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Danielle Vaughan
Algebra and Number Theory
Throughout the course Algebra and Number Theory, a number of topics have
been discussed. The topics include functions, equations, vectors, and number theory
work. It was the goal that through the discussion of all of these topics I would come to
develop an understanding of the fundamental mathematics ideas vital for teaching K-8
mathematics, gain significant mathematical content knowledge in a number of
mathematical areas, acquire strategies for approaching and solving unfamiliar problems,
justify mathematical reasoning, communicate mathematic ideas orally and in writing, and
use technology appropriately to represent my mathematical ideas. Throughout the areas I
have worked on, I have addressed all of these goals and improved in each of them. I have
accomplished much throughout this semester and understand more fully concepts that I
previously did not understand. Although I understand many more concepts, there are still
some that I struggle with. By looking at the concepts I have learned and the ones I still need
to learn, I am able to reflect more clearly on my new-found attitudes towards mathematics.
At the beginning of the semester we began work on equations. An equation occurs
when an algebraic expression (a combination of mathematical symbols, variables and
constants that represents a phrase, not a whole sentence) is set as equal to another
algebraic expression. Right away we began work on the first goal, which was to develop
an understanding of the math we would need to teach in an elementary school classroom
because we learned that to teach students to “solve an equation” we would have to ensure
that they knew how to find the value of an unknown so that there would be a true
statement. When solving an equation one can have one or more solutions, as well as
infinite or no solutions. In order to connect our study of equations to things we would
need to do as teachers, we discussed how to make an equation problem equal a number
we chose. An example of my work during this part of class can be seen when I made an
equation with a solution of 3 by using my algebra skills, as seen below:
x=3
x2=9
x2 +3x=9+3x
x2 +3x+5=9+3x+5
x2 +3x+5=3x+14
2 x2 +6x+10=2(3x+14)
It is skills like this that I will need during my work as a teacher. Additionally, we learned
to solve equations and to know when there was no solution, which is a hard concept for
many students. An equation has no solution when, after simplifying it algebraically, you
can prove that it is actually a false statement. An example can be seen bellow:
x+2-5x-7=10-4x+15
-5-4x=25-4x
-5=25 –false statement so there is no solution
Equations can also have infinite solutions, which occur when the statement will be true
for any number that the unknown can be. An example can be seen bellow:
[(6n+10)/2]-3n=5
(3n+5)-3n=5
5=5 – True statement so there is infinite solutions within the domain.
After working with equations, we moved on to
functions. A function is a relation where each element of a
given set is associated with an element of another set. A
function of some unknown is set as equal to an algebraic
expression. An example of a function can be seen in the graph to
the right, labeled A. In a function there is only one output (f(x))
for every input (x). This means that the graph to the right labeled B is not a function. In a
function the independent variable (often x) determines the domain and the dependent
variable (often y) determines the range.
When working with functions I achieved the third goal and learned how to
acquire numerous strategies for approaching and solving unfamiliar problems. In fact, I
learned how to represent functions in three different ways that would help me to solve a
variety of problems. These different representations included tables, graphs, and
expressions. As an extension of functions we learned how to find their inverses. An
inverse of a function is found when one reverses the dependant and independent
variables. An example of an inverse that I found is shown below:
y=3x
x=3y-1
y-1=(1/3)x
Properties of inverses of linear functions include that the sign of the slope is always the
same, that the steepness is “opposite,” that there is a line of symmetry, and that the
bisector goes through the point of intersection of the inverse and the function. Some
functions will not have inverses, like a constant function and a square root function.
Additionally, some functions will have restricted inverses, like sine
functions and cosine functions. A simple test can be done to find out
if a function has an inverse. This is called the horizontal line test and
it states that if a horizontal line, placed anywhere on the graph, goes
through more than one point, the function does not have an inverse because there will be
more than one output for every input. The quadratic function on the right does not pass
the horizontal line test.
