If $n$ is a natural number then $n$ is a unique product of primes to integer powers
If $n$ is a perfect square then its prime factors will all be to even powers hence when taking the square root the results prime factors will be to integer powers hence the square root is an integer.
If $n$ is not a perfect square then at least one prime factor is to an odd power. Hence its square root will have a prime to a non integer power but all integers can be written as a product of primes to an integer power hence $\sqrt{n}$ cannot be an integer.
Is this sufficient to prove all square roots of non perfect squares is irrational?