Andrea Fischer
MAE 301 Notes
October 24, 2007
Outline
I.
II.
HW Review
Exam Review Session
A. Algebraic Numbers
B. Complex Numbers form a field
C. Function
D. Surjectivity
E. Injectivity
F. Group Law Problems
NOTES OCTOBER 24TH 2007
MAE 301
I. Homework Review-In class we only went over part c, however just to be consistent
there entire solution is included here.
HW#6
Problem#2 The goal of this problem is to define rational functions as equivalence classes
of ratios of polynomials, in analogy with our definition of rational numbers as
equivalence classes of ratios of integers.
(a) Our first definition of a rational number was a ratio of integers. What would be an
analogous definition for a rational function?
A rational function is a ratio of two polynomials.
{f(x)=t(x)/q(x), where t(x) and q(x) are polynomials.}
(b) We decided to explicitly exclude the integer zero from the denominator of a
rational number. What would be an analogous restriction for a rational function?
Restrict the above definition such that q(x) ≠ zero polynomial this is analogous to
explicitly excluding zero from the rationals.
(c) We then defined an equivalence relation on our new set of ratios of integers.
Define an equivalence relation on your set of ratios of polynomials, and prove that
it is an equivalence relation.
Define f(x):={t(x)/q(x),q(x) ≠ zero polynomial and q(x), t(x) are polynomials}
t(x)/q(x)~ g(x)/n(x) t(x)n(x)= q(x)g(x)
Reflexive: f(x)~f(x)
Let f(x)=t(x)/q(x) and f(x)=t(x)/q(x)
t(x)/q(x)= t(x)/q(x), t(x)q(x)= t(x)q(x)
Reflexivity follows from reflexivity of “=”
Symmetric: f(x)~m(x)  m(x)~f(x).
=> Let f(x)=t(x)/q(x) and m(x)=g(x)/n(x)
Let f(x)~m(x) this means t(x)n(x)= q(x)g(x).
If we subtract q(x)g(x) we get
t(x)n(x)- q(x)g(x)=0
If we subtract t(x)n(x) we get
-q(x)g(x)= -t(x)n(x)
Now multiply by negative one
q(x)g(x)=t(x)n(x).
*For proving transitivity one problem with many of our homework was that we used
the property of division to prove transitivity, however we are actually trying to define
division. This is analogous to a mistake we made when going from →. One
student suggested that this was similar to needing to form fields from rings because
we want inverses.*
Transitive: f(x)~m(x) and m(x)~c(x) then f(x)~c(x)
Let f(x)=t(x)/q(x), m(x)=g(x)/n(x), and c(x)=r(x)/w(x)
Let f(x)~m(x) and let m(x)~c(x).
This means t(x)n(x)= q(x)g(x) and g(x)w(x)= r(x)n(x).
If we multiply t(x)n(x)= q(x)g(x) by r(x) then we have
t(x)n(x) r(x) = q(x)g(x) w(x)
Subtract the RHS
t(x)n(x) r(x) - q(x)g(x) r(x)=0
n(x)r(x)=g(x)w(x) so substitute in
t(x)g(x)w(x) - q(x)g(x) r(x)=0
Factor out a g(x)
[t(x)w(x) - q(x)r(x)]g(x)=0
However we know g(x)0
t(x)w(x)=q(x)r(x)
 f(x)~c(x).
II. Exam Review-More Homework Problems
A. HW#7
Problem#2 Algebraic Numbers
Goal: Are the algebraic number countable?
Def An algebraic number is defined to be any root of a polynomial with rational
coefficients.1
To be more concise we can use Melissa’s definition: The set of all algebraic numbers is
the union of all these sets of algebraic numbers of degree n.
Question: What does it mean when we call something the degree of an algebraic number?
What do we mean by degree n?
We mean the degree of “the” polynomial you’re talking about.
Some examples of algebraic numbers:
α= √2 is an algebraic number
It is the root of the polynomial x2-2
However it is also the root of x4-4.
1
Melissa noted that the set of all algebraic numbers is the union, over all n, of these sets
of algebraic numbers of degree n.
This brings up the point that an algebraic number can be the root different polynomials of
different degrees. So in order to only count them once we say:
For an algebraic number α its minimal polynomial, mα(x), is the unique, monic2,
irreducible3 polynomial for which α is a root.
Then the degree of α is the degree of m (x).
Student Conjecture: “The minimal polynomial of given root depends on its index4.”
Counter-Example:
If we take x3-1=(x-1)(x2+x+1) it has the roots 1, 

 One Two 
  roots
 root

3√1) x2+x+1is the minimal polynomial for 
Question: Are the algebraic numbers countable?
