Download The Rational Numbers - Stony Brook Mathematics

Joe Pastore
MAE 301/501
Class Notes 10/10/07
We pick up where we left off from the previous lecture in our discussion of equivalence
classes.
Equivalence Classes: Consider any arbitrary equivalence class. Can it have an infinite number of
elements? Yes, it is possible.
In homework #5, we were asked to find an example of an equivalence relation that would
make sense to a high school calculus student and describe the equivalence classes. For instance,
consider the following equivalence relation:
p(x) ~ q(x)  p(x) = q(x)
Let p(x)  P(R) and let p(x) + c  P(R), cR. Then, p(x) = [p(x) + c]  and thus we
can describe the equivalence class as follows: Given p(x)  P(R),
[p(x)] = {p(x) + c cR}
This suggests that any set of polynomial functions, say p(x), that differ by some constant
c, determines an entire family or class of functions where the derivatives are equivalent. Another
more concrete example is:
f(x) = x^2 + 1
g(x) = x^2 +2
h(x) =x^2 + 3
where f  = g  = h =2x. Hence if we integrate f , g  and h , we get the family of functions x^2
+ c. Thus equivalence classes can indeed contain an infinite number of elements.
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Equivalence classes need not be math related. Consider the ‘Cow Induction Argument’
from a previous class. The set of all blue cows can be considered an equivalence class, or any
other color for that matter.
In everyday life we talk about equivalence classes. The set of all dates on a calendar that
fall on a Monday make up the class of Monday, as do the other days of the week. So, for
example, 10/3/07 ~ 10/10/07  Both dates fall on the same day of the week. Since 10/3/07
and 10/10/07 fall on a Wednesday, they are both elements in the class of Wednesday.
The Rational Numbers
Consider Q, the set of rational numbers. It should not be difficult to see that we are
separating ratios of integers into equivalence classes within Q, where one class of ratios is one
element of Q. Thus classes are defined by fractions in lowest terms. For example, we know that
½ = 3/6 = 10/20 =……= n/2n.
So, if we take an element a/b  Q to be in lowest terms, i.e. gcd (a,b) = 1, then this
provides a distinguished representative of an equivalence class for any element in Q. Thus we
define an equivalence relation simply by stating the definition of the set Q. The definition is
refined by stating conditions for two different ratios of two integers to be considered equivalent.
Q := {a/b a,b  , b  0 and a/b = c/d  ad = bc}
Final thoughts on the set Q



We use the set  to build Q
We want to divide elements p,q  
We want multiplicative inverses of the elements in Q
We have defined a rational number as a ratio of two integers. This idea or definition of
rational numbers has always been presented to math students. However, more often than not, it is
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discussed without mentioning the idea of equivalence relations. Nonetheless, the definition of Q
does indeed define an equivalence relation.
So, for a third representation of Q, we can describe the set as consisting of:
{(a,b)   x , with the equivalence relation (a,b)=(c,d)  ad=bc}.
Rational Functions
There is an interesting connection between ratios of polynomials and integers:

We want to divide polynomials and integers

We want to find multiplicative inverses for polynomials and integers
Let us consider some of the characteristics of group algebra and see what applies to
polynomials. Consider P(F), the set of all polynomials:

Additive Identity – the zero polynomial is the additive identity element in P(R).

Additive Inverse -  polynomial p(x),  -p(x),  p(x) + -p(x) = 0. Thus every
polynomial has a unique additive inverse.

Closure – We can add polynomials. Thus, p(x) + q(x)  P(R) implies that P(R) is
closed under addition.

Associativity – Polynomial addition is associative. Thus [p(x) + q(x)] + r(x) = p(x) +
[q(x) +r(x)]. Similar to , the polynomials form an Additive Group.

Commutativity – Polynomial addition is commutative. Thus p(x) + q(x) = q(x) + p(x)
and the polynomials form an Abelian Group under addition.

Multiplicative Identity – The constant polynomial p(x) = 1 is the multiplicative
identity.
However, the polynomials do not form a multiplicative group, since there does not exist
a multiplicative inverse for polynomials in general.
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Ex. Does there exist q(x)  p(x)q(x)=1? The answer is no, since only constant
polynomials, not equal to the zero polynomial, have inverses.
Let us now define the algebraic structure known as a Ring, which we will use to draw
some interesting analogies.
Definition: A Ring R is a set of elements with two binary operations, Addition (+) and
Multiplication (), satisfying the following three postulates:
i)
R is an abelian group (i.e. commutative) under the operation of addition
ii)
R is associative under the operation of multiplication
iii)
The right and left distributive laws hold; that is, for elements x,y and z of R,
x(y + z)=xy + xz and (x+ y)  z=xz + yz
As a result, the polynomials form a ring, by definition.
Although we have already discussed polynomials, let us give a formal definition:
Definition: A Polynomial is an expression of the form anxn + an-1xn-1 + ... + a1x1 + a0x0
where ai  Q, for i = 0,1,…,n and n is a non-negative integer. This is to say that a polynomial
p(x) is defined over Q.
The polynomials over a ring form a ring, as do the integers. Z and P(R) are special kinds
of rings since they both have a multiplicative identity.
When we consider the roots of a polynomial, something interesting occurs. For instance,
take a real-valued polynomial with rational coefficients. The roots to such polynomials, however,
may lie elsewhere. Some examples:

