Why must $a$ and $b$ both be coprime when proving that the square root of two is irrational?

Question

Suppose we wish to prove that the square root of two is irrational. We begin by assuming that it is rational. Namely, where both $a$ and $b$ are integers $$\frac{a}{b} = \sqrt 2 % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GH9aqpdaGcaaqaaiaaikdaaSqabaaaaa!31D3! $$

Why is it so important that both $a$ and $b$ be coprime and the fraction be irreducible? That is, if $a$ and $b$ are not coprime and the fraction is indeed reducible, then$$\frac{a}{b} \ne \sqrt 2 % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GHGjsUdaGcaaqaaiaaikdaaSqabaaaaa!3294! $$

Answer 1

If you assume that $a,b$ are coprime (which you are entitled to do by properties of the rationals), then you may proceed by contradiction. This is not strictly necessary, it's just simpler than the alternatives, like infinite descent.

Answer 2

It's not that $a$ and $b$ must be coprime; rather, it's that every fraction can be written in lowest terms, and this is one possible way to proceed with the proof. Among all the different ways of writing a fraction, we decide to choose one where the numerator and denominator have no common factors.