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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