurement squared: cm2, square miles, etc.
Rectangles
These formulas about squares were functions of a single variable, s. With rectangles you
have functions of two variables. A rectangle has four right angles, like a square, and its opposite
sides are equal to each other, but all four sides don’t have to be equal (though they can be – a
square is a rectangle). We call the length of one of the pairs of equal sides l and the length of the
other pair w – length and width, with the length belonging to the longer of the pairs, if one pair is
longer than the other:
Can you see that if you started in the upper left corner and went clockwise around the square
you’d go l  w  l  w units before you get back to where you started? You could gather like
terms and write this as 2l  2w . So the formula for the perimeter of a rectangle is
P l , w  2l  2w , or as I like to think of it, P l , w  2 l  w -- go halfway around the rectangle
and then double that distance.
What’s the perimeter of a rectangle with length 5 cm and width 3 cm?
P 5, 3  2  5  2  3  10  6  16 cm, or P 5, 3  2 5  3  2  8  16 cm.
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How about area? Well, the rectangle shown above displays the small squares each of
which has area 1 cm2, and you can see that there are 15 of them enclosed within the rectangle
(three rows of five). So the area of the rectangle is 15 cm2, and the formula for the area of a
rectangle is A l, w  lw .
Parallelograms
Squares, rectangles, and rhombuses are all examples of parallelograms. A
parallelogram is a quadrilateral whose opposite sides are parallel and equal in length, though the
angles they make do not have to be right angles. Here’s one whose bottom side measures b (for
base) units, and whose top and bottom are h (for height) units apart:
As with the rhombus in comparison with the square, it looks like somebody leaned on a rectangle
and collapsed it somewhat.
If I put in another height line, detach the triangle it forms and move it to the left side, I’ve
formed a rectangle with length b and width h, and it’s pretty easy to see that the formula for the
area of a parallelogram is Ab, h  bh :
Triangles
Squares and rectangles are examples of polygons (from the Greek words for “many” and
“angles” or “corners”), which are flat, closed figures made up of line segments. They have the
same number of sides as angles. The smallest number they can have and still enclose some area
is three, and when they do they’re called triangles. Here’s one:
I’ve labelled the lengths of its three sides a, b, and c.
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There are many interesting and useful facts and words about triangles. The lengths of
two sides added together must always be bigger than the length of the third side (otherwise they
wouldn’t be able to meet):
If two sides are equal, the triangle is called isosceles (from the Greek words for “same” and
“sides”), and the angles opposite the equal sides are equal:
Can you see that the bottom angles are equal?
If all three sides of a triangle are equal, it’s called equilateral (equal sides), but I guess it
could also be called “equiangular.” Language has a lot of little quirks. If none of the sides are
equal, it’s called scalene (from the Greek word for “unequal” or “bent”). And of course, if it’s
scalene, none of its angles are equal. If one of its angles is a right angle (90o), it’s called a right
triangle, though there are no left or wrong triangles. No triangle can have two right angles,
because the sum of the degrees in the angles of any triangle is 180o, and because if there were
two right angles the sides would never meet, because they’d be parallel:
It’s pretty easy to see that the formula for the perimeter of a triangle whose sides measure
a, b, and c is P a, b, c  a  b  c . The perimeter is a function of three variables, and its unit is
the same unit of linear measure as the sides themselves.
But what about the area? This is a good deal more complicated, and we’ll look at two
quite different formulas.
The most common one involves drawing an additional line, called the height. It’s as
though you climbed to the top of the triangle and dropped a plumb (vertical) line down to the
side the triangle rests on, which is called the base:
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I’ve labeled the length of the base b and the length of the height h. Then the formula for the area
1
1
of the triangle is A b, h   bh . To see why the is there, look at this:
2
2
Here I’ve taken each of the two triangles within the original triangle and made a copy of them
rotated 180o and bounded by dashed lines. In other words, I’ve doubled the area of the figure.
And behold: it’s a rectangle with length b and width h, so its area is bh .
Sometimes the height is outside the triangle, and sometimes (if it’s a right triangle) the
height can be one of the sides of the triangle:
In the triangle on the left, when we got to the peak of the triangle (called the apex) to drop the
plumb line, it missed the base of the triangle, and we extended the base to the left so that the
plumb line would make a right angle with the extension. In the right triangle, the plumb line
coincided with one of the sides of the triangle.
(But don’t you think it odd that, except in the case of the right triangle, we have to find
the length of something other than the three sides to find the area? I do. Well, there’s a formula
for the area of a triangle that involves only the lengths of the three sides. It’s kind of
complicated, and I’m just going to mention it. It’s called Heron’s Formula, or Hero’s Formula,
after the Greek mathematician Hero of Alexandria, born about 2000 years ago, who thought it
up. To make the formula a little simpler, first we define something called the semiperimeter,
abc
, in other words half of the perimeter. Then the formula looks like this:
s
2
Aa, b, c  s s  as  bs  c . Notice that this isn’t really a function of four variables,
because s itself is determined by a, b, and c, and we could have replaced it in the formula by
abc
each of the four times it occurs -- but that would have looked ghastly! In fact, even
2
181
with s Heron’s Formula is awfully complicated looking, and maybe we should just resign
ourselves to having to find an extra length, the height, to calculate the area of a triangle (though
there is another formula, but you have to use trigonometry, from the Latin words for “triangle”
and “measure”.)
Here’s one more fact about right triangles. It’s called the Pythagorean Theorem after
the Greek philosopher/mathematician Pythagoras who lived about 2500 years ago. He wasn’t
the first person to find it, by any means, but he was a great popularizer, and his name got
attached to it. It was used to construct right angles by the ancient Egyptians, and it’s used in
many architectural and other applications.
Its prominence in human culture is unquestionable. When the Scarecrow meets the
Wizard of Oz in the 1939 classic movie and is given a diploma, the first thing he does is recite
the Pythagorean Theorem (though he gets it wrong!). Watch for yourself:
http://www.youtube.com/watch?v=DxrlcLktcxU
You need a little terminology first. The two sides which connect to form the right angle
are called legs, and we’ll label their lengths a and b. The side opposite the right angle, the
longest side, has that funny name, the hypotenuse, and its length is called c:
The word hypotenuse comes from the Latin words for “stretching beneath,” which can be more
easily visualized if we show the triangle with the hypotenuse as the base, connecting the ends of
the legs:
The Pythagorean Theorem is usually given as an equation, but I’m going to treat it as a
formula, in which you can find the length of the hypotenuse if you know the lengths of the two
legs (and, of course, the fact that the triangle is a right triangle). We’ll call the function H, for
hypotenuse: H a, b  a 2  b 2 .
Let’s say your right triangle has legs with lengths 3 cm and 4 cm. Then the length of the
hypotenuse is H 3, 4  32  4 2  9  16  25  5 cm. A group of three numbers which could
be the lengths of the sides of a right triangle is called a Pythagorean triple, and 3-4-5 is the
simplest Pythagorean triple. It’s how the Egyptians used the Pythagorean Theorem to make right
angles. They would tie twelve equally-spaced knots in a loop of rope, stake a section of five of
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the intervals to the ground, then put a stake at the intersection of three of the remaining intervals
and the other four remaining intervals and pull the stake as far as it could go. That would
produce a right angle because it would make a 3-4-5 right triangle.
Any multiple of a Pythagorean triple is also a Pythagorean triple, like 6-8-10:
H 6, 8  6 2  82  36  64  100  10 . Another Pythagorean triple is 5-12-13, because
H 5, 12  52  12 2  12  144  169  13 . My favorite is 9-40-41:
H 9, 40  9 2  40 2  81  1600  1681  41 Why is it my favorite? Well, because if you
started at your house and rode your bike 40 miles due west, then turned right and rode 9 miles
due north, and then activated the flight capability of your bike and headed straight for home as
the crow flies, your nine-mile detour would add only one mile to your trip home. That seems
amazing to me.
Trapezoids
If a quadrilateral has two opposite sides which are parallel and two which aren’t, it’s
called a trapezoid (from the Latin word for “table”). It looks like a triangle whose top was
removed:
The perimeter of a trapezoid is no problem (just add up the lengths of the four sides), but
for the area we need some more notation and information.
The parallel sides are called bases. Their lengths are labeled b1 and b2. Those little
numbers below and to the right of the b’s are called subscripts. They’re not at all like
exponents. They are distinguishing labels, like first names. Jo and Mo Smith are both Smiths,
but they’re different people. The b’s are both bases, but one’s the top base and the other is the
bottom one.
We also need to know how far apart the two bases are, and we call that the height,
naturally enough, with length h:
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At first the formula for the area of a trapezoid, a function of three variables, looks complicated,
but after we take it apart you’ll see that it isn’t. The formula is Ah, b1 , b2   h b1  b2  .
1
2
1
goes with the b1  b2 , and it’s the length of the line that’s parallel to the
2
bases and lies midway between them. Its length is the average of their lengths, so it equals
1
b1  b2  . Now I’m going to put in this line and then draw height lines at both ends of it.
