2
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ You can apply the Eisenstein criterion to the polynomial $X^2-2$. $\endgroup$ – Olivier Bégassat May 18 '13 at 13:29
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

It has an infinite continued fraction.

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

$\endgroup$
  • 3
    $\begingroup$ Yes, but is your proof that an infinite continued fraction must represent an irrational a direct proof or a proof by contradiction? $\endgroup$ – Old John May 18 '13 at 13:50
  • $\begingroup$ @JohnWordsworth: I don't see how it can be classified under Proof by contradiction?, rather it can't be called direct proof also. $\endgroup$ – Inceptio May 18 '13 at 13:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.