Let assume that [tex]\sqrt3[/tex] is rational. Therefore we can express it as [tex]\dfrac{a}{b}[/tex] where [tex]a,b\in \mathbb{Z}[/tex] and [tex]\text{gcd}(a,b)=1[/tex].
[tex]\dfrac{a}{b}=\sqrt3\\\dfrac{a^2}{b^2}=3\\a^2=3b^2[/tex]
It means that [tex]3|a^2[/tex] and so also [tex]3|a[/tex].
Therefore [tex]a=3k[/tex] where [tex]k\in\mathbb{Z}[/tex].
[tex](3k)^2=3b^2\\9k^2=3b^2\\b^2=3k^2[/tex]
It means that [tex]3|b^2[/tex] and so also [tex]3|b[/tex].
If both [tex]a[/tex] and [tex]b[/tex] are divisible by 3, then it contradicts our initial assumption that [tex]\text{gcd}(a,b)=1[/tex]. Therefore [tex]\sqrt3[/tex] must be an irrational number.
