# Prove the square root of 2 is irrational. [duplicate]

Possible Duplicate:
How can you prove that the square root of two is irrational?

I know there are many, many ways to prove that the square root of 2 is irrational. I was wondering what you think is the best way, or your favorite way. I want a good understanding of several ways to prove it so the more answers, the better.

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I voted to close as "this question will likely solicit debate, arguments, polling, or extended discussion". –  MJD Nov 7 '12 at 3:23

## marked as duplicate by Pedro Tamaroff, Ｊ. Ｍ., MJD, Old John, amWhyNov 7 '12 at 4:00

Here are some of my favorite (sketches) of proofs for the irrationality of $\sqrt{2}$.
• Using Newton's method to approximate roots of the polynomial $f(x) = x^2 - 2$, then showing that the sequence does not converge to a rational number.
• Proof by contradiction, assume that $\sqrt{2} = \frac{n}{m}$ for some $n,m \in \mathbb{Z}$ with $m \neq 0$, then $2m^{2} = n^2$, hence $n$ must be even and we can let $n = 2k$ for some $k \in \mathbb{Z}$, but then $m^2 = 2k^2$ will also be even, which is impossible if $\frac{n}{m}$ is reduced. Therefore, $\sqrt{2}$ cannot be expressed as a ratio of integers.
• Since $f(x) = x^2 -2$ is irreducible over $\mathbb{Q}[x]$, its roots must lie in some finite extension field $\mathbb{Q}(\sqrt{2})$ over the rationals.