证明 √2 是无理数

反证法。

假设 $\sqrt2=\frac{m}{n}$，其中 $m,n$ 互素整数。

则 $m^2=2n^2$，所以 $m^2$ 为偶数，进而 $m$ 为偶数，设 $m=2k$。

代入得 $4k^2=2n^2\Rightarrow n^2=2k^2$，因此 $n$ 也为偶数。

这与 $m,n$ 互素矛盾，故 $\sqrt2$ 为无理数。

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Original Explanation

Proof by contradiction: Assume $\sqrt{2} = \tfrac{m}{n}$ where $m,n$ are coprime integers. Then $m^2 = 2n^2$, so $m^2$ is even, so $m$ is even. Let $m=2k$. Then $4k^2 = 2n^2$, so $n^2=2k^2$, meaning $n$ is also even. Hence $m,n$ share 2 as a factor, contradicting the assumption that they are coprime. Thus $\sqrt{2}$ must be irrational.