Expression for Integer as Product of Primes is Unique

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Theorem

Let $n$ be an integer such that $n > 1$.


Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.


In the words of Euclid:

If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it.

(The Elements: Book $\text{IX}$: Proposition $14$)


Proof 1

Aiming for a contradiction, suppose the supposition false.

That is, suppose there is at least one positive integer that can be expressed in more than one way as a product of primes.

Let the smallest of these be $m$.

Thus:

$m = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$

where all of $p_1, \ldots p_r, q_1, \ldots q_s$ are prime.

By definition, $m$ is not itself prime.

Therefore:

$r, s \ge 2$

Let us arrange that the primes which compose $m$ are in order of size:

$p_1 \le p_2 \le \dots \le p_r$

and:

$q_1 \le q_2 \le \dots \le q_s$

Let us arrange that $p_1 \le q_1$.

Suppose $p_1 = q_1$.

Then:

$\dfrac m {p_1} = p_2 p_3 \cdots p_r = q_2 q_3 \cdots q_s = \dfrac m {q_1}$

But then we have the positive integer $\dfrac m {p_1}$ being expressible in two different ways.

This contradicts the fact that $m$ is the smallest positive integer that can be so expressed.

Therefore:

$p_1 \ne q_1 \implies p_1 < q_1 \implies p_1 < q_2, q_3, \ldots, q_s$

as we arranged them in order.

From Prime not Divisor implies Coprime:

$1 < p_1 < q_j: 1 < j < s \implies p_1 \nmid q_j$

But:

$p_1 \divides m \implies p_1 \divides q_1 q_2 \ldots q_s$

where $\divides$ denotes divisibility.

Thus from Euclid's Lemma for Prime Divisors:

$\exists j: 1 \le j \le s: p_1 \divides q_j$

But $q_j$ was supposed to be a prime.

This is a contradiction.

Hence, by Proof by Contradiction, the supposition was false.

$\blacksquare$


Proof 2

Aiming for a contradiction, suppose $n$ has two prime factorizations:

$n = p_1 p_2 \dots p_r = q_1 q_2 \dots q_s$

where $r \le s$ and each $p_i$ and $q_j$ is prime with $p_1 \le p_2 \le \dots \le p_r$ and $q_1 \le q_2 \le \dots \le q_s$.

Since $p_1 \divides q_1 q_2 \dots q_s$, it follows from Euclid's Lemma for Prime Divisors that $p_1 = q_j$ for some $1 \le j \le s$.

Thus:

$p_1 \ge q_1$

Similarly, since $q_1 \divides p_1 p_2 \dots p_r$, from Euclid's Lemma for Prime Divisors:

$q_1 \ge p_1$

Thus, $p_1 = q_1$, so we may cancel these common factors, which gives:

$p_2 p_3 \cdots p_r = q_2 q_3 \dots q_s$


This process is repeated to show that:

$p_2 = q_2, p_3 = q_3, \ldots, p_r = q_r$

If $r < s$, we arrive at $1 = q_{r + 1} q_{r + 2} \cdots q_s$ after canceling all common factors.

But by Divisors of One, the only divisors $1$ are $1$ and $-1$.

Hence $q_{r + 1}, q_{r + 2}, \ldots, q_s$ cannot be prime numbers

From that contradiction it follows that $r = s$.

Thus:

$p_1 = q_1, p_2 = q_2, \ldots, p_r = q_s$

which means the two factorizations are identical.

Therefore, the prime factorizations of $n$ is unique.

$\blacksquare$


Proof 3

The proof proceeds by strong induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

the prime decomposition for $n$ is unique up to order of presentation.


Note that it has been established in Integer is Expressible as Product of Primes that $n$ does in fact have at least $1$ prime decomposition.


Basis for the Induction

$\map P 2$ is the case:

$n = 2$

which is trivially unique.


Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.


This is the induction hypothesis:

the prime decomposition for all $j$ such that $0 \le j \le k$ is unique up to order of presentation.


from which it is to be shown that:

the prime decomposition for $k + 1$ is unique up to order of presentation.


Induction Step

Either $k + 1$ is prime or it is composite.

If $k + 1$ is prime, there is only one way to express it, that is, as the prime $k + 1$ itself.


So suppose $k + 1$ is composite.

Aiming for a contradiction, suppose $k + 1$ has the two prime decompositions:

$k + 1 = p_1 p_2 \dotsm p_r = q_1 q_2 \dotsm q s$

where:

$p_1 \le p_2 \le \dotsb \le p_r$

and:

$q_1 \le q_2 \le \dotsb \le q_s$

Because $q_1$ is a divisor of $k + 1$:

$q_1 \divides p_1 \le p_2 \le \dotsb \le p_r$

where $\divides$ indicates divisibility.

Thus by Euclid's Lemma for Prime Divisors:

$\exists p_i \in \set {p_1, p_2, \ldots, p_r}: q_1 \divides p_i$

But as $q_1$ and $p_i$ are both primes, it follows by definition that $q_1 = p_i$.

In a similar way it can be shown that:

$\exists q_j \in \set {q_1, q_2, \ldots, q_s}: p_1 = q_j$

So we have:

$p_1 = q_j \ge q_1$

and:

$q_1 = p_i \ge p_1$

Thus:

$p_1 \ge q_1 \ge p_1$

and so:

$p_1 = q_1$

Thus $\dfrac {k + 1} {p_1}$ is an integer such that:

$\dfrac {k + 1} {p_1} \le k$

and so:

$p_2 p_3 \dotsm p_r \dfrac {k + 1} {p_1} = q_2 q_3 \dotsm q s$

But by the induction hypothesis:

$p_2 = q_2, p_3 = q_3, \dotsc, p_r = q_s$

where furthermore $r = s$.

Therefore the prime decomposition for $k + 1$ is unique.


So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction:

$\forall n \in \Z_{\ge 2}$, the prime decomposition for $n$ is unique up to order of presentation.

$\blacksquare$


Also see


Historical Note

This theorem is Proposition $14$ of Book $\text{IX}$ of Euclid's The Elements.


Sources