This is a proof by contradiction.

First, let's prove the square root

Let P/Q=the square root of 2, where P and Q are positive integers and P/Q is fully simplified.

P^2/Q^2=2

P^2=2Q^2

This means P^2 is even, meaning P is even, as P^2=2* A number.

Let 2K=P, as P is even.

4K^2=2Q^2

2K^2=Q^2

This means Q^2 is even, meaning Q is even.

Since P and Q are both even, that means the fraction is unsimplified, contradicting our beginning statement, which means the square root of 2 is irrational!

Replace P/Q=sqrt2 with P/Q=sqrtN, where N is a prime. Going through the same logic and replacing words like even and 2 with 0 mod N and N, we arrive that every sqrt(Prime) is irrational!

Now, take a non-perfect square. Replace factors until this number is square free, resulting in a numer Nsqrt(K) where K is not a square. This means Sqrt(K) is irrational and NsqrtK is also irrational