How do I prove that numbers like $\sqrt{2}$ or $3 + \sqrt{5}$ are irrational? I always get stuck on the contradiction part.

The standard method is proof by contradiction. For $\sqrt{2}$, assume it's rational, so it can be written as $p/q$ where $p, q$ are coprime integers and $q \ne 0$. Squaring both sides gives $2 = p^2/q^2$, so $p^2 = 2q^2$. This means $p^2$ is even, implying $p$ must be even. Let $p = 2k$. Substituting gives $(2k)^2 = 2q^2$, so $4k^2 = 2q^2$, which simplifies to $2k^2 = q^2$. This means $q^2$ is even, so $q$ must also be even. This contradicts our initial assumption that $p$ and $q$ are coprime. Thus, $\sqrt{2}$ must be irrational. For $3 + \sqrt{5}$, assume it's rational, say $x$. Then $x - 3 = \sqrt{5}$. Since $x$ and $3$ are rational, their difference $x - 3$ must also be rational. However, we know $\sqrt{5}$ is irrational. A rational number cannot equal an irrational number, leading to a contradiction. Therefore, $3 + \sqrt{5}$ is irrational.