x^2 – 2 has coefficients in Z  Q. However, the roots are irrational.

x^2 + 1 also has coefficients in Z  Q. However, the roots are complex.
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Primality
Let us now examine the idea that just as we have integer values that are defined as prime
numbers, we can also have prime polynomials. First we define what it is to be a prime integer.
Definition: Any integer with exactly two distinct factors, which are either +1 or -1 and
itself, is considered to be prime.
However, can we consider –2 to be prime? It can be factored as +2 and –1. It can also be
factored as –2 and +1, which contradicts our definition.
As a result, we will now refine the above definition of prime by defining a unit.
Definition: An integer p is prime if whenever p=ab then either a or b is a unit. An
element x in a ring R is a unit if it is invertible with respect to multiplication, meaning  element
yR such that xy=yx=1. Thus y is the unique inverse of x and the set of all units in a ring R
form a group under ring multiplication, denoted by U(R).
When considering the prime integers, both 1 and –1 have a unique multiplicative inverse
in Z. Hence 1 is its own inverse since 11=1 as is –1 since (-1)(-1)=1. Consequently, these are the
only units in Z, since no other elements have multiplicative inverses. To find a multiplicative
inverse of other elements nZ, we would need to find an element in the set Q that is not an
element in Z.
Ex. 3n =1  n =1/3  Z.
In secondary school settings, the definition of a prime polynomial is usually generalized.
For example, consider the following description.

A polynomial that cannot factor is prime (consider coefficients in Q and rational
factorization)
Now, consider the following example: (2x+4) = 2(x+2)
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Is 2 or x+2 a unit (i.e. an invertible element in the ring Q)?
As a result of the preceding example, we formalize our definition of prime polynomials
as follows:
Definition: A polynomial is said to be prime or irreducible if whenever f(x)=g(x)h(x),
either g or h is a unit. Thus when a polynomial is irreducible this is analogous with an integer
being prime.
By our refined definition of prime, this suggests that whenever a prime polynomial has
two factors, one of the factors is a unit, meaning it has a multiplicative inverse. Therefore,
p(x)=2(x+2) is prime or irreducible, since ½ is a multiplicative inverse of 2. In the more general
case, f(x)=c, where c is any constant, is invertible, with inverse g(x)= 1/c.
Notice the following correlation between the prime decomposition of an integer and the
factorization of a polynomial:
Ex. 1) 35 = 75, where 7 and 5 are measurably smaller in terms of absolute value.
2) (x^2 – 4) = (x+2)(x-2), where (x+2) and (x-2) are measurably smaller in terms of
degree.
We know any integer (n1) may be factored uniquely as a product of primes:
i)
Up to rearrangement
ii)
Up to multiplication by a unit
However, this leads us to an interesting question: Is the factorization of polynomials
unique? Consider the above example:
(x^2 – 4) = (x+2)(x-2)
=1/2(2x+4)(x-2)
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Note that the two representations of x^2 – 4 on the right hand side are not distinct, since
they differ only by multiplication by a unit. This exemplifies the fact that polynomial
factorization is unique, as the different representations of our given polynomial are essentially
equivalent.
One student presented the following question: Based on the definition, should constants
be considered irreducible polynomials?
Since any polynomial f(x)=c, where c0, is invertible, we can consider all such
polynomials as being irreducible. Based on our intuition, it might seem fair to suggest that we
should redefine what it means to be an irreducible polynomial, since the set Z has only two units,
1 and –1, while the set P(R) has an infinite number of units. Thus these two rings are not
isomorphic, simply because we do not have a one to one correspondence between units.
Question from the note-taker: Is it possible to redefine irreducibility, or primality, within the ring
of polynomials so that we may obtain an isomorphism with the ring of integers?
Sources: Abstract Algebra and Solutions by Radicals, John. E., Margaret W. Maxfield
Lecture Notes, MAE301/501, Professor Lisa Berger
Numbers, Groups and Codes, J.F. Humphreys & M.Y. Priest
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