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Measuring with body parts
In this exercise you will make some measurements using
your own body parts. Find two parts of your body with
which to measure lengths. The longer measure will be
called a “Limb” and a smaller measure will be called a
“Digit”. Choose your body parts so that your “limb” is ten
times as long as your “digit”.
1. Measure the length of your table in limbs only.
2. Measure the length of your table in digits only.
3. Measure the length of your table in limbs and digits,
use as many limbs as possible.
4. Measure your partner’s height in limbs and digits,
use as many limbs as possible.
5. Discuss with your partner how you would convert
your measurement of the table in limbs and digits to
your friend’s measurement of the table in his/her
limbs and digits.
The meaning of decimal numbers,
expanded form.
Consider the whole number 3025 .
We know that this number can be written in expanded form
as follows:
3025 = 3000 + 0 + 20 + 5
3025 = 3 • 1000 + 0 • 100 + 2 • 10 + 5 • 1
3025 = 3 • 10 3 + 0 • 10 2 + 2 • 101 + 5 • 10 0
Any whole number can always be written as a sum
of multiples of powers of 10.
Similarly, “decimal numbers” are also written using powers
of 10. Consider the number 52.306 .
This number can be written in expanded form as follows:
3
0
6
+
+
10 100 1000
1
1
1
52.306 = 5 • 10 + 2 • 1 + 3 • + 0 •
+6•
10
100
1000
1
1
1
52.306 = 5 • 101 + 2 • 10 0 + 3 • 1 + 0 • 2 + 6 • 3
10
10
10
52.306 = 5 • 101 + 2 • 10 0 + 3 • 10 −1 + 0 • 10 −2 + 6 • 10 −3
52.306 = 50 + 2 +
( Recall that
10 −2 =
by definition ! )
1
10 2
, and that in general
10 − n =
1
10 n
A decimal number is one which is written in the form
…C • 10 2 + B • 101 + A • 10 0 + a • 10 −1 + b • 10 −2 + …
or in its short-hand form using the “decimal point” to
separate the non-negative powers of 10 from the negative
powers of 10 .
In words, a decimal number is a sum of multiples of powers
of 10 where the powers may be any integer.
For example,
3 • 10 2 + 2 • 101 + 1 • 10 0 + 4 • 10 −1 + 5 • 10 −2 + 7 • 10 −3 + 1 • 10 −4 = 321.4571
is a decimal number. The left hand side is the expanded
form and the right is the short-hand form.
Finding the powers of ten on the number line
The non-negative real numbers, with convenient markers
0
1
10
10
0
10
100
1
10
2
zoom in
1
10
100
0
11
100
100
11
10
10
-2
10
0.01
0.1
10
-1
1
10
0
Note that 10 raised to a negative power is still a positive number, just a very small positive number !
Rational numbers and Decimal numbers
A rational number is a number which can be written as
a
a ratio of two numbers: /b , where a is any integer
and b is a positive integer.
Consider the decimal number 0.25 .
It’s “easy” to find a rational representation (fraction form)
for this number simply by adding the terms.
0.25 is simply
2
5
+
10 100
which is
25
100
The fact that 0.25 “terminates” after the hundredths place
means that we can use ordinary arithmetic to find the
fraction representation. We can use a similar procedure for
any decimal number which eventually terminates.
So,
every decimal number which
terminates is a rational number.
Example:
3
1
5
315
63
0.315 =
+
+
=
=
10 100 1000 1000 200
But in one sense, decimal numbers such as 0.25 can be
considered as “repeating”; that is, a pattern exists in the
sequence of digits beginning somewhere to the right of the
decimal point and this pattern continues “forever” to the
right. For the number 0.25 , after the first two places, each
subsequent digit is zero.
Of course when we add infinitely many zeroes, we get zero.
That is . . .
