There are plenty of simple proofs out there that $\sqrt{2}$ is irrational. But does there exist a proof which is not a proof by contradiction? I.e. which is not of the form:

Suppose $a/b=\sqrt{2}$ for integers $a,b$.

[deduce a contradiction here]

$\rightarrow\leftarrow$, QED

Is it impossible (or at least difficult) to find a direct proof because ir-rational is a negative definition, so "not-ness" is inherent to the question? I have a hard time even thinking how to begin a direct proof, or what it would look like. How about:

$\forall a,b\in\cal I \exists \epsilon$ such that $\mid a^2/b^2 - 2\mid > \epsilon$

There are plenty of simple proofs out there that $\sqrt{2}$ is irrational. But does there exist a proof which is not a proof by contradiction? I.e. which is not of the form:

Suppose $a/b=\sqrt{2}$ for integers $a,b$.

[deduce a contradiction here]

$\rightarrow\leftarrow$, QED

Is it impossible (or at least difficult) to find a direct proof because ir-rational is a negative definition, so "not-ness" is inherent to the question? I have a hard time even thinking how to begin a direct proof, or what it would look like. How about:

$\forall a,b\in\cal I \;\exists\; \epsilon$ such that $\mid a^2/b^2 - 2\mid > \epsilon$.