2
In fact, the
The final step is to take the two triangles sticking out on the bottom and rotate them to complete
1
b1  b2  and whose height is h, so that its area is in fact the
2
product of those two factors, as the formula indicates:
the rectangle whose length is
Of course, there are polygons with more than four sides, like the pentagon (five), the
hexagon (6), the octagon (8), etc., and they come in various shapes. If all the angles of a
polygon are equal, and also the sides are equal, the polygon is called regular. Thus a square
could be called a regular quadrilateral. A stop sign is in the shape of a regular octagon.
Circles
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Of course we can’t leave the subject of plane figures (flat, two-dimensional ones)
without considering that most perfect of plane figures, the circle.
A circle is the set of points in a plane that are the same distance, r (called the radius from
the Latin word for “rod” or “spoke of a wheel”), from a point called the center:
Imagine a bicycle wheel with an infinite number of spokes, each one r units long. If you start
somewhere on the circle and follow a radius through the center and straight out to the edge of the
circle, you have the diameter, labeled d, from the words in a variety of languages meaning
“measure across.” Of course its length is twice the radius, so we can make the formula for the
diameter D r   2r .
The distance around the circle is not called the perimeter; it’s called the circumference,
from the Latin “to lead around,” and it involves the wonderful number  , the Greek p,
pronounced “pi.” It turns out that for a circle of any size, if you take the a string the length of the
diameter, and lay it out around the circle, marking off where it ends and starting again, it will
reach the starting point after an exact number of times, a little more than three, which we call  .
This number is an irrational number, never ending or repeating as a decimal, and it starts out
3.1415926, but for most purposes 3.14 is a close enough approximation:
Here I’ve started at the right side of the circle and marked the end of each diameter with a cross
hatch. The part left over is about 0.14 times the diameter.
So the formula for the circumference of a circle is C d    d , and since d  2r it can
also be written C r    2r   2 r . (The 2 and the  are written first because they are both
constants.)
It makes sense that  should be less than 4, because I could enclose the circle within a
square whose side measures d units, for a perimeter of 4d, and the circle cuts off all the corners:
185
How about the area of the circle? Here I can cover the circle with four squares, each r units on a
side, each with an area of r2, for a total of 4r2 square units, and again the circle cuts off the
corners:
So it shouldn’t be a surprise that the effect of cutting the corners again leads to the number  ,
and that the formula for the area of a circle is A r    r 2 .
Three-Dimensional Shapes
Having looked at all the common plane figures, we now turn to the solid ones. Perhaps
the most basic one is the cube. It has 12 edges, all equal, and 6 faces. Think of a die (singular of
“dice”), a sugar cube, or an alphabet block. Here’s one whose edge measures s units:
We don’t talk about perimeters of three-dimensional shapes, because that would depend
on which path you chose to measure. Instead, we’re interested in their surface area, the total
area of all the faces. If the edges measure 1 unit, each face is a square whose sides measure 1
unit and whose area is 1 unit2, so with six faces, the surface area is 6 unit2. The function for the
surface area of a cube whose edges measure s units is thus SA s   6 s 2 . (I’m calling this
function SA to make a distinction between the surface area of a solid figure and the area of a
plane figure, though I don’t like using two letters to name a function.)
186
The other thing we’re interested in with three-dimensional shapes is their volume, which
is how much space they enclose. If you have a cube whose edge measures 1 unit, we call its
volume 1 cubic unit, or 1 unit3:
You can see why the exponent 3 has the nickname “cubed.”
If the cube has edges which measure s units, you could fit s2 1-cubic-unit cubes in its
bottom, and since there would be s such layers, the volume of the cube is given by the formula
V s   s 3 .
Boxes
A more general shape similar to the cube has many names, but let’s call it a box. (The
most amusing of its other names is rectangular parallelepiped, from the Greek words for
“plane surface” and “ground.”) Its six faces are rectangles. So a cube is a box. We label the
lengths of its three dimensions l, w, and h, for length, width, and height:
Its surface area is the sum of the areas of six rectangles: the front and back, each with
area lh; the two sides, each with area wh; and the top and bottom, each with area lw. So the
formula for the surface area of a box is SA l, w, h  2lh  2wh  2lw . The unit of surface area
will be unit2.
You can fit lw 1-cubic-unit cubes on the bottom of the box, and there will be h such
layers, so the formula for the volume of a box is V l , w, h  lwh . The unit of volume will be
unit3.
Cylinders
A cylinder (from the Greek word for “roller” or “tumbler”), or, to use its proper name, a
right circular cylinder, is what we might think of as a can. We specify it by the radius of its top
and bottom, whose length we call r, and its height h:
187
Its surface area is made up of the area of the top and bottom and the area of the side (the label of
the can). Imagine making a vertical cut in the label and spreading it flat. You’ll get a rectangle:
Its width is h, and its length is the circumference of the top and bottom, or C  2 r . Thus the
area of the side/label is 2 r h . (If you’re not sure about the length being C  2 r , try cutting
the label off a can yourself.) And the area of the top (and the bottom) is of course the area of a
circle with radius r, A r    r 2 . So the formula for the surface area of a cylinder is
SA r , h  2 r 2  2 r h .
To understand the formula for the volume of a cylinder, let’s go back to the volume of a
box. Instead of talking about 1-cubic-unit cubes filling the bottom layer, we could say that the
area of the bottom of the box is lw, and we multiply this by the height h to get the formula for the
volume to be V l , w, h  lwh . We do the same thing with the cylinder, only here the area of the
bottom is  r 2 . Multiplying this by the height h, we get the formula V r , h    r 2 h .
Spheres
There are many other solid shapes, but we’re going to consider only one more, the
sphere, from the Greek word for “ball.” Just as the circle is the most perfect plane figure, the
sphere is the most perfect three-dimensional one. It’s defined the same way as the circle (points
a certain distance r from a certain point, the center), but now the points are not confined to a
plane – they can be anywhere in three-dimensional space. Here’s a sphere:
In practice you won’t use the formulas for its surface area and volume (unless you paint
balls or fill balloons), but they’re so amazing that I just can’t omit them.
188
The formula for the surface area is SA r   4 r 2 . This means that you could take four of
the circles shown as shaded in the diagram above and exactly cover the outside of the sphere.
(Of course you’d have to do quite a bit of cutting and arranging.) I don’t know why I find this
formula so amazing, but I do. Why exactly four?
4
The volume of a sphere is V r    r 3 , not quite as remarkable as the surface area
3
formula but still surprising. If you know calculus, you can actually figure out the formula for the
volume from the formula for the surface area, by a process which essentially has you (mentally)
add on layers to the outside of the sphere, making it slightly bigger.
Formulas
Linear:
Perimeter of a square: Ps   4s
Perimeter of a rectangle: P l, w  2l  2w  2 l  w
Perimeter of a triangle: P a, b, c  a  b  c
Length of the hypotenuse of a right triangle: H a, b  a 2  b 2
Diameter of a circle: D r   2r
Circumference of a circle: C d    d , or C r   2 r
Area:
Volume:
Area of a square: As   s 2
Area of a rectangle: A l, w  lw
Area of a parallelogram: Ab, h  bh
1
Area of a triangle: A b, h   bh
2
Area of a circle: A r    r 2
Surface area of a cube: SA s   6 s 2
Surface area of a box: SA l, w, h  2lh  2wh  2lw
Surface area of a cylinder: SA r , h   2 r 2  2 r h
Surface area of a sphere: SA r   4 r 2
Volume of a cube: V s   s 3
Volume of a box: V l , w, h  lwh
Volume of a cylinder: V r , h    r 2 h
4
Volume of a sphere: V r    r 3
3
189
Worksheet for Lecture #21
Find each of the following, rounding to the nearest tenth if necessary. Be sure to include
appropriate units.
1. Given a rectangle with length 6.7 cm and width 3.1 cm, find
(a) Perimeter
(b) Area
2. Given a Circle with radius 4.9 in., find.
(a) Circumference
(b) Area
3. Given a cylinder with radius 2.5 feet and height 3.2 feet, find
(a) Volume
(b) Surface Area
190
Assignment for Lecture #21: Formulas
If necessary, round answers to the nearest
tenth.
1) Find the perimeter of a square with side
4.7 cm.
9) Find the length of the hypotenuse of a
right triangle with legs 7 in. and 24 in.
2) Find the area of a square with side 4.7
cm.
10) Find the diameter of a circle with radius
42.5 cm.
3) Find the perimeter of a rectangle with
length 3.2 ft and width 1.3 feet.
11) Find the circumference of a circle with
radius 42.5 cm.
4) Find the area of a rectangle with length
3.2 ft and width 1.3 feet.
12) Find the area of a circle with radius 42.5
cm.
5) Find the area of a parallelogram with
length 13.2 m and height 4.1 m.
13) Find the surface area of a cube with edge
9.1 in.
6) Find the perimeter of a triangle with
sides 19.2 in., 14.3 in., and 6.5 in.
14) Find the volume of a cube with edge 9.1
in.
7) Find the area of a triangle with base 2.7
yds and height 6.5 yards.
15) Find the surface area of a box with
length 6.4 m, width 5.9 m, and height
8.5 m.
8) Find the length of the hypotenuse of a
right triangle with legs 10 mm and 24
mm.
Continued on p. 192
191
16) Find the volume of a box with length 6.4
m, width 5.9 m, and height 8.5 m.
17) Find the surface area of a cylinder with
radius 2.9 feet and height 1.1 feet.
