# Expression for Integer as Product of Primes is Unique/Proof 1

## 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.

## Proof

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.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 13.2$: The fundamental theorem of arithmetic - 1979: G.H. Hardy and E.M. Wright:
*An Introduction to the Theory of Numbers*(5th ed.) ... (previous) ... (next): $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic