I am trying to prove that $\sqrt{2}$ is irrational, assuming the fundamental theorem of arithmetic is true. Here is my attempt.
Suppose for the sake of contradiction $\sqrt{2}$ is rational, and write $\sqrt{2} = \frac{a}{b}$ for $a,b \in \mathbb{Z}$, $b \neq 0$, where $a,b$ have no common factors. As $\left(\frac{a}{b}\right)^2 = \left(- \frac{a}{b}\right)^2$, we can assume without loss of generality that $a,b$ are both nonnegative. By the fundamental theorem of arithmetic, there exist a unique prime factorization (up to ordering of factors) of $a$ and $b$. Let $m$ and $n$ be the number of prime factors, with multiplicity, of $a$ and $b$, respectively. Then $2 = \frac{a^2}{b^2}$, so $2b^2 = a^2$. As $a$ has $m$ prime factors, $m^2$ has $2m$ prime factors. Similarly, $b^2$ has $2n$ prime factors as $b$ has $n$ prime factors. Therefore, $2b^2$ has $2n+1$ prime factors. But $2n+1$ is odd and $2m$ are even, contradicting the fact that $2b^2 = a^2$ has a unique prime factorization. So $2b^2 \neq a^2$, and no such quotient of rationals exists.
I feel like there are a number of holes in this proof and things I could've said more succinctly. How does it look?