18) Find the volume of a cylinder with
radius 2.9 feet and height 1.1 feet.
19) Find the surface area of a sphere with
radius 16.3 mm.
20) Find the volume of a sphere with radius
16.3 mm.
192
Lecture #22: Symbolic Logic
In algebra, the variables stand for numbers. But in a branch of philosophy called
symbolic logic, they stand for statements. A statement is a sentence which expresses a fact. It
can be either true or false, but not both true and false, and not neither true nor false. It’s one
or the other. We call “true” and “false” truth values.
Let’s call the statement “It rained yesterday” p, and statement “I watered the garden
today” q. (We generally use the letters p, q and r to represent statements.) Unlike variables in
algebra, which we try to solve for, i.e. find out what they’re equal to, what we’re interested in in
symbolic logic is deciding on the truth or falsehood of various statements based on the truth or
falsehood of p, q, etc., and we do this through what are called truth tables.
Truth Tables
Here’s the simplest example. Say we consider the statement “It didn’t rain yesterday.”
We call it not p, because it’s the opposite of p, also known as its negation. We then make a
table. The column on the left is headed p, and underneath it are written all possible truth values
for p, namely T (for true) and F (for false). The column on the right is headed not p, and under it
are written the truth values for it which result from the truth values for p. The top row tells us
that when p is true, not p is false, and the bottom row contains the information that when p is
false, not p is true:
p
not p
T
F
F
T
Now let’s consider a truth table that involves two statements, joined by the word “and”:
“Yesterday it rained, and I watered the garden today.” Under what conditions would this
statement be true? I think it’s pretty obvious that both individual statements would have to be
true. The truth table for “and” needs four rows, because there are four different arrangements of
T’s and F’s for p and q: p and q could both be true; p could be true and q false; p could be false
and q true; they could both be false:
p
q
p and q
T
T
F
F
T
F
T
F
T
F
F
F
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The fact that the only T in the last column occurs when p and q are both true corresponds to our
idea of the meaning of “and.”
How about “or”? What would have to be true for the statement “It rained yesterday or I
watered the garden today” to be true? This is slightly complicated, because there are actually
two distinct uses of the word “or.” One is the “either/or,” which is the more common usage in
daily speech, and the other is the “and/or,” which is used in statistics. The “either/or” statement
would be written “Either it rained yesterday or I watered the garden today.” The “and/or”
statement would be “It rained yesterday or I watered the garden today or both.” The difference is
that the either/or statement is false if both of the separate statements are true, but the and/or
statement is true if both are true. So we need two truth tables:
p
q
p either/or q
p
q
p and/or q
T
T
F
F
T
F
T
F
F
T
T
F
T
T
F
F
T
F
T
F
T
T
T
F
The only difference comes in the top row, where both p and q are true.
There are many other truth tables, and they can have more than two statements. If there
are three statements, say p, q, and r, then there were will be eight rows:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
In fact, if there are n statements, there will be 2n rows.
We won’t look at any other truth tables here; you’ve seen the important ones: not, and,
either/or, and and/or.
194
If-Then Statements
In everyday speech, we often make statements that include the words “if…then….” For
instance: If it’s raining, then your garden will get wet. If we call the statement “it’s raining” p,
and “your garden will get wet” q, then we can symbolize our first statement as if p then q. This
can also be written p  q , which can be pronounced “p implies q,” or simply “if p then q.” The
tenses of the verbs in p and q can be changed to sound better in various contexts. “Your garden
will get wet” could be “Your garden is wet” if that sounds better.
In looking at the statement if p then q, we’re not interested in the truth of p or the truth of
q. We’re interested in the truth of the statement as a whole, which says that as a consequence of
the current rain your garden will get wet. For the discussion that follows, we assume that this is
a true statement.
There are three statements that we can make by changing the order of the p and the q and
by putting not in front of the statements. Each of these altered statements has a name, and the
truth of the statement “if p then q” may or may not imply the truth of the altered statement.
First, you can reverse the p and the q. If p then q becomes if q then p, or q  p . In
words the latter statement would be “If your garden gets wet, then it is raining.” We call this
reversal of p and q the converse of the original if-then statement. Is the statement “If your
garden gets wet, then it is raining” true if the statement “If it’s raining, then your garden will get
wet” is true? Not necessarily. It might be that your garden is getting wet because you’re
watering it, not because it’s raining.
Thinking that the truth of an if-then statement implies the truth of its converse can be
misleading. Let’s say that p is “I’m lucky,” and q is “I will pass the test.” Then if p then q is “If
I’m lucky, then I will pass the test.” This statement does not imply the converse, that if you
passed the test you were lucky. You could have studied, and then the outcome wouldn’t have
been due to luck. Or you could have cheated, in which case the outcome was due to dishonesty,
not luck at all.
An even more definite example is if p is “That animal is a dachshund,” and q is “That
animal is a dog.” Then if p then q is “If that animal is a dachshund, then it’s a dog.” I think
we’d all agree that this is a true statement. But how about if q then p, which is “If that animal is
a dog, then it’s a dachshund”? That huge elkhound?? I don’t think so. The converse in this case
is definitely not true. Even if that dog is a dachshund, that doesn’t mean that the statement “If
that animal is a dog, then it’s a dachshund” is true, because the animal being a dog doesn’t imply
that it’s a dachshund.
195
Now let’s leave p and q in their original order, but put not in front of each: if not p then
not q. This statement is called the inverse of the original statement, if p then q. “If it’s not
raining, then your garden won’t get wet.” This is obviously not true, because, again, you could
be watering it with your hose.
There’s one rearrangement of if p then q which is true if the original statement is true. It
is called the contrapositive, and it occurs when you both switch the order of p and q and
precede them with not. In other words, if not q then not p. “If your garden didn’t get wet, then
it wasn’t raining.” Had it been raining, your garden would have gotten wet; it didn’t get wet, so
it wasn’t raining. “If you didn’t pass the test, then you weren’t lucky.” If you’d been lucky you
would have passed. “If that animal isn’t a dog, then it’s not a dachshund.” How could it be? If
it were a dachshund, it would be a dog.
Even though we can’t in general say that the converse of a true if-then statement is true,
there are cases in which it is true, i.e. in which “if p then q” and “if q then p” are both true. Let
p be “A triangle is equilateral” and q be “A triangle is equiangular.” Then “if p then q” is “If a
triangular is equilateral, then it’s equiangular,” and “if q then p” is “If a triangular is equiangular,
then it’s equilateral.” The truth of both of these statements is established by the laws of
geometry (and also by common sense).
Syllogisms
If you string together two if-then statements, with the “then” part of the first being the
“if” part of the second, you get a syllogism (from Greek words meaning “think” and “together.”
For instance, if p is “It’s raining,” q is “Your garden gets wet,” and r is “Your lettuce will
sprout,” then “if p then q” is “If it’s raining, then your garden will get wet,” and “if q then r” is
“If your garden gets wet, then your lettuce will sprout.” And from this we can conclude that “If
it’s raining, then your lettuce will sprout.” In other words, “if p then q” and “if q then r” imply
“if p then r,” or symbolically, p  q and q  r imply p  r . This is also an illustration of the
transitive property of implication, similar to the transitive property of equality (if a  b and
b  c , then a  c ) and the transitive property of inequality (if a  b and b  c , then a  c ).
You can string together as many if-then statements as you like, as long as the “then” part
of each statement is the “if” part of the next, and your conclusion will be an if-then statement in
which the “if” part is the “if” part of the first statement, and the “then” part is the “then” part of
the last statement.
Looking again at the example in which p is “I’m lucky,” and q is “I will pass the test,”
let’s add an r: “I will pass the course.” Then the statements “if p then q” and “if q then r” are “If
196
I’m lucky, then I will pass the test,” and “If I pass the test, then I will pass the course.” From
them we can conclude “If I’m lucky, then I will pass the course.”
Syllogisms also work when the if-then form is replaced by an “is” form and an “all-are”
form: “This figure is a square” and “All squares are rectangles” imply “This figure is a
rectangle.”
(But you have to guard against the mistake of thinking that “if p then q” and “if r then q”
imply “if p then r,” or symbolically that p  q and r  q imply p  r . For instance, to the
statements p, “It’s raining”, and q, “Your garden will get wet,” let’s add r, “You water your
garden.” Then “if p then q” is “If it’s raining, then your garden will get wet,” and “if r then q” is
“If you water your garden, then it will get wet.” It would be ridiculous to assume that these
statements imply that “If it’s raining, then you will water your garden.” There’s a name for this
sort of mistake: the fallacy of the undistributed middle.)
Or, to look at the examples about squares and rectangles, the statements “This figure is a
rectangle” and “All squares are rectangles” by no means implies “This figure is a square.” The
if-then statements must link together in the way described – with the “then” statement of each
“if-then” statement being the “if” part of the next.
Venn Diagrams
Venn diagrams were thought up (and named for) John Venn, a British philosopher who
lived from 1834 to 1923. They involve drawing circles and labeling each one. Circles can
overlap, or one circle can be entirely within another. Let’s go back to our dog example. We can
make this Venn diagram:
In the circle on the left are all animals, on the right all things with four legs. Contained in both
circles are dogs – they are animals with four legs. In the animal circle but outside the dog area
are fish, snakes, insects, etc. Inside the things-with-four-legs area but outside the dog area are
chairs, tables, etc. If we want to include cats, we’d put them in the center along with the dogs.
