Let assume that [tex]\sqrt7[/tex] is a rational number. Therefore it can be expressed as a fraction [tex]\dfrac{a}{b}[/tex] where[tex]a,b\in\mathbb{Z}[/tex] and [tex]\text{gcd}(a,b)=1[/tex].
[tex]\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2[/tex]
This means that [tex]a^2[/tex] is divisible by 7, and therefore also [tex]a[/tex] is divisible by 7.
So, [tex]a=7k[/tex] where [tex]k\in\mathbb{Z}[/tex]
[tex](7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2[/tex]
Analogically to [tex]a^2=7b^2[/tex] ------- [tex]b^2[/tex] is divisible by 7 and therefore so is [tex]b[/tex].
But if both numbers [tex]a[/tex] and [tex]b[/tex] are divisible by 7, then [tex]\text{gcd}(a,b)=7[/tex] which contradicts our earlier assumption that [tex]\text{gcd}(a,b)=1[/tex].
Therefore [tex]\sqrt7[/tex] is an irrational number.
