TITLE: Irrationality of $\sqrt{2}$: why can we assume lowest terms?
QUESTION [6 upvotes]: In all textbook proofs of the irrationality of $\sqrt{2}$, one will inevitably see something to the effect of "we assume that $\frac{p}{q}$ is in lowest terms", or that numerator and denominator do not have 2 as a common factor etc.
But why can we do this? I'm starting to self study math but given that I am embarking on a journey of rigor, this seems a decidedly un-rigorous introduction to the subject. Is there more behind this just being an assumption, or does it necessarily follow from the way we've set it up? If the latter, I honestly do not see it. Please advise.

REPLY [4 votes]: If the argument was fleshed out more, then it would go something like this:

Assume, for the sake of contradiction, that $\sqrt{2}$ is rational. This means that it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, and $b\neq0$. If $a$ and $b$ have any common factors, then cancel them, and denote the lowest terms fraction as $\frac{p}{q}$.

The phrasing 'assume that $\frac{p}{q}$ is in lowest terms' is a little misleading, in my opinion. It is true that if $x$ is a rational number, and $p$ and $q$ are integers such that $\frac{p}{q}=x$, then there is no reason to think that $\frac{p}{q}$ is in lowest terms.
What the authors of the proof are trying to say is that any rational number has a unique lowest terms representation. And from this point onwards we will choose to denote that lowest terms representation as $\frac{p}{q}$.