Arithmetic Theorem
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
![arithmetic theorem _gr_1.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_1.gif)
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
![arithmetic theorem _gr_30.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_30.gif) (ii) If
is a prime number,
are integers, and
then
for some
![arithmetic theorem _gr_37.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_37.gif) (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.
![arithmetic theorem _gr_68.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_68.gif) (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
![arithmetic theorem _gr_131.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_131.gif) 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
![arithmetic theorem _gr_151.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_151.gif)
Solution.
We assume there are only a finite number of primes of the form
say
is all of them; and we consider the integer
![arithmetic theorem _gr_154.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_154.gif) 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
![arithmetic theorem _gr_163.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_163.gif)
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
Proposition (Greatest Common Divisor) Let
and
be positive integers.
(i) Any two nonzero integers
and
have a greatest common divisor, which can be expressed as the smallest positive linear combination of
and
![arithmetic theorem _gr_183.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_183.gif) (ii) An integer is a linear combination of
and
if and only if it is a multiple of their greatest common divisor. (iii) There exists integers
and
such that
if and only if
![arithmetic theorem _gr_189.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_189.gif) (iv) If
and
then
![arithmetic theorem _gr_192.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_192.gif) (v) If
and
then
![arithmetic theorem _gr_196.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_196.gif) (vi)
if and only if
and
![arithmetic theorem _gr_199.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_199.gif) (vii) If
and
is a common divisor of
and
then
if and only if
![arithmetic theorem _gr_205.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_205.gif) (viii) If
and
then
![arithmetic theorem _gr_208.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_208.gif) (ix) If
then
![arithmetic theorem _gr_210.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_210.gif)
Proof.
(i) For every positive integer there are only a finite number of divisors and thus given two integers there are only a finite number of common divisors.
Therefore, the greatest common divisor of every pair of integers must exist and be unique.
The fact that
is the smallest linear combination of
and
follows from the Euclidean Algorithm. (ii) Let
then since
the Well-Ordering Axiom yields a smallest positive integer, say
Then
for some integers
and
By the Division Algorithm
with
and so
Whence
Therefore,
and similarly,
So
is a common divisor of
and
Let
be a common divisor of
and
Then
for some integers
and
Then
as desired.
(iii) Apply (ii). (iv) If
then there exist integers
and
such that
Thus
for some
since
Therefore,
![arithmetic theorem _gr_243.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_243.gif) (v) If
then there exist integers
and
such that
Thus
for some
and
since
and
Therefore,
![arithmetic theorem _gr_253.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_253.gif) (vi) If
then there exist integers
and
such that
Thus,
and so
Also,
and so
Conversely, if
then there exists integers
and
such that
If
then there exists integers
and
such that
Thus,
for some integers
and
Therefore,
as desired.
(vii) If
then there exist integers
and
such that
and so
since
and
Thus,
Conversely, if
and
then there exists integers
and
such that
Therefore,
for some
and
Thus
and so
![arithmetic theorem _gr_294.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_294.gif) (viii) If
then there exist integers
and
such that
If
then
for some integer
Thus,
and so
as desired. (ix) Suppose
and let
be the set of common divisors of
and
and let
be the set of common divisors of
and
.
We will first show that
If
then
and
for some
and
Thus,
Hence,
so that
Conversely, suppose
is a common divisor of
and
so that
and
Thus,
Thus
so that
Therefore,
Hence the largest element in
namely
is the same as the largest element in
namely
Proposition (Euclidean Algorithm) Let
and
be positive integers with
If
then
If
then apply the division algorithm repeatedly as follows:
![arithmetic theorem _gr_345.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_345.gif)
![arithmetic theorem _gr_346.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_346.gif)
![arithmetic theorem _gr_347.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_347.gif)
![arithmetic theorem _gr_348.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_348.gif)
![arithmetic theorem _gr_349.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_349.gif)
![arithmetic theorem _gr_350.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_350.gif)
This process ends when a remainder of 0 is obtained.
This must occur after a finite number of steps; that is, for some
![arithmetic theorem _gr_351.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_351.gif)
![arithmetic theorem _gr_352.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_352.gif)
![arithmetic theorem _gr_353.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_353.gif)
Then
the last nonzero remainder, is the greatest common divisor of
and
![arithmetic theorem _gr_356.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_356.gif)
Proof.
The proof follows from the property: if
then
Thus, if
then
and so
Also, as in the statement of the proposition,
Example (Euclidean Algorithm) Use the Euclidean Algorithm to find
By the Division Algorithm,
![arithmetic theorem _gr_371.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_371.gif)
![arithmetic theorem _gr_372.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_372.gif)
![arithmetic theorem _gr_373.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_373.gif)
Thus,
Proposition (GCD and Linear Combinations) There are an infinite number of ways to write
as a linear combination of
and
![arithmetic theorem _gr_378.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_378.gif)
Proof.
Let
There exists integers
and
such that
and thus, for any integer
we have
![arithmetic theorem _gr_384.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_384.gif)
which expresses
as a linear combination of
and
Definition (Least Common Multiple) The least common multiple of two nonzero integers
and
is the smallest positive integer that is divisible by
and
Example (Least Common Multiple) The least common multiple of
and 21 is
The least common multiple of
and 36 is
The least common multiple of 2 and 20 is
The least common multiple of 7 and 11 is
One immediate result from the Fundamental Theorem of Arithmetic is the ability to find the GCD and LCM from a factorization of two given integers.
Say we are given
and
and we are able to find the unique factorization of each (and assume that all exponents are nonnegative and the
are all primes in both
and
), namely
![arithmetic theorem _gr_405.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_405.gif)
then
![arithmetic theorem _gr_406.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_406.gif)
because for each prime
in
and
have exactly
factors of
in common.
Proposition (Least Common Multiple and Greatest Common Divisor) If
and
are positive integers, then
![arithmetic theorem _gr_414.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_414.gif)
Proof.
Let
and
have prime factorization; namely
![arithmetic theorem _gr_417.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_417.gif)
Now we can form the set of primes
consists of all the
and
So we can write
![arithmetic theorem _gr_421.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_421.gif)
where the exponents are zero where necessary.
Now let
and
Then, we have,
![arithmetic theorem _gr_424.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_424.gif)
![arithmetic theorem _gr_425.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_425.gif)
Example (Least Common Multiple) Show that if
and
are integers, then
if and only if
and
![arithmetic theorem _gr_431.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_431.gif)
Solution.
Suppose
Since
and
it follows that
and
Conversely, suppose
and
Using the unique factorization theorem we can write
and
Then
for
and because
and
Hence,
Example (Least Common Multiple and Greatest Common Divisor) Show that if
and
are positive integers, then
![arithmetic theorem _gr_450.gif]](pages/arithmetic-theorem/Images/arithmetic-theorem_gr_450.gif)
Solution.
Let
be a prime that divides
or
Then
divides
and
Hence
divides both sides of the equation
Define,
and
by
and
say
and
Without loss of generality, suppose
Then
so
Also,
But
so
Therefore,
But
so the same power of
divides both sides of the equation.
Therefore, the two sides must be equal.
Cite this as: Arithmetic Theorem Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/arithmetic-theorem.html
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