Proving a Number Is Irrational

"Prove $\sqrt{2}$ is irrational" is a staple interview question, not because the result is useful day-to-day, but because it tests whether a candidate can execute a clean proof by contradiction under pressure — a reasoning pattern that recurs constantly in quantitative work.

The template: assume the opposite of what you want to prove, then show that assumption leads to something impossible. Suppose $\sqrt{2}$ were rational — writable as $p/q$ in lowest terms, where $p$ and $q$ share no common factor. Squaring both sides gives $2 = p^2/q^2$, so $p^2 = 2q^2$. This means $p^2$ is even, and since only an even number squared can be even, $p$ itself must be even — write $p = 2k$. Substituting back: $(2k)^2 = 2q^2$, so $4k^2 = 2q^2$, giving $q^2 = 2k^2$ — meaning $q^2$ is even too, so $q$ is also even. But that contradicts the starting assumption that $p$ and $q$ share no common factor, since both are now even (sharing a factor of 2). The assumption breaks, so $\sqrt{2}$ cannot be written as $p/q$ at all — it is irrational.

The same skeleton — assume rationality as a fraction in lowest terms, derive a shared-factor contradiction — extends directly to proving $\sqrt{n}$ is irrational for any $n$ that isn't a perfect square, and interviewers sometimes push candidates to generalize the argument on the spot rather than just reciting the $\sqrt{2}$ case, which is really the point of asking it.