Proof Outline:
First take a polynomial
p(x)=a0+ a1x+ a2x2+…+ anxn, where degree of p=n
1. There are n+1 rational coefficients of any such polynomial, which can be put into
one-to-one correspondence with x x…x
N+1 times
(which is countable)
2. This polynomial can have at most n roots. Therefore there are countably many
roots of any such polynomial.
3. The set of all of these roots is the union over, n of the polynomials with degree n.
(the union of countable sets is countable)
B. HW#1/HW#2
Problem#4/Problem#3
Definition: An ordered field is a set F closed under two operations, + and ·, satisfying the
field axioms stated in class, with a relation <, that satisfies the following properties:
1. For every x and y in F, exactly one of the following is true: x < y, y < x, or x = y.
2. The relation is transitive: x < y and y < z imply that x < z.
3. x < y implies that x + z < y + z.
4. x < y and 0 < z implies xz < yz.
(a) Let −x denote the additive inverse of x. Use the defining properties of an ordered
field to prove that: if x < y then −y < −x.
2
3
4
5
A monic polynomial means the leading coefficient is a one
We defined an irreducible polynomial to be analogous to being prime f(x)=g(x)h(x) g or h is a unit
The student defines: index- the integer n in a radical
Where denotes a primitive cube root of one.
defining the n-th root:
Given x,y є x<y and –x denoted as the additive inverse of x.
Add –x to each RHS
x+-x<y+-x, add additive inverse of y to each LHS
-y+0<-y+y-x (By Axiom Three)
∴-y<-x.
(b) Show 0 < 1.
Given 0,1∊
Either 0<1,1=0, or 1<0 (By Axiom One)
We know 1≠0.
If we assume 1<0, then let 0<x
Hence x(1)<x(0) (By Axiom Four)
⇒x<0↯
∴0<1.
(c) If 0 < x < y, then 0 < 1/y < 1/x .
Given x,y∊ ,0<x<y
First we must show 0<1/x, so we can multiply by 1/x.
Assume for purpose of contradiction 1/x<0.
Then multiply both sides by x on LHS
(x)1/x<(x)0
1<0↯.
∴0<1/x.
Now multiply through by 1/x.
0<1<y(1/x).
Do the same approach with 1/y.
We then get 0<1/y<1/x.
(d) Prove that the complex numbers do not form an ordered field.
Obviously 0≠¡, therefore we are left with two cases:
From previous we know 0<1.
Case One: Suppose 0<¡
Multiply each side by ¡ (Axiom 4)
¡(0)<¡(¡)
=>0<-1 Therefore we can multiply
by -1 because “zero precedes -1”
Now multiply by -1
Then –1(0)<-1(-1)
⇒0<1
0+1<-1+1
⇒1<0↯
Case Two: Suppose ¡<0
can’t multiply by ¡ because
“¡precedes zero”
-¡+¡<-¡+0
0<-¡ this shows for this case
“0 precedes -¡”
therefore we can multiply by it
Take ¡<0
multiply each side by -¡
-¡(¡)<(¡)0
⇒1<0↯
∴The Complex numbers do not form an ordered field because they do not satisfy
axiom one.
C. Function
Def- Every element in the domain maps to a unique element in the range
Ex: y=x2
D. Surjectivity
Def A function f:A→B is surjective if for every yB there exists an xA such that
f(x)=y.
An example of a surjectivity proof
Proof Plan: To prove that a function is f:A→B is surjective:
 take an arbitrary y in B to prove (construct) that there exists an x s.t. f(x)=y
Let f: → be defined by f(x)= x3-1. Prove surjectivity:
1. Let y denote an arbitrary element of .
2. Set x=3√(y+1)
3. f(x)=f(3√(y+1))=( 3√(y+1))3-1
4. f(x)=y. As Required.
E. Injectivity
Def For every f(x)=f(y) in the range x=y in the domain.
Proof Plan: Use definition:
Contrapositive
1. Suppose f(x)=f(y)
1.Suppose x≠y
2. Prove x=y
2.Prove f(x) ≠f(y)
An easy way to think about injection: If a function spits out same number twice, it needs
to be from the same number as its input.
F. Group Law Problems
On previous homework there has been problems involving group laws of multiplication
and addition.
Why does 3x+5x=8x?
This is just repeated Addition
We can rewrite this problem as : (x+x+x) + (x+…+x) = ( x+.. +x)
“x” 3 times “x” 5 times “x” 8 times
Similarly why does x2 x3= x5
We can rewrite this problem as : (xx)  (xx x) = (xx x xx x)
“x” 2 times “x” 3 times “x” 5 times