Irrationality

Irrationality of 2 :-

Let, the converse √ 2 = m / n is where m is Integer and n is Natural and m/n is irreducible over Integer .
Now m² / n² = 2
=>  m² = 2* n²
=> 2 | m²
=> 2 | m  [As only squire of  an even number is even in Integer ]
=> 4 | n²
=> 2 | n
=> 2 | m,n so m/n is not irreducible over Integer . (Contradicts)

Similarly to prove that √p is irrational  where p is prime we will follow the same path .
Let, Irreducible over Integer m/n = √ p  where m is Integer n is Natural
=>   m² = p * n²
=>   p | m => p | n => m/n is reducible (Contradict)

Similarly for p is prime and p^1/a is irrational [ we will assume the same and m^a = p . n^a → p | m → p | n then m/n is reducible so contradicts ]

We can further say if n = p₁^α₁ . p₂^α₂.. [ ∀p_i are primes ] then n^1/a will be irrational
if  ∃α_i can s.t a ∤ α_i.