197
Now, how about those dachshunds? Let’s make a Venn diagram for dogs and
dachshunds.
The dachshund category is totally contained in the dog category. And the mutts? Another circle
within the dog category:
We can even put individual dogs within their proper circles. Here’s the diagram,
including Nolan the dachshund and Gus the mutt:
Venn diagrams are closely related to syllogisms. We can easily see that since Nolan is a
dachshund and all dachshunds are dogs, it follows that Nolan is a dog. He is within the dog
circle. And it’s equally clear from our Venn diagram of animals and things with four legs that
just because all dogs are animals and all dogs are things with four legs, we can’t conclude that
all animals are things with four legs or that all things with four legs are animals. The circles for
animals and things with four legs overlap, but neither is wholly contained within the other.
For a very entertaining set of Venn diagrams, look at this Google doodle which appeared
to honor John Venn on his 180th birthday:
http://www.google.com/doodles/john-venns-180th-birthday
198
Worksheet for Lecture #22
If p is “I have a laptop” (a true statement), and q is “I have a tablet” (a false statement),
1) Write in words the statement not q, and state if it’s true or false.
2) Write in words the statement p and q, and state if it’s true or false.
3) Write in words the statement p either/or q, and state if it’s true or false.
4) Write in words the statement p and/or q, and state if it’s true or false.
If p is “My shoes are too tight” and q is “My feet hurt”
5) Write in words the statement “If p then q.”
Assuming the statement in #5 is true,
6) Write its converse and state if it’s true or false.
7) Write its inverse and state if it’s true or false.
8) Write its contrapositive and state if it’s true or false.
199
200
Assignment for Lecture #22: Symbolic Logic
The statement p is “My dog is black.” This is a false statement.
The statement q is “My cat is white.” This is a true statement.
The statement r is “My hamster is brown.” This is a true statement.
1) Write out the statement not p. Is this statement true or false?
2) Write out the statement not r. Is this statement true or false?
3) Write out the statement p and q. Is this statement true or false?
4) Write out the statement q and r. Is this statement true or false?
5) Write out the statement p either/or q. Is this statement true or false?
6) Write out the statement p and/or q. Is this statement true or false?
7) Write out the statement q either/or r. Is this statement true or false?
8) Write out the statement q and/or r. Is this statement true or false?
The statement p is “x minus 12 equals  4 .”
The statement q is “x equals 8.”
9) Write out the statement if p then q. Is this statement true or false?
10) Write out the converse of if p then q. Is this statement true or false?
11) Write out the inverse of if p then q. Is this statement true or false?
12) Write out the contrapositive of if p then q. Is this statement true or false?
201
The statement p is “x equals 8.”
The statement q is “x plus 2 is greater than 9.”
13) Write out the statement if p then q. Is this statement true or false?
14) Write out the converse of if p then q. Is this statement true or false?
15) Write out the inverse of if p then q. Is this statement true or false?
16) Write out the contrapositive of if p then q. Is this statement true or false?
17) Write the conclusion of this syllogism:
If x  12 , then x is greater than 11.
If a number is greater than 11, then it is a solution of the inequality x  4  14 .
18) Write the conclusion of this syllogism:
1 is a fraction.
33
All fractions can be written as decimals which either terminate or repeat.
19) Make a true Venn diagram using all these terms:
Quadrilateral
Square
Rectangle
Trapezoid
Parallelogram
20) Make a true Venn diagram using all these terms:
Triangle
Polygon
Plane figure
Pentagon
Circle
202
Sample Test on Assorted Topics in Algebra and Logic
Covers Lectures and Assignments #17-22
1) Graph y  3x  2 .
1
2) Graph y  x .
3
3) Graph y  4 .
203
4) Draw the line containing the points  2, 0 and 0, 2 . What is its slope?
5) If f x   x 1 , find f 4.7 to the nearest thousandth.
6) If f x  x  5 , find f  3 .
7) If f x   x 2  7 , find f  4.
3
8) If f x   2 , find f   .
4
9) If f  x   x 4 , find f 10 to the nearest tenth.
1
From this list, pick the best word to describe the functions in #10 – 12.
Linear function
Constant function
Absolute-value function
Quadratic function
Polynomial function
Exponential function
Logarithmic function
10) y  3 x  4
11) y  x 5  4 x 3  2 x  1
12) y  2 x  3
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From this list, pick the best word to describe the relation of the second function to the first
in #13-15.
Vertical translation
Horizontal translation
Reflection across the x-axis
Reflection across the y-axis
Shrinking/Stretching
13) y  x 2 , y   x 2
14) y  x , y  x  2
15) y  2 x , y  2  x
16) Find log 10 0.01
17) Find log 7 1
18) Find log 81 9
19) Find log 8 8
20) Find log 2 64
21) If f x, y   x  y , find f  6,2 .
22) If f x, y, z   x  y  z , find f  3, 5,  4 .
In #23-30, round the answer to the nearest tenth if necessary.
23) Find the area of a square with side 6.4 m.
24) Find the perimeter of a rectangle with length 7.3 in. and width 4.1 in.
25) Find the area of a triangle with base 4.6 cm and height 9.0 cm.
26) Find the length of the hypotenuse of a right triangle with legs 9 ft and 12 ft.
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27) Find the circumference of a circle with radius 3.4 yd.
28) Find the area of a circle with radius 2.7 m.
29) Find the surface area of a cube with edge 4.3 mm.
30) Find the volume of a box with length 8.5 in., width 2.4 in., and height 3.7 in.
For #31-34, use these statements:
The statement p is “I live in Ukiah.” This is a true statement.
The statement q is “I work at Schat’s.” This is a false statement.
The statement r is “I am a student.” This is a true statement.
31) Write out the statement not r. Is this statement true or false?
32) Write out the statement p and q. Is this statement true or false?
33) Write out the statement p either/or r. Is this statement true or false?
34) Write out the statement p and/or r. Is this statement true or false?
For #35-38, use these statements:
The statement p is “x equals 5.”
The statement q is “x minus 4 is less than 2.”
35) Write out the statement if p then q. Is this statement true or false?
36) Write out the converse of if p then q. Is this statement true or false?
37) Write out the inverse of if p then q. Is this statement true or false?
38) Write out the contrapositive of if p then q. Is this statement true or false?
39) Write the conclusion of this syllogism:
75 is a positive number.
All positive numbers have a square root.
40) Make a true Venn diagram using all these terms:
Quadrilateral
Square
Triangle
Polygon
Parallelogram
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Answers
1)
11) Polynomial function
12) Absolute-value function
13) Reflection across the x-axis
2)
14) Vertical translation
15) Reflection across the y-axis
16)  2
17) 0
3)
18) 1
2
19) 1
20) 6
4)
21) 8
22)  23
23) 41.0 m2
24) 22.8 in.
Slope = 1
25) 20.7 cm2
5) 0.213
26) 15 ft
6) 8
27) 21.4 yd
7) 9
8) 2
28) 22.9 m2
9) 1.8
29) 110.9 mm2
10) Exponential function
30) 75.5 in.3
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31) I am not a student. False.
32) I live in Ukiah and I work at Schat’s. False
33) Either I live in Ukiah or I am a student. False.
34) I live in Ukiah or I am a student, or both. True.
35) If x equals 5 then x minus 4 is less than 2.
36) If x minus 4 is less than 2 then x equals 5. False.
37) If x does not equal 5 then x minus 4 is not less than 2. False
38) If x minus 4 is not less than 2 then x does not equal 5. True
39) 75 has a square root.
40)
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Lecture #23: Summation
Sometimes we just want to add up a bunch of numbers, and we don’t want to write them
all out. We need a more compact notation, and fortunately there is one. It’s called sigma
notation, because it uses the Greek capital S, written  and pronounced “sigma.”
Let’s say we want to add up all the whole numbers from 1 to 100. We could write
1  2    99  100 , or, as my niece used to say, “One, two, skip a few, ninety-nine, one
hundred.” But there’s a better way. We write
100
 i , pronounced “the sum of i
from i  1 to 100,
i 1
and it means exactly the same as 1  2    99  100 . You start by letting i be 1, then kick it up
to 2, then 3, and so on, until you reach 100, at which point you stop. The 1 is called the lower
limit of summation, the 100 is called the upper limit of summation, and the i is called the
dummy variable, not because it’s stupid but because it just holds a place for the numbers.
100
And what does
i
equal? Well, you could go ahead and actually add the 100 numbers,
i 1
but there’s a story about the German mathematician Carl Friedrich Gauss (1777-1855), who,
when he was a boy, was told by his teacher to find this answer. The teacher hoped to keep the
class busy for a while, but Gauss almost immediately answered, “5,050.” He didn’t actually add
the numbers, but he saw a pattern:
He saw that the two numbers on the ends, 1 and 100, added up to 101. And so did the next
numbers in, 2 and 99. And so did….well, you get the idea. The numbers paired off and had a
sum of 101. And how many pairs were there? Well, 100 numbers make 50 pairs. And since the
product of 50 and 101 is 5,050, Gauss got the answer very quickly, probably much to the
teacher’s chagrin.
