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: 



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> $\forall a,b\in\cal I \;\exists\; \epsilon$ such that $\mid a^2/b^2 - 2\mid > \epsilon$.
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Direct proof of irrationality?
Below is a simple direct proof that I found as a teenager:


**THEOREM** $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.


Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout


so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$


This idea [immediately generalizes](https://math.stackexchange.com/a/91595/242) to a proof by induction on degree that $\Bbb Z$ is integrally closed (i.e. the monic case of the rational root test).


Nowadays my favorite proof is the 1-line gem using Dedekind's **conductor ideal** - which, as I explained at length [elsewhere](https://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313), beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs.


