In section 6.1 of Richard Hammack's Book of Proof, they use the following proof (by contradiction) that $\sqrt{2}$ is irrational.
Suppose that $\sqrt{2}$ is rational, such that
$$
\exists a,b \in \mathbb{Z}, \sqrt{2} = \frac{a}{b} \tag{1} \label{eq1}
$$
Let the statement $C$ be defined as
$$
C : a \ \text{is odd and} \ b \ \text{is odd},
$$
and suppose that $C$ is true. Squaring both sides of equation \ref{eq1} and re-arranging yields
$$
a^2 = 2b^2,
$$
which means that $a^2$ is even, and hence $a$ is even. Does this not contradict the original assumption that $C$ is true, and hence complete the proof? Instead, the proof continues to show that $b$ is also even, which then completes the proof, but I am not sure why. I know that the negation of $C$ is
$$
\sim C : a \ \text{is even or} \ b \ \text{is even},
$$
so if $a$ is even, then both $C$ and $\sim C$ are true, which is a contradiction.
Best Answer
If a fraction $\frac{a}{b}$ is fully reduced then $a$ is odd or $b$ is odd, and so the statement $C$ would actually be $$ C : a \ \text{is odd or} \ b \ \text{is odd}, $$ and its negation would be $$ \sim C : a \ \text{is even and} \ b \ \text{is even}. $$ Since both $C$ and $\sim C$ being true implies a contradiction, then $\sim C$ is only true when both $a$ is shown to be even and $b$ is shown to be even.