# 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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