In February we moved on to systems of equations and worked further on using
multiple strategies because any system of equations can be solved by using algebra
(substitution or algebraic addition/subtraction), as well as graphically. An example of a
system solved all three ways is shown below:
x+y=3
y-3=2x
Substitution:
x+y=3
y=3-x
y-3=2x
(3-x)-3=2x
-x=2x
3x=0
x=0
y=3-0
y=3
(0,3)
Addition:
2x+2y=6
+ -2x+y =3
0+3y=9
y=3
x+3=3
x=3
(0,3)
Subtraction:
x+y=3
- -2x+y=3
3x=0
x=o
0+y=3
y=3
(0,3)
Graphically:
(0,3)
When working with systems of equations it is also important to remember that some
systems might have no solutions (if they form parallel lines) and some might have infinite
solutions (if they form the same line).
After working with linear equations and linear functions, it was only natural to
move onto quadratic equations and functions. A quadratic equation would be represented
by the generalization ax2+bx+c=0 and a quadratic function would be represented by the
generalization F(x)= ax2+bx+c. There are also certain special cases that we discussed in
their general forms including when x2 =m. In this case the solution would be x=±√m
(quadratic equations usually have two solutions). We also discussed the case of factoring,
which is demonstrated by the example:
x2-3x=0
x(x-3)=0
x1=0 x-3=0
x2=3
We then discussed the general formula for solving quadratic equations, which is
represented by
where ax2+bx+c=0. In order to make sense of this
formula a proof was done, which I have replicated below:
ax2+bx+c=0
ax2+bx= -c
x2+(b/a)x= -(c/a)
x2 +2(b/2a)x+(b/2a)2= -(c/a)+(b/2a)2
(x+(b/2a))2 = -(c/a)+(b2/4a2)
(x+(b/2a))2 = (-4ac/4a2)+(b2/4a2)
(x+(b/2a))2= (-4ac+b2)/(4a2)
x+(b/2a)= ±√(-4ac+b2)/(4a2)
x=-(b/2a)±√(-4ac+b2)/(4a2)
x= -b±√(-4ac+b2)
2a
When working on this proof we satisfied the fourth goal, which was to demonstrate the
ability to justify mathematical reasoning and construct formal proofs, exactly like the one
above. While working on this proof, another necessary reasoning was discussed: why one
cannot divide by 0. This is a much shorter proof and is shown below:
a/b=c so c x b=a (this is true for any multiplication problem)
When disproving something one only needs one counter example, so here is one:
3/0=C but C x 0 ≠ 3
We also examined the movement of different quadratic functions if you change
the constants. If a function was represented by f(x)=ax2+bx+c then the constant c will
control the vertical movement of the function and a will change the width. By knowing
how the constants change the functions’ position and shape, one can make accurate
predictions about what a function looks like.
We also predicted the movement of exponential growth and decay functions. If a
function is represented by y=ax+m+n, then if a is greater than one, the function is
increasing (growth) and if a is less than one, the function is decreasing (decay) (p.s. a
cannot be one because this would just be a constant function and well, that not helpful at
all when we’re talking about exponential growth and decay). Also, if n is greater than 0
the function moves up from where is would be if n was 0, and if n is less than 0 it moves
down. If m is greater than 0 than the function moves left from where it would be if m was
0, and it moves right if m is less than 0.
After this work on predicting function appearance we worked for a brief period of
time on logarithmic functions. The unique thing about logarithmic functions is that if the
equation is logba=n, then bn=a. It was during this section of class that we worked
explicitly on using appropriate technology (goal six) for creating tables, graphs, and
representations because we represented logographic functions in geogebra as a graph,
then as an expression, and as a table. All three of these representations are shown below
(the graph is done using the technology available by geogebra).
y=log2x
x
1
2
4
y
0
1
2
A vector is a geometric object that has both a magnitude (length) and direction
(often represented by arrows). A vector is expressed with small arrows above the variable
and can be represented by coordinates. For example, F could be expressed as (F1,
F2), which means the vector moves F1 to the right and F2 up (if these values were
negative then the vector would move to the left and down). The norm
of a vector can also be known as the length of the vector. To find the length of a
vector one could apply the Pythagorean Theorem. This means that the norm (length)
of F is √F12 +F22.
