Is my proof of irrationality correct?

Question

There is a lot of proof on this site about different numbers being irrational. This is just a generalized version.

THE STATEMENT:-

If $\forall c,k\in\mathbb{N},N\in\mathbb{N}:N\neq c^k$

then $\sqrt[k]{N}$ is irrational.

MY PROOF:-

Let $a,b\in\mathbb{Z}:\gcd(a,b)=1$

$\implies \gcd(a^k,b^k)=1$, call it $eq(1)$

Assume $\sqrt[k]{N}=\frac{a}{b}$

$\implies N=\frac{a^k}{b^k}$

$\implies a^k=Nb^k$

$\implies \gcd(a^k,b^k)=\gcd(Nb^k,b^k)=b^k$, call it $eq(2)$

Now, combining $eq(1)$ and $eq(2)$ would give

$b^k=1$

$\implies b=1$

$\implies \sqrt[k]{N}=a$

$\implies non-integer=integer$, which is a Contradiction

Which means our assumption was wrong and $\sqrt[k]{N}$ is irrational.$QED$

MY QUESTION:-

Is my proof entirely valid or is there a problem and if there is a problem then what is it?

Answer 1

At the moment you say

Assume $\sqrt[k]{N}=\frac{a}{b}$

you should already mention that $b=1$ is ruled out, because you have $N\neq c^k$ in the initial conditions. This would let you finish the proof earlier, when you arrive at $b=1$ later, without having to use vague terminology like "non-integer = integer".

Answer 2

The proof is entirely valid. Although you are using the fact that $(a,b) = 1 \implies (a^k,b^k) = 1$. Which of course is easy to prove with prime factorization.