So what does 1 + 2 + … +509 + 510 equal? (We could write this as
510
 i .)
The pairs from
i 1
outside in add up to 511, and there are 255 such pairs, since the quotient of 510 and 2 is 255. So,
since the product of 255 and 511 is 130,305, the numbers add up to 130,305.
n
The general formula is
 i  2 n  1 .
n
It doesn’t even matter if n is an odd number, so
i 1
that not all numbers pair off – the middle one is just half of a pair.
We can add up other things than just the first n whole numbers. For instance we could
add up the first ten perfect squares: 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100. Using
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10
summation notation, this is
i
2
. We let i equal 1 and square it, getting 1, then let i equal 2 and
i 1
square that, getting 4, and so on till we get to 10 and square it, getting 100.
You could easily do this sum on a calculator, and the answer would be 385. But there’s a
formula for this, just like our previous example, and indeed there are formulas for lots of
n
n n  12n  1
interesting sums. The general formula is  i 2 
. It looks complicated, but it
6
i 1
10
10 10  12  10  1 10  11  21
really isn’t – just put 10 in for n:  i 2 

 385 . It’s amazing!
6
6
i 1
There are many interesting summation formulas like these, but I’ll resist the urge to
present more of them. Oh, well, maybe one more. Look at this pattern:
2 0  1  2  1  21  1
2 0  21  1  2  3  4  1  2 2  1
2 0  21  2 2  1  2  4  7  8  1  23  1
2 0  21  2 2  23  1  2  4  8  15  16  1  2 4  1
2 0  21  2 2  23  2 4  1  2  4  8  16  31  32  1  25  1
2 0  21  2 2  23  2 4  25  1  2  4  8  16  32  63  64  1  26  1
2 0  21  2 2  23  2 4  25  26  1  2  4  8  16  32  64  127  128  1  27  1
2 0  21  2 2  23  2 4  25  26  2 7  1  2  4  8  16  32  64  128  255  256  1  28  1
Do you see the pattern? Let’s just look at the beginning and the ending of the last line:
20  21  22  23  24  25  26  27  28  1
7
Using summation notation, we could write this as
2
 28  1 . The previous line would be
i
i 0
6
2
i
 2 7  1 . And the general formula would be
i 0
n
2
i
 2 n1  1 . In words we would say that
i 0
0
the sum of the powers of 2 beginning with 2 is equal to one less than the next higher power of 2.
Expressing Sums of Empirical Numbers
The word empirical comes from the Greek word for experience, and is defined as “based
on, concerned with, or verifiable by observation or experience rather than theory or pure logic.”
Basically, it’s numbers that just happen rather than ones you can figure out. Let’s say I ask six
people to count the change they have in their wallets at the present moment. There’s no way I
could get those numbers correct just by thinking.
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If I want to find the total amount of change the six people have, I would add up the
amounts. Sometimes we use the summation symbol  just to indicate adding things up,
without limits, a dummy variable, or an expression like i , i 2 , or 2 i . I give a name to the
variable representing the amount of change a person has (how about x?) and put the amounts in a
list:
x
1.35
0.75
0.42
3.07
1.27
2.43
If I want to indicate adding the numbers up, I can simply write
x.
In this case,
 x  1.35  0.75  0.42  3.07  1.27  2.43  9.29 .
The six people have a total of $9.29 in change.
If I wanted to be a little more precise about this process, I could designate the amount of
change individuals have by using subscripts. So Person #1’s amount of change, $1.35, would be
labeled x1; Person #2’s, which is $0.75, would be x2; and so on. Then I could write the total
using both limits and a dummy variable, but not an expression like i 2 , more of a name really, xi:
6
x
i 1
i
. This equals x1  x2  x3  x4  x5  x6 and is exactly the same as the sum
which equaled 9.29.
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 x above,
Lecture #24: What Is Calculus?
The branch of mathematics known as calculus (from the Latin word for “a pebble or
stone used for counting”) has many applications in our modern technologically-oriented world.
It is based on algebra and geometry, and at least 95% of success in learning calculus depends on
how fluent a person is in algebra. You now know enough algebra to be able to understand at
least the two basic ideas of calculus.
Differential Calculus – the Study of Slopes
When we looked at the slope-intercept equation of a straight line, y  mx  b , we saw
that the slope m is a measure of how the line slants, and it doesn’t matter which two points on the
line you choose to determine it. The equation y  mx  b is a linear function because its graph is
a straight line.
But how about functions which aren’t linear? What can we say about their slopes.
Consider our old friend y  x 2 . It looked like this when graphed:
We can’t really talk about the slope of this parabola, because its slant and direction change as
you go from left to right. At first it’s moving downward, then at the origin it’s flat, and then it
begins to rise. So instead, we imagine drawing a straight line which just grazes the parabola,
touching it at one point. I’ve drawn two such lines, and labeled the points they touch A and B:
We call these lines tangent lines, from the Latin word for “touching.” It’s the slope of these
tangent lines that we are interested in. You can see that the tangent line at point A has a negative
slope and the one at B a positive slope. If we drew the tangent line that grazes the parabola at the
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point 0, 0 , its slope would be 0. The slopes of the tangent lines change in a systematic way as
we move from left to right.
And we can state precisely how they change and what they equal. The slope of a tangent
line at a point on the parabola with x-coordinate x is 2x . The process of finding this formula for
the slope of the tangent in terms of x is called differentiating the function y  x 2 , and the
function y  2 x is called the derivative of x 2 . At the point  3, 9 the tangent line has slope
1 1
2  3  6 ; at the point 0, 0 the slope is 2  0  0 , as mentioned, and at the point  ,  the
2 4
1
slope is 2   1 .
2
Applications of Differentiation
This is all very interesting, but how is it used? Well, let’s say you go for a drive, and we
graph your progress, putting the time (in minutes) on the horizontal axis, and the distance
traveled (in miles) on the vertical axis. The graph might look like this:
Even without the axes having numbers marked off on them, you can see that you started off at a
good pace (because you covered a fair amount of ground at first), then slowed down and stopped
for a while, then went on, then stopped again, then went on, then stopped again, and finally went
on. It was a go-and-stop-and-go kind of trip.
The results of calculus show that, if the graph depicts the total distance traveled at any
point in time (in other words, the graph is that of the distance function d t  , the number of
miles traveled after t minutes have elapsed), then the slope of the tangent line at any point on the
graph is the speed you’re going at that exact moment. It’s called the instantaneous rate of
change. It’s what you would see on your speedometer. I’ve shown several of these tangent lines
in these graphs:
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In the first graph, the point where the tangent line is drawn is in a part of the trip where
you’re making steady progress. The slope is fairly large. In the middle, you’re in a part of the
trip where you’re stopped, and, sure enough, the slope of the tangent line, which is your speed, is
0. The tangent line on the right comes at a time when you’re slowing to a stop, and its slope (and
hence your speed) is smaller than the one depicted on the left.
So the derivative of the distance function d t  could be called the velocity function v t 
. Similarly, if you’re a manufacturer, and you graph the number of items you produce on the
horizontal axis and the total cost of producing them on the vertical axis, thus making a cost
function, the derivative of this function, which shows what it costs to make the next item, is
called the marginal cost function, which is important in business applications. Like the
distance function, the cost function never dips down, because the costs mount as you make more
items.
This is a very characteristic shape for a cost function. At first, when the production process is
just getting started, each item is fairly expensive to make, and the total cost increases quickly.
Then the process hits its stride; everything is working smoothly. It doesn’t cost much to make
additional items. But then equipment starts to wear out, and costs rise rapidly again.
Here’s a growth curve with a common shape. It’s called a logistic curve and looks like
this:
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Say you’re growing lettuce in a small area. At first the plants are tiny and grow slowly. The
slopes of the tangent lines, which indicate the rate of growth, are very small. The tangent lines
are almost flat. Then the plants start growing more and more rapidly, and the tangent lines have
larger slopes. But eventually the lettuce exhausts all the space (and possibly the nutrients)
available, and it ceases to grow. It’s reached what’s called the carrying capacity of its
environment. The tangent lines are virtually horizontal.
Many other physical situations exist in which a function has a physical interpretation, and
its derivative, the function that gives the slopes of its tangent lines, has a related physical
interpretation.
Integral Calculus
What if instead of graphing the distance function (how far you’ve gone since you started)
you graphed the velocity function? It would be like having your speedometer hooked up to a
graphing device. This is the subject matter of integral calculus.
It turns out that if you find something called the area under the curve (the velocity
function in this case) you get the total distance traveled. Let’s start with an easy example. What
if you travel at a constant speed of 60 miles per hour for half an hour? It’s pretty obvious that
you’ll have gone 30 miles. Here’s a graph of this example:
The grey shaded area is indeed equal to 30. It’s a rectangle with dimension 0.5 and 60, for a
product of 30. The area under the velocity curve gives the distance. We say that the distance
function is the integral or antiderivative of the velocity function.
What if you start with a speed of 0 miles per hour and speed up steadily until you’re
going 60 miles per hour, and it takes you half an hour to reach that speed? How far will you
have gone? Here’s a graph:
215
The grey shaded area will be equal to the distance you traveled. It’s a triangle with base 0.5 and
1
height 60, so the formula for the area of a triangle, A b, h   bh , gives us
2
1
A 0.5, 60  0.560  15 . You went 15 miles, exactly half of the distance you went when you
2
traveled at 60 miles an hour for half an hour. This makes sense, because your average speed in
this second example, as you accelerated from 0 mph to 60 mph, was 30 mph.