Vectors, like scalars (only magnitude) can be added and subtracted. To add
vectors algebraically you simply add the horizontal and vertical displacements. This
means that if U= (U1,U2) and V=(V1,V2) then U+V=(U1+V2,U1+V2). The addition of
these vectors can be seen geometrically by the dotted line in the image to the right. When
one is adding vectors, the commutative property (order
does not matter) and the associative property (parentheses
do not matter) exists. You can see in the picture that
whether or not you add vector u to vector v, or vector v to
vector u, the resulting vector is still AD. To show this
algebraically one would have to use the commutative
property of addition because if U+V=(U1+V2,U1+V2) and V+U=(V1+U2,V1+U2) then the
sum is the same since the commutative property of addition tells us that U1+V2=V2+U1. If
one was adding three vectors he or she could see how the
associative
property
works
with
vectors.
(U+V)+W=(u1+v1+w1,u2+v2+w2), which has the same values of
U+(V+W) because the sum of this is (u1+v1+w1,u2+v2+w2).
Because the order in which you add these numbers does not
matter (due to the associative property of addition) the order in
which you add these vectors also does not matter. This is also illustrated in the picture to
the right.
When one is subtracting vectors it is slightly more complicated. To make is
simpler one can think about subtracting two vectors as finding what one
would need to add to the subtrahend to get the minuend. When one looks
at it this way he or she can picture fairly simply how to find the difference
geometrically. This is shown on the right. To solve vector subtraction
algebraically one follows the same algorithm that is done for addition. This means that UV=(U1-V2,U1-V2). The same properties, however, do not apply to subtraction. For
example, the commutative property does not apply to subtraction. This can be shown
algebraically because that if U= (U1,U2) and V=(V1,V2) then U-V=(U1-V2,U1-V2);
however, V-U=(V1-U2,V1-U2). Because V and U are numbers and numbers do not have a
commutative property of subtraction U-V does not equal V-U. This means that there is
also no associative property of vectors. To find if there is an associative property of
vectors one could again to look towards algebra. If U= (U1,U2), V=(V1,V2), and
W=(W1,W2) then (U-V)-W=(U1-V1-W1,U2-V2-W2) and U-(V-W)=(U1-(V1-W1),U2-(V2W2))=(U1-V1+W1,U2-V2+W2). This means there is no associative property for subtraction
of
vectors.
An
example is shown
below:
You can also perform scalar multiplication of vectors. In scalar multiplication the
vector is multiplied by a magnitude. This means that every component is multiplied by
that number. If the number’s absolute value is less than one, then the length will be
smaller than the original vector. If the absolute value of the number is more than one,
then the length will be greater than the original vector. kV = (kV1,kV2). If the scalar value
is negative, then the direction will switch. An example of the effects of scalar
multiplication is shown below:
During the last month of class we worked on number theory. During our work on
number theory we examined prime factorization. The prime factorization of a composite
number is found by breaking down that number into only prime numbers, which when
multiplied together are equal to the composite number. Prime numbers are numbers that
are only divisible by one and itself. For example the prime factorization of 1092 is
22·3·7·13 and its factor tree is shown below:
We also worked with prime factorization to make generalizations about how the
prime factorization could be used to determine the number of factors a number would
have. When making the generalization we determined that any number represented by
p1ap2bp3c will have (a+1)(b+1)(c+1) factors. In order to explain why this works one needs
to first look at the number of factors represented by a number pn, whose number of
factors is represented by n+1. This is because the factors of pn can be represented by
{p0p1,p2,…,pn} so the number of factors in this list is the number that n represents
(because the values of P to the 1-n are included) and then you need to add one because p0
or 1 will always be a factor of any number and is not included in this list. Now that we
have determined that pn has n+1 number of factors we can show that p1ap2bp3c will have
(a+1)(b+1)(c+1) factors because we need to find all of the combinations of the factors of
each of the numbers that these primes are made up of. When combining sets one can just
multiply them together because one is determining the number that will be represented if
all of the number in a were combined with all the numbers in b and then all of the
numbers in a were combined with all the numbers in c, and so forth, until all of the
combinations are found; this type of repeated addition is the definition of multiplication.
