Prove that √2 is irrational

Suppose $p^2=2q^2$ with $p,q$ positive integers. Among all such pairs there is one with $q$ smallest — the positive integers are well-ordered, so this is legitimate.

Since $p^2=2q^2$ is even, $p$ is even; an odd $p$ would give an odd $p^2$. Write $p=2k$:
$$4k^2=2q^2⟹q^2=2k^2.$$

So $(q,k)$ is another pair of exactly the same kind, and $k<q$. That contradicts the minimality of $q$.

The part worth noticing

This is nearly always dressed up as "assume $p/q$ is in lowest terms", which makes it look like a proof about fractions. It is not. It is a descent: from any solution you can manufacture a strictly smaller one, and then another, which no set of positive integers tolerates.

The only number-theoretic input is that $p^2$ even implies $p$ even. No greatest common divisors, no unique factorization. That economy matters — the same descent runs for $\sqrt{3}$ and $\sqrt{5}$ with almost no change, whereas the lowest-terms framing tempts you into machinery the problem never asked for.

The same descent, carried out with no arithmetic at all.

Suppose $m^2=2n^2$ in positive integers, and take the solution with $m$ smallest. Read it as an area: one square of side $m$ has exactly the area of two squares of side $n$.

Lay the two $n$-squares inside the $m$-square, one in the top-left corner and one in the bottom-right. Because $2n^2=m^2$, they have precisely enough area to cover it — so whatever they overlap in the middle, they must leave uncovered in the other two corners.

• The overlap is a square of side $2n-m$.
• The two uncovered corners are squares of side $m-n$.

Equal areas give
$$(2n-m{)}^2=2(m-n{)}^2.$$

Both sides are built from positive integers: $n<m<2n$ follows from $n^2<m^2=2n^2<4n^2$. And $2n-m<m$. A strictly smaller solution, from an assumed smallest one. Contradiction.

Why I would post this one and not the parity proof

The parity proof establishes a negative. This one shows the mechanism: every square you try to build has a smaller one hiding inside it, and you can carry the construction out with a ruler. The impossibility stops being an accident of even numbers and becomes something you can see happening.

It costs exactly what the parity proof costs — well-ordering, nothing more. It is the same argument. It simply has a body.

Stanley Tennenbaum found it in the 1950s as an undergraduate and never published it; it circulated by word of mouth for decades before John Conway wrote it down.

$\sqrt{2}$ is a root of
$$x^2-2=0.$$

The rational root theorem says that any rational root $p/q$ in lowest terms must have $p\mid 2$ and $q\mid 1$. The only candidates are therefore $\pm 1$ and $\pm 2$, and none of them works: $1\ne 2$ and $4\ne 2$.

What this actually buys you

On its own, nothing the descent did not already give — and it costs a good deal more, because the rational root theorem is proved from unique factorization, or at minimum from Euclid's lemma. Both are far stronger than "$p^2$ even implies $p$ even".

It earns its place by generality rather than by insight. The same three lines show that $\sqrt{n}$ is irrational for every $n$ that is not a perfect square, that $\sqrt[3]{2}$ is irrational, and that $x^2-x-1$ has no rational root at all.

So the descent proof is beautiful and specific; this one is plain and general. Which you prefer probably says more about what you want a proof to do than about either argument.