This is from Discrete Mathematics and its Applications
Example Prove that $\sqrt{2}$ is irrational by giving a proof by contradiction.
Solution Let $p$ be the proposition "$\sqrt{2}$ is irrational." To start a proof by contradiction, we suppose that $\neg p$ is true. Note that $\neg p$ is the statement "It is not the case that $\sqrt{2}$ is irrational," which says that $\sqrt{2}$ is rational. We will show that assuming that $\neg p$ is true leads to a contradiction.
If $\sqrt{2}$ is rational, there exist integers $a$ and $b$ with $\sqrt{2}=a/b$, where $b\ne 0$ and $a$ and $b$ have no common factors (so that the fraction $a/b$ is in lowest terms.) (Here, we are using the fact that every rational number can be written in lowest terms.) Because $\sqrt{2}=a/b$, when both sides of this equation are squared, it follows that
For the phrase "$a$ and $b$ have no common factors" , does that actually mean $a$ and $b$ have no common factors other than $1$? I feel like this would throw some readers off.
From what I read on http://www.infoplease.com/ipa/A0933353.html, $1$ can be a common factor of two numbers.