Question on proof of irrational of $\sqrt{2}$

Question

I am able to follow every step in the standard proof by contradiction that $x^2 = 2$ has no solutions in $\mathbb{Q}$ except for one line (paraphrasing):

Assume there is such an $x$, and write $x = \frac{m}{n}$ where $m$ and $n$ have no common factors.

I know this is trying to say "$m$ and $n$ are relatively prime,'' $\gcd(m,n) = 1$, etc. But $1$ is a common factor even if that is the case. Is it convention to not consider $1$ a "common factor"?

I apologize if this is trivial. I am trying to understand the language used.

Answer 1

My guess is that the author meant that $m$ and $n$ have no common prime factors.

Answer 2

A very good spot. Mathematical proof is all about precision in the details, and this is a detail that has been missed by the author.

The way I've known the proof to be written is that $m$ and $n$ are chosen such that the fraction $\frac mn$ is in simplest form (I.e. can't be cancelled down), which essentially conveys that they must have no common factor except $1$.