What if the area under the velocity curve isn’t something simple like a rectangle or a
triangle? Then a clever trick is used, which I’ll illustrate with our triangle example:
You make rectangles which more or less have the same total area as the area under the velocity
curve. You can see that the grey shaded rectangles have some parts that stick up above the
curve, and also that some of the area under the curve isn’t in the rectangles. But you add up the
areas, using the  notation, and get an approximation of the area. And then you make
skinnier rectangles, and obviously more of them, and you’ll get a closer and closer
approximation.
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You can see that the sum of the areas of these rectangles will be closer to the actual area under
the curve than when there were fewer rectangles. Calculus uses the concept of a limit to mean
“more and more and more,” and the limit of the sum of the areas of the rectangles is defined as
the area under the curve.
Here’s an illustration of a rather irregular curve with the rectangles drawn in:
Just as the distance function is the integral of the velocity function, the total cost function
is the integral of the marginal cost function, and many other useful functions pair off in this way,
which is called the Fundamental Theorem of Calculus: if f x  is the derivative of g x  , then
g x  is the integral of f x  .
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Lecture #25: Quantitative Language
You might wonder why there’s a separate lecture on quantitative language. Hasn’t this
whole course been about precisely that? Aren’t algorithms just directions in words? A glance at
the Index of Lectures shows the vast variety of words we’ve used to discuss numbers.
The reason for this special emphasis on language is that math students often have trouble
translating words and phrases into mathematical symbols and sentences. In fact, they may resent
the “intrusion” of language into mathematics. Even if they find math unpleasant, at least, they
believe, it doesn’t involve writing.
How wrong they are! As you’ve seen, mathematics is a language, a very general
language that can concisely describe many different phenomena in the same terms. So I’ll just
point out some particular trouble spots.
First, there’s the tendency to pronounce expressions incorrectly. If you say “5, 4,” for 54
you are cheating yourself of the meaning of the expression. Say “5 to the 4th power,” as it should
be said. You’ll be helping yourself to conceptualize the nature of exponential expressions.
Likewise, if you say “log 3, 81” for log 3 81 instead of “log base 3 of 81” you are making it less
likely that you will comprehend the true meaning of logarithms.
Another common mistake is to say things like “5 and 2 make 7” instead of “5 plus 2
equals 7.” “And” is not the operator word for addition, and 5 and 2 “make” 7 only when added.
The more specific and precise your quantitative language is, the better you’ll be able to use
mathematics, rather than have it use you.
The area of quantitative language which students find the most difficult by far is that
which involves comparing the size of numbers. The Trichotomy Property from Lecture # 13 is
easy enough in theory: Given any two numbers, either the first is equal to the second, less than
it, or greater than it.
Here are some words and phrases that translate into these relations:
Is: Though it obviously doesn’t always mean “equals,” it’s one distinct possibility, as in
“Jo is 26 years old.”
Has: Ditto, as in “Jo has 3 cats.” Of course you have to realize that the variable in the
resulting equation J  3 refers not to Jo, but to the number of cats she has.
Exceeds: If a number x exceeds 20, it is greater than 20, or x  20 . Think of exceeding
the speed limit, which means going faster than the limit posted. If you’re going 65 mph in a 65mph zone, you’re not exceeding the speed limit, as in “But, officer, I was only going 65.”
“Exceeds” has no common antonym, but if it did, the word would mean “is less than.”
An antonym is a word that means the opposite of another word. It was actually coined (made
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up) in 1867 to be the opposite of “synonym,” which means a word whose meaning is the same
as another word’s. (So “antonym” is the antonym of “synonym”….)
Positive: If a number x is positive, it is greater than 0, or x  0 .
Negative: If a number x is negative, it is less than 0, or x  0 .
At least: Without a doubt, the phrase that students have the most trouble translating is
“at least,” as in “I have at least 10 hours of work to do today.” If x stands for the number of
hours of work I have to do today, this translates into x  10 ; in other words x is greater than or
equal to 10. “Greater than or equal to” was one of the combinations that two of the symbols
from the Trichotomy Property, > and =, made when the symbols were paired. So “at least” is a
synonym for “greater than or equal to.” I think the problem arises because of the L-E at the
beginning of “least,” which makes people think of “less.” But you know better: “You must be at
least 21 to purchase alcohol.” “I’m 16, does that work?” “No.” But on your 21st birthday you
are officially at least 21.
At most: This doesn’t bother students as much, for some reason. It means “less than or
equal to.” If a number x is at most 45, then x  45 . Like “at least,” it combines two of the three
Trichotomy Property options, < and =.
This final group of three words pertains to a situation described in the next lecture,
Lecture #26: Variables in Statistics. If a group of people is asked whether they will vote for
Candidate A or Candidate B, and if more people pick Candidate A than pick Candidate B, we
say that Candidate A has a majority. In other words, the fraction of people voting for Candidate
1
A is greater than . If fewer people pick Candidate A than pick Candidate B, we say that
2
Candidate A has a minority of the vote. The fraction of people voting for Candidate A is less
1
than (and thus Candidate B has a majority). If the votes are evenly split, neither candidate has
2
a majority.
What if there are three candidates, Candidate C having joined in the race? A candidate
with more than half of the total vote has a majority, and a candidate with less than half has a
minority, but if one candidate has more votes than the other two, even if his/her votes aren’t
more than half of the total votes, we say that this candidate has a plurality. Depending on the
election laws, either the candidate with the plurality is the outright winner, or the two top votegetters face off in a future election to see who wins.
219
Lecture #26: Variables in Statistics
So far we’ve dealt with two kinds of variables: the ones in algebra which stand for
numbers and the ones in symbolic logic which stand for statements. In statistics we encounter a
third kind of variable, something that describes a person, a dog, a college, a city, a nation, a
business, or pretty much anything.
A variable could be a person’s gender or age, a dog’s breed or weight, a college’s
enrollment, a city’s population, a nation’s type of government or gross national product, or a
business’s number of employees. It’s not that these variables necessarily vary; most people’s
gender stays constant throughout their life, and a dog’s breed certainly does. It’s that the
variables vary from person to person, from dog to dog, etc.
We want to classify these variables according to their type. The two basic types are
qualitative (also called categorical) and quantitative. Qualitative variables deal with some
characteristic of the group being studied. People are either male or female. Dogs are of many
breeds, not only dachshunds and mutts. Colleges can be private or public, and can be two-year
institutions, four-year institutions, or have graduate programs and thus be termed universities.
Cities in the United States can be in one of fifty states (plus the District of Columbia). A nation
can be a democracy, a republic, a monarchy, or something else. A business can be retail or
wholesale.
But in none of these cases can we assign a number to the variable. We can only put the
group’s members into categories. We can’t talk about the average gender of people, the average
breed of dogs, or the average state that U.S. cities belong to. The very concept makes no sense.
So if we want to report on what we’ve discovered, all we can do is say what fraction (or percent)
of the group belongs to which category. We can say what fraction of a group of people are male,
what fraction of a group of dogs are dachshunds, what fraction of institutions of higher learning
in a certain state are public, what fraction of the nations of the world have a democratic form of
government, and what fraction of businesses in a certain city are retail. Pretty much anything
that describes some kind of group can be a qualitative variable, and we can say what fractions of
the group belong to which category. And that’s as far as we can go in describing that variable.
The fraction describes the whole group, not the individual member.
These qualitative variables are classified according to whether there are two categories or
more than two categories. When there are exactly two categories, we call the variable binomial
(from the Greek words for “two” and “name”). People’s gender is a binomial variable, as are
also the private/public nature of an institution of higher learning and the retail/wholesale nature
of a business.
But some qualitative variables clearly have more than two possible categories. There are
many other kinds of dogs than just dachshunds and mutts. There are fifty states for a city to be
located within. Institutions of higher learning can be two-year, four-year, or universities. There
are many forms of government available to nations. Variables like this are called multinomial
(many names). Sometimes we know how many categories a qualitative variable has, for instance
220
if the variable is which of the fifty states a city is located within, and sometimes we don’t (try
listing all breeds of dog!). Either way, the variable is multinomial.
If you look at the fractions of a group in each category of a qualitative variable, I hope
it’s obvious to you that the fractions have to add up to 1. Everything in the group is in one
category or another. Say you’re looking at institutions of higher learning in a certain city, and
there are 3 two-year colleges, 5 four-year colleges, and 2 universities. The fraction of the group
3
5
2
which are two-year colleges is
, four-year colleges
, and universities
, and
10
10
10
3 5 2 3  5  2 10
  

 1.
10 10 10
10
10
There’s a special relationship between the fractions of a group if the variable is binomial.
If you know one fraction you can figure out the other by subtracting the fraction from 1. Say
you’ve got a group of 20 people, of whom 13 are female. So the fraction of the group which is
13
7
female is
. Since 20  13  7 , the fraction which is male is
, and indeed
20
20
13 20 13 20  13 7
.
