Medium№ 002 · Number theory · Foundations · posted by The Thinking Board

Prove that √2 is irrational

Not for the board · 0 of 5

The problem

Show that there is no pair of integers $p, q$ with $q \neq = 0$ such that
$(\frac{p}{q})^{2} = 2.$

Equivalently: the diagonal of a square is incommensurable with its side. That older phrasing is worth keeping in view, because the discovery landed as a fact about geometry a long time before it was a fact about numbers.

The standard proof is four lines of parity. It is correct, and it is also faintly unsatisfying — it tells you no such fraction exists without showing you what goes wrong when you try to build one.

So post the argument you find most illuminating, and be explicit about what you are assuming. Unique factorization is a heavy tool to bring to a problem this small.

The older statement — side and diagonal share no common measure

1 proof posted

0 votes cast

Most convincing Newest

0votes

0beautiful

0laurels

A proof

Clave Razer · 13 Oct 2026

First at the board

Let's assume that $2$ is rational such that
$\frac{p}{q}$
is in it's simplest/irreducible form.
$\frac{p}{q} = 2$
Then taking the square of both sides.

(\frac{p}{q})^{2} = 2 = \frac{p ^{2}}{q ^{2}}

$p^{2} = 2 q^{2}$
since $p^{2} = 2 q^{2}$ the $p$ must be even, hence $p = 2 k$
$(2 k)^{2} = 2 q^{2}$

$4 k^{2} = 2 q^{2}$
$2 k^{2} = q^{2}$
therefore $q$ must be even too.

Since both $p$ and $q$ share a common factor 2, $\frac{p}{q}$ cannot be the simplest form. This forms a contradiction with primary assumptions therefore $2$ cannot be expressed in terms of $p$ and $q$ .

Prove that √2 is irrational

1 proof posted

A proof

No replies yet

Add a proof