# Proving that every integer greater than or equal to $2$ can be uniquely factored into primes

I need to prove:

Every integer $$n$$, $$n \ge 2$$, can be factored uniquely into primes. (By "unique," we mean unique up to the order in which the primes are listed.)

I assume I need to use induction, but I'm unsure of how to prove the $$n+1$$ case.

Any assistance would be greatly appreciated!

• unique factorization is easily researched online (or in your text). – lulu Apr 3 at 0:08

Assume the statement is false. Because the positive integers are well ordered, there is a smallest $$n$$ that cannot be uniquely factored into primes. Then $$n$$ cannot itself be prime or $$n=n$$ is the unique prime factorization of $$n$$. Thus, $$\exists a, b \lt n$$ such that $$n=ab$$.
But $$n$$ is the smallest number that cannot be uniquely factored into primes, so $$a, b \lt n \Rightarrow a \text{ and } b$$ can be factored into primes (and indeed, we can do so uniquely). That demonstrates that $$n$$ is a product of primes, so $$n=pm$$ for some prime $$p, m =n/p \lt n$$. Because $$n/p$$ can be factored into primes uniquely, it follows that the prime factorization of $$n$$ is likewise unique, contradicting our original assumption that $$n$$ cannot be uniquely factored into primes.