Expression for Integer as Product of Primes is Unique/Proof 3

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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.


Proof

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.

$\blacksquare$


Sources

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