By using prime factorizations we also explored the GCF and the LCD of sets of
numbers. The GCF is the largest factor that can be divided into both numbers in order to
get a whole number and the LCD is the smallest number that both numbers can be
divided into to get a whole number. The prime factorization is useful for finding these
two aspects of math because they break up a number is such a way that it is easier to see
its components. If one knew the prime factorizations of two numbers than she could
simply take the prime numbers that were in both the first number and the second number.
She would then have to multiply the similar prime numbers together to get the GCF. If
one took those same numbers and found the smallest number of primes that could
represent both numbers within it, then she would have found the LCM.
Let’s look at an example, 12 and 15. The prime factorizations are shown below:
12
2
15
6
3
5
2 3
12=22·3
15=3·5
By looking at these factorizations we can tell that the LCM is 22·3·5 because this is the
smallest number of primes that could have both numbers within it. We can also tell that
the GCF is 3 because this is the only prime that is similar in both.
The final thing we worked on regarding number theory was divisibility rules.
These are ways of discovering whether a given number is divisible by a certain number
without performing the division. The divisibility rule for two is often the first one learned
by students and states that a number is divisible by 2 if the last digit is 0 or an even
number. The reason that this is true is because we are in the base 10 system and 10 is
divisible by 2. This means any place, except the first, is always divisible by 2 so you only
have to look at the ones place and determine if that number is divisible by 2. The rule for
4 follows similar reasoning. A number is divisible by 4 if the last 2 digits represent a
number divisible by 4 because every place greater than the tenth is divisible by 4 (100 is
divisible by 4 meaning that 100 times anything will be divisible by 4 and all of the places
greater than 100 can be represented by 100 times something). A number is divisible by 8
if the last 3 digits represent a number divisible by 8 for the same reason, every placevalue greater than the hundredth is divisible by 8. This means you only have to look
towards the hundredth place and the numbers below it to ensure divisibility.
The 5 and the 10 divisibility rules go hand in hand. A number is divisible by five
if the last digit is divisible by 5 or is given not value (has a zero in it). This is because 10
is divisible by 5 and therefore any number that is also divisible by ten is divisible by five
(the value of any place value greater than one is always divisible by 10). A number is
divisible by 10 if the first place is given no value (a zero) because a ten will never fit into
the ones place, but all the other place values are made up of it so it will always fit into
them.
The divisibility rule for 9 and 3 is unique. Any number is divisible by three if you
find sum of all the digits and it is divisible by three and any number is divisible by nine if
you find the sum of all the digits and it is divisible by nine. The reason this is true is
shown below algebraically:
any number abc= a(100)+b(10)+c
a(99+1)+b(9+1)+c
a·99+a+b·9+b+c
Divisible by 3 and 9 so it does not have to be factored into our
problem when determining divisibility.
a+b+c-if this is divisible by 3 or 9 then everything is divisibly
by 3 or 9.
While I feel that I have made great strides in the area of number theory and algebra, I
feel that I still need to work on my algebra skills. I find that when I approach a problem I will
rush through it, and then it loses its meaning. I need to work more on stepping back and
allowing myself to interpret the problem. I need to first find out what it is asking before I try
to use an algorithm I have memorized. By taking a breath and stepping back to examine the
problem at a broader level I should be able to plan more clearly and to reason about the
solution more logically. I believe that this step is just as important as knowing how to
perform certain algorithms, like the quadratic formula, because even if I can find and answer
to the problem, I cannot be sure it is correct if I cannot truly understand what the problem is
asking.
I feel that through this course I have come to learn a number of strategies to use when
I am in the subject of algebra, geometry, number theory, and calculus, which is goal number
two. Additionally, through reflective papers like this and my final presentation on number
theory problems on the MTEL and MCAS, I have also been able to practice the ability to
communicate mathematical ideas in new ways: orally and linguistically. By looking at math
like this I was able to think more critically and not feel that I can only rely on the algorithms I
know; instead I am now confident enough to believe that I can create the algorithm I need
using the logic I already know. Most of the concepts in this class are ones that I have already
been taught. I do not feel like I learned them in this class. Instead, I feel as though I have
refined my knowledge of them in a way that will allow me to apply my skills on a broader
level.