1
1
+5•
10
100
1
1
1
1
0.25 = 2 • + 5 •
+0•
+0•
+…
10
100
1000
10000
0.25 = 2 •
0.25 = 0.25000…
the decimal representation for this number repeats zeroes
beginning with the third (thousandths) place.
Notation, there are at least two ways to denote a repeating
decimal: using ellipsis, or using a “bar” to denote the
repeating digits
0.25000… = 0.250
1.0373737… = 1.037
****************************
Now:
Consider the decimal 0.3333 . . . where it is understood
that there is a three in every decimal place.
This decimal represents a non-trivial infinite sum.
0.3333… = 3 •
1
1
1
1
+ 3•
+ 3•
+ 3•
+…
10
100
1000
10000
The problem is to find a rational expression (a fraction) for
this decimal, if possible. This is challenging because we
have to add up infinitely many non-zero numbers. It’s an
interesting philosophical question. (Zeno’s Paradox
anyone ? )
In the first place, the only way that infinitely many (nonnegative) numbers can sum to a number OTHER than
infinity, is if the terms in the sum get pretty small pretty
fast.
In our number, the terms in the sum are getting small
pretty fast:
3
3
3
3
,
,
,
, …
10 100 1000 10000
In fact, each term is one-tenth of the previous term.
In analysis, we handle expressions such as these using a
general theory of infinite series. We won’t use those
techniques.
We will use algebra to find a fraction
to represent 0.3333 . . .
In particular we will need to solve a linear equation.
So here goes . . .
First name the quantity,
Let
x = 0.3333 . . .
Multiply both sides by 10
(why?)
10•x = 10•( 0.3333 . . . )
10x = 3.3333 . . .
( short cut: to multiply by a
power of 10, move the decimal )
10x = 3 + ( 0.3333 . . .)
( rewrite, for clarity )
Subtract the same quantity, x ,
from both sides of this equation
10x – x = 3 + ( 0.3333 . . . ) – x
since x is the same as 0.3333 . . .
we can substitute 0.3333 . . . for x
10x – x = 3 + ( 0.3333 . . . ) – ( 0.3333 . . . )
combine like terms
10x – x = 3 + ( 0.3333 . . . ) – (0.3333 . . . )
0
9x
9x = 3 + 0
9x = 3
9x 3
=
9 9
9x 3
=
9
9
3 1
x= =
9 3
divide both sides by nine
So we did it! x = 0.3333 . . . =
1
3
The decimal number 0.3333 . . . is rational,
and it is equal to one-third.
Claim: every decimal number which
“repeats” is a rational number.
Another example: Find a fraction for 0.5721721 . . .
Where it is understood that each decimal place beyond the
tenths is 7, 2 or 1 in cyclical order. In other words, the
decimal “repeats” 721. Notice that the 5 in the tenths
place is not part of the repeating cycle. This is important.
Let
x = 0.5721721 . . .
Multiply both sides by 10
Then
10•x = 10•(0.5721721 . . . )
10x = 5.721721 . . .
(rewrite)
10x = 5 + (0.721721 . . . )
Also, since
x = 0.5721721 . . .
We can multiply both sides by 10,000
10,000•x = 10,000•( 0.5721721 . . . )
10,000x = 5721.721721 . . .
10,000x = 5721 + (0.721721 . . . )
now from the equation
10,000x = 5721 + (0.721721 . . . )
subtract 10x from both sides
10,000x – 10x = 5721 + (0.721721 . . . ) – 10x
(rewrite)
10,000x – 10x = 5721 + (0.721721 . . . ) – 10x
on the right hand side, replace 10x
with { 5 + (0.721721 . . . )}
since these are the same
10,000x – 10x = {5721 + (0.721721 . . . )} –
{ 5 + (0.721721 . . . )}
remove groupings on the right hand side and
combine terms
9990x = 5721 – 5 + (0.721721 . . . ) – (0.721721 . . . )
9990x = 5716
divide both sides by 9900
5716
x=
9990
so the decimal number 0.5721721 . . .
