Proof that $\sqrt{2}$ is irrational, using the prime factorization

Question

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?

Answer 1

It looks fine. Note than you technically do not need to count all prime factors of $a,b$ but only the exponent of $2$. (Your proof then basically relies on $2b^2=a^2$, so the exponent of $2$ on the right is even and on the left it’s odd).

But you can use the fundamental theorem more elegantly. Try to proof that any root of a positive integer is either integral or irrational. To do this take any rational number $p/q$ and try to deduce from the factorisations of $p,q$ when $p/q$ is integral, and when it isn’t. Then try to prove that $p^2/q^2$ (or even $p^n/q^n$) does not change anything about this property. (You can also trivially extend the FT to the rational numbers, giving each rational number a unique prime factorisation with arbitrary integral exponents.)

EDIT: Also you do not really need the FT for your prove. Suppose $a,b$ are coprime. Then $2b^2=a^2$ implies $2|a$ by the primality property $p|ab\Leftrightarrow p|a\lor p|b$. But then $a=2c$ and thus $2b^2 = 4c^2$, so $b^2=2c^2$. Thus $2|b$, which contradics $a,b$ coprime. This prove does not work for roots composite numbers though (it does work if the number is divided by some prime $p$, but not by $p^2$. So the FT gives us a much stronger result.