1




20 20 20
20
20
A common use of statistics involves taking surveys or polls, and sometimes the questions
asked lead to qualitative variables. “In the upcoming election, are you going to vote for
Candidate A, or are you going to vote for Candidate B, or are you undecided?” Three categories
here, thus multinomial. “Are you a resident of an incorporated city?” This is a very common
situation in surveys – the answer is either yes or no, and the variable is binomial. “What do you
think is the biggest challenge facing our country today, unemployment, income inequality, global
warming, crime, health care costs, or something else?” Multinomial.
Quantitative Variables
Any variable that assigns a number to each member of a group is called a quantitative
variable, and obviously these are of great interest in a variety of fields, especially the physical
and social sciences. But it turns out that different kinds of quantitative variables have vastly
different characteristics, which means that the mathematical operations we can perform on them
also differ.
This topic is called levels of measurement, and there are four of them.
The lowest, most primitive level of measurement is the nominal level, in which the
numbers aren’t really numbers at all in the sense of being able to have computations done on
them. They are actually names, which is what the word “nominal” means. Think of 007 –
Double-O-Seven. Your Social Security number is a ten-digit number the government uses to
identify you. Adding up the Social Security numbers of your family would be ridiculous, as
would be finding the average of those numbers. Bragging that your SS number is bigger than
someone else’s would be laughable. There are many other examples of quantitative variables at
the nominal level of measurement – phone numbers, bus routes, and ZIP codes, to list a few.
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The next lowest level, in which the numbers have a little mathematical meaning, is called
the ordinal level of measurement, and it applies to questions like, “On a scale of 1 to 10…..” A
10 is certainly better than a 9, a 9 is more positive than an 8, and so on, which is why this level is
called ordinal, referring to order. So the values of the variable can be compared to each other
and ranked (though you could argue that one person’s 8 is really like another person’s 5, because
the first person tends to be very enthusiastic and supportive – but let’s not go there). But what
you can’t do (though it’s done all the time) is to take an average rating, or to subtract the ratings
(which is done less often), as in the example about the 8 and the 5, and say that the first person is
3 more favorable than the second one. The intervals between the ratings are meaningless. Some
other examples of variables at the ordinal level are movie and restaurant stars, and grades in
courses.
I’m going to mix together the highest two levels but will distinguish between them in the
Statistics course. They share many characteristics anyway. The levels are called the interval
level and the ratio level. Variables at these levels involve numbers that can be averaged and
subtracted. Think of weight. Each person has a weight; we can compute the average weight of a
group of people; and we can say that one person weighs a certain number of pounds (or
kilograms) more or less than another person. These are the variables we really like in statistics,
because we can do so much with the numbers! Other examples of variables at these levels are
temperature, age, height, number of TVs in a household, number of children a person has, and
the number of movies you watched last year.
Do you notice anything different between the first three examples – temperature, age and
height – and the last three – number of TVs, children and movies watched? Consider the
discussion in Lecture #7 about the distinction between measuring and counting. You measure
the former and count the latter.
We have a name for quantitative variables which involve measuring: They’re called
continuous. That’s because you can always find a value of the variable between two values; in
other words the values are dense in the sense we discussed in Lecture # 13. You might be said
to weigh 150 lbs., but if you had a more accurate scale your weight might register as 149.7 lbs.,
and if you had an even more accurate scale it might say…..well, you get the idea. We measure
continuous variables with some sort of instrument – scales, thermometers, yard stick, etc.
You obviously don’t measure the number of TVs, children or movies watched – you
count them. You can’t have 2.6 TVs, children or movies watched. The numbers for these
variables aren’t dense; they have gaps. We call these discrete variables (from the Latin word for
“separated,” as opposed to concrete, which comes from the Latin word for “grown together”). It
makes perfect sense to average a set of discrete numbers, though there are many jokes about the
family which has 2.6 children. We’re not saying that there is such a family, or such a 0.6 child,
rather that we added up all the children in the families and divided by the number of families and
got 2.6.
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Lecture #27: Visual Display of Data
When you have a group of numbers, or “yes’s” and “no’s” or any other values of
statistical variables, this resulting group is called a data set. Data is the plural of datum, from
the Latin word meaning “thing given.” Often data sets are overwhelming to look at, and the
important facts about the sets are obscured by their very size. So we make pictures of the sets,
and these pictures help us to grasp their essence.
The kind of picture you make depends on the kind of variable the data set displays. In
the case of qualitative or categorical data sets, you have a couple of choices.
First there’s the ever-popular pie chart. Here are a couple of examples:
The one on the left is about movies, and the one on the right depicts ways that people in the
United Kingdom get to work.
Pie charts are effective because they use the simple-to-view circle, which is divided into
labeled slices according to what fraction of the data set falls into the various categories. The
number in each category is called its frequency, and the fraction of the set which the category
takes up is called the relative frequency. The size of each slice is determined by what’s called
its central angle, the angle the slice makes at the center of the pie. You’ll construct one in
Statistics, but for now you can just appreciate how compactly the pie chart conveys the nature of
the data set.
Another way to represent categorical data is by using a bar graph. In this kind of graph,
the height of the bars is determined by the frequency of the category represented. Here is the
favorite-movie data set put into a bar graph:
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In this graph, as in the pie chart, it’s very easy to see that the most popular type of movie
is the romance, trailed slightly by action films, with drama coming in last.
If you arrange the bars from left to right (or top to bottom) from the one with the highest
frequency to the one with the lowest frequency, you get a Pareto chart, which gives added ease
to understanding which categories have the most responses and which have the least.
Again, the length of the bar shows the frequency of the category.
Here are a couple:
As you can see, these graphs are often used in business contexts.
The reason we can move the bars around and arrange the categories any way we like is
that the variable depicted is qualitative or is quantitative but at the nominal level of
measurement. If the variable is quantitative, and the numbers have some mathematical meaning,
you can’t just manipulate the picture to show decreasing categories: You have to maintain the
order of the variable involved.
The resulting bar graphs are called histograms. The etymology of this word is uncertain.
It might come from the Greek word meaning “set upright,” or it might have been invented by the
great English mathematician and biometrician Karl Pearson (1857-1936), using the beginning
and the end of the phrase “historical diagram.” The height of the bars again corresponds to
frequencies on the vertical axis, and the values of the variable are grouped from smallest to
largest on the horizontal axis:
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Looking at the pattern of the bars leads us to consider what’s called the shape of the
distribution. Some, like the middle one, are symmetrical, with the tallest bars in the center,
tapering off to the sides. This resembles what in statistics we call a normal distribution, one of
the most common in nature and human affairs. The ones on the sides are called skewed. On the
left, large values of the variable predominate; on the right, small values are the most common.
The last kind of visual representation of data we’ll consider concerns bivariate data
(literally “two variables”). In these situations, each member of the group has two numbers
associated with him, her, or it – one value for each of two variables. Pictures of bivariate data
are commonly called scatter plots. One of the variables is represented on the horizontal axis
and thus is the x-coordinate of a dot on the graph; the other is represented on the vertical axis and
is the y-coordinate. Here’s one:
Each dot shows the weight and height of an individual person. Not surprisingly, heavier people
tend to be taller, and vice versa. We call this a positive relationship, in which larger values of
one variable are associated with larger values of the other.
We can imagine a straight line running among these points, and its slope would be
positive:
225
In statistics, we study the characteristics of this line, which is called, impressively enough, the
least-squares best-fit regression line.
Here’s a scatter plot that shows a negative relationship:
Smaller values of the process input (on the horizontal axis) are associated with larger values of
the quality characteristic (on the vertical axis). If we drew a straight line amid the points, it
would have a negative slope.
Some scatter plots show a definite pattern of points which a straight line would by no
means do justice to. These various patterns correspond to some of the functions we looked at in
Lecture #20. Here’s one:
226
I hope this reminds you of a quadratic function, whose graph is a parabola (in this case a
downward-opening one). At first the relation is positive, but after the function peaks at about 0.5
on the x-axis, the relation becomes negative. Many examples from natural and human
experience fit this model.
Here’s another, depicting the price of stamps over time:
The curve that fits the points best is shown also, and it’s an exponential function, conveying the
idea that the price of stamps increased gradually at first but began rising more and more rapidly.
As you may remember, exponential growth may not be sustainable over the long run, and
a more realistic view might be to view the situation as a logistic function, where the growth rate
tapers off and the size of the variable on the y-axis stabilizes. This is illustrated in this scatter
plot with the logistic curve included:
Or we might have a situation in which the growth rate slows as the x-variable increases,
resulting in a scatter plot best approximated by a logarithmic function:
227
Lastly, here’s a recognizable pattern which we didn’t discuss in Lecture #20:
The dots rise and fall, and go on rising and falling, in cycles. This happens in many situations
which are dependent on the seasons. They are best described using trigonometric functions,
also called circular functions, and that’s all I’m going to say on that subject.