5716
can be written as the fraction 9990
In Summary,
Every decimal number which repeats or terminates
represents a rational number.
To convert a repeating decimal to a fraction, use the
method shown above, to convert a terminating decimal to a
fraction, write the decimal in expanded form with fractions
and do the addition.
Convert a fraction to a decimal.
Let’s convert a fraction to a decimal.
Example: convert
1
4
to a decimal number.
We need to find the whole numbers a, b, c . . .
In the following expansion:
1 a
b
c
=
+
+
+…
4 10 100 1000
✫ Starting with the tenths (the largest unit) find the whole
number a .
Set
1 a
=
if we’re lucky, a will be a whole number,
4 10
in which case we will be finished. Solve for a :
1 a
=
4 10
1
a
= 10 •
4
10
10
a
= 10 •
4
10
10
=a
4
10 •
multiply both sides by 10
and simplify.
a=
10
= 2 24
4
The solution, a , is not a whole number (shucks).
The number a which solves the proportion is a bit bigger
than two. This means that we can put a = 2 for the tenths
but that we must continue on to the hundredths (at least) to
find the complete decimal representation. In other words,
2
10
is a first approximation for
1
4
.
So now we write
1 2
b
c
=
+
+
+…
4 10 100 1000
In preparation for finding the whole number b ,
Subtract
2
10
from both sides of this equation and simplify.
1 2
b
c
−
=
+
+…
4 10 100 1000
5
4
b
c
−
=
+
+…
20 20 100 1000
1
b
c
=
+
+…
20 100 1000
✫ Now find the whole number b .
1
b
=
By setting 20 100
and solving for b .
If the solution is a whole number, we are finished,
otherwise we keep going to the next decimal place.
1
b
= 100 •
20
100
100
=b
20
100 •
multiply both sides by 100
and simplify
b = 5 exactly! So we are finished.
1 2
5
=
+
4 10 100 = 0.25
Example: convert
3
8
to a decimal number.
We need to find the whole numbers a, b, c . . .
In the following expansion:
3 a
b
c
=
+
+
+…
8 10 100 1000
3 a
As before, begin with the tenths by setting 8 = 10
and solving for a. If a is a whole number, we are
finished, if a is not a whole number, we use the whole
number part and then go to the next place value.
3
a
10 • = 10 •
8
10
30
=a
8
multiply both sides by 10
and simplify
30
a=
= 3 68 , not a whole number.
8
In the decimal expansion we put a = 3, and go to the next
place value.
So
3 3
b
c
=
+
+
+…
8 10 100 1000
In preparation for finding the number b ,
3
10
Subtract
from both sides of the equation and simplify
3 3
b
c
−
=
+
+…
8 10 100 1000
15 12
b
c
−
=
+
+…
40 40 100 1000
3
b
c
=
+
+…
40 100 1000
3
b
=
and set 40 100 to find the number b
If b is a whole number, we are finished, if b is not a
whole number, we use the whole number part and then go
to the next place value.
100 •
3
b
= 100 •
40
100
300
=b
40
multiply both sides by 100
and simplify
300
b=
= 7 24
40
Then put b = 7 and go to the next place value.
3 3
7
c
=
+
+
We now have 8 10 100 1000 + …
In preparation for finding the number c ,
Subtract
simplify.
3
10
and
7
100
from both sides of the equation and
3 3
7
c
−
−
=
+…
8 10 100 1000
3
7
c
−
=
+…
40 100 1000
30
28
c
−
=
+…
400 400 1000
2
c
=
+…
400 1000
and then set
2
c
=
400 1000
and solve for c
If c is a whole number, we are finished, if c is not a
whole number, we use the whole number part and then go
to the next place value, as before.
1000 •
2
c
= 1000 •
400
1000
2000
=c
400
multiply both sides by 100
and simplify
c = 5 exactly, so we are finished and
3 3
7
5
=
+
+
= 0.375
8 10 100 1000
2
11
Example: convert
to a decimal number.
