Prove that `sqrt(7)` is irrational.

Hello!

We'll prove this by a contradiction. Let `sqrt(7)` is a rational number, that means `sqrt(7)=m/n` where m and n are natural numbers.

Also let m and n be coprime. If they aren't, divide both by their GCF.

Now we square the equality and obtain `7n^2=m^2.` Therefore 7 is a factor of `m^2.` Because 7 is a prime number, 7 is also a factor of m itself. So `m=7k` for some natural k.

Thus `7n^2=7^2*k^2,` or `n^2=7k^2.` So 7 is a factor of `n^2` and therefore of n. But this means m and n have a common factor 7, which is a contradiction. This contradiction proves that `sqrt(7)` is irrational.