# 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