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.
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2You can apply the Eisenstein criterion to the polynomial $X^2-2$. – Olivier Bégassat May 18 '13 at 13:29
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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.
Dedalus
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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
arax
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It has an infinite continued fraction.
$$ \!\ \sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}. $$
Inceptio
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4Yes, 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
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1@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