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The standard proof that $\sqrt{2}$ is irrational (for example, the one in Baby Rudin) says that, upon assuming for a contradiction that $\sqrt{2}$ is rational, that we can write $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ "have no common factors."

This does not seem completely precise to me. What we want to say is that $p$ and $q$ are relatively prime, so that they no common factors other than $1$. Or that they cannot both be even.

Am I correct in this? Is it standard to define relatively prime this way? (In other words, it's so obvious that $1$ is a common factor that we can just discount it?)

John P.
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    Yes, the standard definition for relatively prime is that the greatest common divisor is $1$. That s, no common factors other than $1$. – QC_QAOA Apr 13 '20 at 05:53
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    Rudin says (talking about the existence of a rational number $p$ such that $p^2=2$) "If there were such a $p$, we could write $p = m/n$ where $m$ and $n$ are integers that are not both even." – Magdiragdag Apr 13 '20 at 06:30

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this says that if p and q has common factor lets divide both of them by that factor.so at the end we have p and q which doesn't have any common factor.so if we square the equation we can see that p*p=2*q*q, so p is divisible by 2,so if it is divisible by 2 we have that right hand side is divisible by 4.if right hand side is divisible by 4, left is divisible by 4, so q is also divisible by 2,here we have a contradiction that p and q have no prime divisors.

dato nefaridze
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