Direct proof for the irrationality of $\sqrt 2$. [duplicate]

Prove that $\sqrt 2$ is irrational using direct proof.

I have seen TONS indirect proofs (e.g. proof by contradiction) for it, and people say that it's difficult to proof this directly. So is this impossible?

Thank you.

marked as duplicate by Namaste, Pedro Tamaroff♦, azimut, clark, ZanderMay 18 '13 at 14:00

• You can apply the Eisenstein criterion to the polynomial $X^2-2$. – Olivier Bégassat May 18 '13 at 13:29

What do you exactly mean by a "direct proof"?

The most direct argument I can think of for showing that $\sqrt{2}$ is irrational uses continued fractions. $\sqrt{2}$ has an infinite continued fraction (namely: $[1,2,2,2,...,]$) and can as such not be rational.

1) wikipedia has given a constructive proof, see http://en.wikipedia.org/wiki/Square_root_of_2

2) all rational numbers have a finite continued fraction expression, but $\sqrt2$ doesn't

It has an infinite continued fraction.

$$\!\ \sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.$$

• Yes, but is your proof that an infinite continued fraction must represent an irrational a direct proof or a proof by contradiction? – Old John May 18 '13 at 13:50
• @JohnWordsworth: I don't see how it can be classified under Proof by contradiction?, rather it can't be called direct proof also. – Inceptio May 18 '13 at 13:55