228
Changing a fraction to a percent, 18
Changing a percent to a decimal, 13
Changing a percent to a fraction, 18
Circle, 184
Circular functions, 228
Circumference of a circle, 185
Closed circle, 101
Closed geometric figure, 177
Closed interval, 102
Coefficient, 137
Combining like terms, 73
Commutative property of addition, 90
Comparison, words of, 219
Compound inequalities, 102
Constant, 138
Constant functions, 163
Constant terms, 75
Contents of absolute value, 109
Continuous variables, 222
Contradictions, 84
Contrapositive of a statement, 196
Converse of a statement, 195
Conversion factor, 46
Converting an improper fraction into a
mixed number, 23
Coordinates, 144
Cost function, 214
Counting, 44, 222
Cube, 186
Cube root, 126
Cubit, unit of measurement, 43
Cylinder, 187
Data, 223
Data set, 223
Datum, 223
Decider, 28
Decimal denominator, 14
Decimal fractions, 11
Decimal notation, 119
Default index, 126
Default placement of the decimal point, 12
Default setting of signs, 55
Default sign for operations, 64
Degree of a polynomial, 168
Denominator, 1
Dense, 100, 222
Index
Absolute value, 55
Absolute value equations, 109
Absolute-value functions, 163
Absolute value operator, 67
Acre, 45
Addition of signed number, 56
Addition rule (for equations), 79
Algorithm, 1
Al-Khwarizmi, 1
All real numbers, R, 85
Amount, in percent problems, 38
And, 193
And/or, 194
Antiderivative of a function, 215
Antonym, 215
Apex of a triangle, 181
Applications, 89
Area, 44, 178
Area under the curve, 215
Argument of a function, 157
Bar graph, 223
Base, in exponential expressions, 67
Base, in percent problems, 38
Base of a triangle, 180
Bases of a trapezoid, 183
Between, 23
Binomial, as a polynomial, 168
Binomial variables, 220
Bivariate data, 225
Boundaries, 44
Box, 187
Brackets, 64
Building factor, 3
Calculus, 212
Caret, 68
Carrying capacity, 215
Cartesian plane, 146
Categorical variables, 220
Center of a circle, 185
Centi-, prefix, 43
Central angle, 223
Changing a decimal to a fraction, 17
Changing a decimal to a percent, 12
Changing a fraction to a decimal, 13
229
Derivative of a function, 213
Descartes, René, 143
Descending order, 170
Diameter of a circle, 185
Diamond, 177
Difference, 4
Differential calculus, 212
Differentiating a function, 213
Dimensional analysis, 45
Dimensionality, 45
Discount, 94
Discrete variables, 222
Distance function, 213
Distribution, 73
Division of signed numbers, 59
Divisor, 2
Domain, 158
Dummy variable, 209
Either/or, 99, 194
Empirical, 210
Empty set, Ø, 84
Equations, 79
Equilateral, 180
Equivalency, 44
Equivalent, 64
Equivalent measures, 44
Etymology, 1
Euclid, 148
Evaluating algebraic expressions, 71
Expanded notation, 27
Exponential decay, 170
Exponential expression, 68
Exponential functions, 169, 227
Exponential growth, 170
Exponents, 67
Expression, 63
Factor, 2
Formulas, 177
Fraction bar, 66
Frequency, 223
Fuel efficiency, 33
Function notation, 157
Functions, 157
Functions of several variables, 172
Fundamental Theorem of Calculus, 217
Furlong, unit of measurement, 43
Gauss, Carl Friedrich, 209
Graph of an equation, 146
Grouping symbols, 64
Guess-and-check, 89
Half-open interval, 102
Hand, unit of measurement, 43
Height of a triangle, 180
Hero of Alexandria, 181
Hero’s formula, 181
Heron’s formula, 181
Hexagon, 184
Higher roots, 126
Histogram, 224
Horizontal lines, 152
Horizontal translation, 164
Hypotenuse of a right triangle, 182
Identities, 84
Identity element for multiplication, 66
Identity element for addition, 79
If-then statements, 194
Implicit multiplication, 64
Improper fraction, 3, 22
Index of a radical, 126
Inequalities, 99
Inequalities with absolute value, 110
Infinity,  , 101
Inspection, solving an equation by, 79
Instantaneous rate of change, 213
Integers, 55
Integral calculus, 215
Integral of a function, 215
Interval level of measurement, 222
Interval notation, 100
Inverse of a statement, 195
Inverse functions, 159
Inverse operations, 125
Irrational numbers, 125
Isolate a variable, 81
Isosceles, 180
Kilo-, prefix, 43
Leading coefficient of a polynomial, 168
Least common denominator, 5
Least-squares best-fit regression line, 226
Leg of a right triangle, 182
Lemniscate, 101
Length, 44
230
Levels of measurement, 221
Light-year, unit of measurement, 43
Like terms, combining, 74
Limit, 217
Linear, 79
Linear equations in one variable, 99
Linear equations in two variables, 136
Linear functions, 163
Linear measurement, 177
Logarithmic functions, 171, 227
Logistic curve, 214, 227
Lower limit of summation, 207
Majority, 219
Making a 1, 80
Making a 0, 80
Marginal cost function, 214
Measuring, 44, 222
Meter, unit of measurement, 43
Milli-, prefix, 43
Minority, 219
Mixed numbers, 3, 16
Modeling situations, 89
Monomial, 168
Multinomial variables, 220
Multiple, 2
Multiplication of signed numbers, 59
Multiplication rule (for equations), 80
Multiplication rule of counting, 12
Natural numbers, 55
Negation, not p, 193
Negative exponents, 117
Negative relationship, 226
Nominal level of measurement, 221
Non-negative, 136
Normal distribution, 225
Notation, 64
Number line, 21
Numerator, 1
Octagon, 184
One-step equations, 81
Open circle, 100
Open interval, 102
Opposites, 58, 66
Or, 194
Order of magnitude, 118
Order of operations, 63
Ordered pair, 136
Ordinal level of measurement, 222
Origin, 144
Parabola, 167
Parallelogram, 179
Parenthesis, parentheses, 64
Pareto chart, 224
Pearson, Karl, 224
Pentagon, 184
Percent, 20, 38
Percent-base-amount problems, 37
Perfect cubes, 126
Perfect squares, 125
Perimeter, 177
Periods, 27
Pi,  , 185
Pie chart, 223
Place values, 27
Plane figure, 184
Plurality, 219
Polygon, 179
Polynomial, 168
Polynomial functions, 168
Positive relationship, 225
Power, 67
Power Rule of exponents, 116
Pressure, 33
Prime factorization, 14
Product, 3
Product Rule of exponents, 115
Proportion, 37
Pythagoras, 182
Pythagorean Theorem, 182
Pythagorean Triple, 182
Quadrant, 145
Quadratic functions, 166
Quadrilateral, 177
Qualitative variables, 220
Quantitative variables, 220
Quotient, 2
Quotient Rule of exponents, 116
Radical sign, 67
Radicand, 67
Radius of a circle, 185
Raising a fraction to higher terms, 3
Range, 158
231
Rates, 33
Ratio level of measurement, 222
Ratios, 33
Reciprocals, 63
Rectangular coordinate system, 143
Rectangle, 178
Rectangular parallelepiped, 187
Recursive, 1
Reducing a fraction to lowest terms, 1
Reflection across the line y = x, 160
Reflection across the x-axis, 165
Reflection across the y-axis, 170
Regular polygon, 184
Relative frequency, 223
Repeated addition, 59
Repeated multiplication, 67
Repeater bar, 15
Repeating decimal, 15
Rhombus, 177
Right circular cylinder, 187
Right triangle, 180
Rigid transformation, 165
Rise, 149
Rounding numbers, algorithm, 28
Run, 149
Scalene, 180
Scatter plot, 225
Scientific notation, 118
Semiperimeter, 181
Set brackets, 163
Shape of a distribution, 225
Shrinking a graph, 165
Sigma notation, 209
Signed numbers, 55
Significant digits, 120
Simplifying algebraic expressions, 73
Skewed, 225
Slope, 148
Slope fraction, 149
Slope-intercept form, 150
Solve an equation, 79
Solved for y, 138
Solving linear equations in one variable,
79
Speed, 33
Sphere, 188
Square, 177
Square root, 67
Standard form of a linear equation with two
variables, 137
Statement, 193
Stretching a graph, 165
Subscripts, 183
Subtraction of signed numbers, 58
Sum, 4
Summation, 209
Surface area, 186
Syllogism, 196
Symbolic logic, 193
Synonym, 219
Syntax, 56
Systems of equations, 90
T-table, 146
Tangent line, 212
Tax, 93
Terms, of a fraction, 1
Three-dimensional shapes, 186
Transformations, 165
Transitive Property of Equality, 116
Transitive Property of Implication, 196
Transitive Property of Inequality, 102
Trapezoid, 183
Trial-and-error, 89
Triangle, 179
Trichotomy Property, 99
Trigonometric functions, 228
Trigonometry, 182
Trinomial, 168
Truth table, 193
Truth values, 193
U. S customary units of measurement, 43
Undefined, 59
Union,  , 102
Unit fraction, 126
Unit price, 34
Upper limit of summation, 209
Variables, 71
Variables in statistics, 219
Velocity function, 214
Venn diagram, 197
Venn, John, 197
Vertex, 168
232
Vertical translation, 161
Volume, 44
Volume of three-dimensional shapes, 187
Whole numbers, 2, 55
Word problems, 89
X-axis, 143
X-coordinate, 144
X-intercept, 151
XY plane, 146
Y-axis, 143
Y-coordinate, 144
Y-intercept, 148
Zero as an exponent, 115
233