Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime or a product of a finite number of primes and this factorization is unique except for the rearrangement of the factors. This result is one of the most fundamental results in all of mathematics. Its importance was not always know; indeed before its importance was known there were a few famous mathematicians who assumed unique factorization for others sets of numbers to yield untrue results.
In this topic we start out with Euclid’s Lemma which states that if a prime divides a product then it must divide one of the factors. This lemma is very useful and will be used to show there are infinitely many primes and that every integer greater than 1 is a product of primes. Euclid’s Lemma can be found in Euclid's book The Element, and is instrumental in proving the Fundamental Theorem of Arithmetic.
Before we prove the Fundamental Theorem of Arithmetic, it is important to realize that not all sets of numbers have this property, (it fact many do not have this property). Let's consider the set of even integers, namely
![]()
We can add, subtract, even multiply elements of
and obtain elements in
so we say the set
is closed under addition, subtraction, and multiplication. However, this is not he set of integers so we should define what we mean by divisibility; namely for any two elements of E we say
means there
such that
The importance of
should not be overlooked. For example,
because
where
But
because there is no
such that
So what are some primes in
For example, 2 is prime in
because there does not exist
in
such that
Similarly, 6 is prime in
because there does not exist
in
such that
Also, 10 and 30 are primes, and so
is not. But since
we conclude that factorization into primes in
is not unique.
The first step in showing the Fundamental Theorem of Arithmetic is usually proving Euclid's Lemma.
Proposition (Euclid's Lemma)
(i) An integer
is a prime number if and only if
for any integers
and
![]()
(ii) If
is a prime number,
are integers, and
then
for some
![]()
(iii) Every integer (except
) is a product of primes.
(iv) There are infinitely many prime numbers.
Proof. (i) If
is prime and
then
or
If
then
or if
then
In either case,
or
Similarly, when
Conversely, suppose
for any integers
and
To show that
is prime let
be a divisor of
Then
for some
and so
or
Thus, if
for some
then
and so
If
for some
then
and so
as desired.
![]()
(ii)
The statement is clearly true when
and
is (i). Assume the statement is true for
and suppose
Then by (i),
or
So by the induction hypothesis there is some
such that
or it may be the case that
Therefore, there is some
such that
as desired.
(iii) Consider the set
consisting of all positive integers greater than 1 that are not a product of primes. Assume for a contradiction that
is not empty, then by the Well-Ordering Axiom there is a least element say
Then
is itself not a prime number and so,
where
and
Since
is the least element in
and
are products of primes; and thus so is
This contradiction shows that
is empty and so every integer greater than
is a product of primes. Therefore, every integer (except
) is a product of primes because if
and
then
is a product of primes as desired.
(iv) Assume, as Euclid did, that there are only finitely many prime numbers say,
Consider
which by (iii) must be a product of primes. Let's say
and since
we have
Therefore,
which can not be true. Therefore, there can not be a finite number of prime numbers, as desired.
Proposition (Fundamental Theorem of Arithmetic) Every integer
can be written uniquely in the form
where
are prime numbers and
are positive integers.
Proof. Every integer has a prime factorization by Euclid's Lemma. If there is an integer greater than 1 for which the factorization is not unique, then by the Well-Ordering Axiom there exists a smallest such integer, say
Assume that
has two factorizations say
and
where
and the
and
are all positive for
and
By Euclid's Lemma,
for some
and
for some
Since all the numbers
and
are prime, we must have
and
Then
since
Let
be the integer defined as
![]()
If
then
has a unique factorization. If
then
and
has two factorizations. Both cases reveal that
can not exist as desired.
Example (Unique Factorization) Find the unique prime factorization of
Solution. We have,
Example (Unique Factorization) Find the unique prime factorization of
Solution. We have,
Example (Unique Factorization) Factor
and
into a product of primes with the same set of primes with (possibly zero) nonnegative exponents.
Solution. We have, the set of primes in both integers is
and so we have
and
Example (Unique Factorization) There are infinitely many primes of the form
![]()
Solution. We assume there are only a finite number of primes of the form
say
is all of them; and we consider the integer
![]()
Notice that the prime factorization of
must contain at least one prime of the form
because otherwise
would be of the form
In general, integers of the form
have products of the form
which can be seen by letting
and
then
![]()
Thus, there is one prime in the prime factorization of
that is of the form
say
Either
or
If
then
which means
and so
which is absurd. If
then
implies
which is also absurd. Therefore, there are infinitely many primes of the form
Fundamental Theorem Of Arithmetic
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/fundamental-theorem-of-arithmetic.html


