√2, by descent

Suppose $\sqrt2$ were a fraction, written in lowest terms as $p/q$. Then a second representation exists:

which is checked by cross-multiplying: $(2q-p)^2-2(p-q)^2 = 2q^2-p^2 = 0$ whenever $p^2 = 2q^2$. But from $1 < p/q < 2$ we get $0 < p-q < q$: the new denominator is strictly smaller than the smallest possible one. No such fraction can exist.1

This is infinite descent in its purest form: from any counterexample, a smaller one; hence none at all.

The continued fraction

The identity $\sqrt2 = 1 + \dfrac{1}{1+\sqrt2}$ feeds itself forever:

Truncating gives the convergents $1,\ \tfrac32,\ \tfrac75,\ \tfrac{17}{12},\ \tfrac{41}{29},\ \tfrac{99}{70},\ldots$ — each the best rational approximation for its size, each satisfying the Pell alternation $p^2-2q^2=\pm1$. The descent above is the continued fraction running backwards.

One consequence sits in every office. A paper size with aspect ratio $r$ keeps its proportions when halved exactly if $r^2 = 2$; that is why A4 measures $210 \times 297$ millimeters, with $297/210 = 1.4142\ldots$ The sheet is the irrationality proof, laminated.