Proof that $\root \of k$ is irrational. There is a proof in my math textbook about the fact that:

If $k$ is a positive integer that is not a perfect square , then $\root \of k$ is irrational.

Proof:
It is easy to see that there exists a positive integer $n$ such that:
$$n^2 < m < (n+1)^2 \to n < \root \of m < (n+1)$$
So , $\root \of m$ is not an integer. $....(1)$
Lets assume that $\root \of k$ is in the form $\frac{p}{q}$ where $p$ and $q$ are coprime integers and $q > 1$. Therfore:
$$ \root \of k = \frac{p}{q} \to k = \frac{p^2}{q^2} \to kq = \frac{p^2}{q}$$
Obviously , $kq$ is an integer but $\frac{p^2}{q}$ is not a integer , because $p$ and $q$ are natural numbers , coprime and $q > 1$. So because of the contradiction,
$$kq \neq \frac{p^2}{q} \to \root \of k \neq \frac{p}{q}$$
So , $\root \of k$ cant be written of the form $\frac{p}{q}$ where $p$ and $q$ are coprime integers and $q > 1$ $ ..... (2)$
From $(1)$ and $(2)$ we can conclude that $\root \of k$ is irrational. $~~~~~~~~~~~\square$

Is this a valid proof?


This part is not necessary:

It is easy to see that there exists a positive integer $n$ such that:
$$n^2 < m < (n+1)^2 \to n < \root \of m < (n+1)$$
So , $\root \of m$ is not an integer. $....(1)$

You only need:
By assumption, $k$ is a positive integer that is not a perfect square, i.e. $\ \nexists m\in\mathbb{N}\ $ such that $\ m^2 = k\ $ and by taking the positive square root of both sides, we see that  $\ \nexists m\in\mathbb{N}\ $ such that $\ m = \sqrt{k}. $
$$$$
I actually think the rest of the proof is fine, although [this is where you wrote "obviously"], if you want to be more rigorous, you should justify why

$p$ and $q$ are natural numbers , coprime and $q > 1$

implies that

$\frac{p^2}{q}$ is not a integer.

You can do this, for example, by writing both $p$ and $q$ in prime factorisation form and applying Euclid's lemma.