We need to find the whole numbers a, b, c . . .
In the following expansion:
2
a
b
c
=
+
+
+…
11 10 100 1000
2
a
=
Set
and solve for a
11 10
2
a
= 10i
11
10
20
=a
11
10i
a=
multiply both sides by 10
and simplify
20
= 1 119 not a whole number.
11
We set a = 1 and continue to the hundredths.
So
2
1
b
c
=
+
+
+…
11 10 100 1000
In preparation for finding b subtract
2
1
b
c
−
=
+
+…
11 10 100 1000
20
11
b
c
−
=
+
+…
110 110 100 1000
9
b
c
=
+
+…
110 100 1000
1
10
from both sides.
9
b
=
Now set
and solve for b
110 100
9
b
= 100i
110
100
900
=b
110
100i
b=
multiply both sides by 100
and simplify
900
= 8 112
110
2
11
Notice the appearance of the fraction
again. Hmmm.
Anyway, put b = 8 and go to the thousandths.
2
1
8
c
=
+
+
+…
We have
11 10 100 1000
In preparation for finding c , subtract
both sides and simplify.
2 1
8
c
− −
=
+…
11 10 100 1000
9
8
c
−
=
+…
110 100 1000
90
88
c
−
=
+…
1100 1100 1000
2
c
=
+…
1100 1000
1
10
and
8
100
from
2
c
=
To find the number c we solve
1100 1000
2
c
1000i
= 1000i
multiply both sides by 1000
1100
1000
and simplify
2000
=c
1100
c=
2000 20
=
= 1 119
1100 11
since c is not a whole number, the decimal expansion does
not stop with the thousandths, we must “keep going”.
But again, notice the reappearance of the fraction
9
11
.
Here is what we’ve got so far:
2
1
8
1
d
=
+
+
+
+…
11 10 100 1000 10000
Two-elevenths is a rational number. So its decimal will
either terminate or repeat. One may speculate that the next
number d will be 8 , and that the digits 1 8 will repeat.
Let’s check it out.
Following the same procedure, subtract
8
100
and
1
1000
from both sides.
1
10
,
2
1
8
1
d
− −
−
=
+…
11 10 100 1000 10000
2
1
d
−
=
+…
1100 1000 10000
20
11
d
−
=
+…
11000 11000 10000
9
d
=
+…
11000 10000
9
d
=
To find the number d we solve
11000 10000
10000i
9
d
= 10000i
11000
10000
90000
=d
11000
d=
multiply both sides
by 10000 and simplify
90000
90
==
= 8 112
11000
11
so d is 8 . And again, notice the re-appearance of the
2
11
fraction
, our original fraction. This is evidence that our
decimal is repeating.
Answer:
2
11
= 0.181818 . . .
Claim: this procedure is equivalent to the
customary “long division” algorithm.
Zeno’s paradox
(after wikipedia)
“That which is in locomotion must arrive at the half-way
stage before it arrives at the goal.”
—Aristotle
----------------------------------------------------------------------Zeno Argued that motion is impossible. (!)
Suppose Tom wants to catch a stationary bus two blocks
away. Before he can get to the bus, he must get halfway
there (one block away). Before he can get halfway there, he
must get a quarter of the way there ( one-half a block
away). Before traveling a quarter of the way, he must
travel an eighth of the way (one-fourth of a block); before
an eighth, he must travel a sixteenth of the way, etc. The
distance he must go, in blocks is given by the following
series:
1
1 1 1 1
…
+ + + + +1
32 16 8 4 2
This description requires Tom to complete an infinite
number of tasks, which Zeno maintained was impossible.
This sequence also presents a second problem in that it
contains no “first” distance to run, for any possible (finite)
first distance could be divided in half, and hence would not
be first after all. So, the trip cannot even begin!
The paradoxical conclusion then would be that travel over
any finite distance can neither be completed nor begun, and
hence all motion must be an illusion.