The square root of 2 is irrational

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


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 \(\sqrt 2\),


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


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.