# The square root of 2 is irrational

$$\sqrt 2$$, the unique pos­i­tive real num­ber whose square is 2, is not a ra­tio­nal num­ber.

# Proof

Sup­pose $$\sqrt 2$$ is ra­tio­nal. Then $$\sqrt 2=\frac{a}{b}$$ for some in­te­gers $$a$$ and $$b$$; with­out loss of gen­er­al­ity let $$\frac{a}{b}$$ be in low­est terms, i.e. $$\gcd(a,b)=1$$. We have

$$\sqrt 2=\frac{a}{b}$$

From the defi­ni­tion of $$\sqrt 2$$,

$$2=\frac{a^2}{b^2}$$
$$2b^2=a^2$$

So $$a^2$$ is a mul­ti­ple of $$2$$. Since $$2$$ is prime, $$a$$ must be a mul­ti­ple of 2; let $$a=2k$$. Then

$$2b^2=(2k)^2=4k^2$$
$$b^2=2k^2$$

So $$b^2$$ is a mul­ti­ple of $$2$$, and so is $$b$$. But then $$2|\gcd(a,b)$$, which con­tra­dicts the as­sump­tion that $$\frac{a}{b}$$ is in low­est terms! So there isn’t any way to ex­press $$\sqrt 2$$ as a frac­tion in low­est terms, and thus there isn’t a way to ex­press $$\sqrt 2$$ as a ra­tio of in­te­gers at all. That is, $$\sqrt 2$$ is ir­ra­tional.

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