So we know that $\pi$ is irrational, that's fact! So we can't write it as $\frac{p}{q}$ where $p$ and $q$ are integers.
We also know that the square root of a prime number is irrational/
But what if $\pi$ can be written as the square root of $\frac{p}{q}$ where $p$ and $q$ are integers? Since$\frac{p}{q}$ would surely be some non integer number and its square root would surely be irrational?
So, is my intuition wrong?
On
If the following is true: $$\pi=\sqrt{\color{white}{\overline{\color{black}{\dfrac pq\,}}}}\tag{1}$$ then it would mean that $\pi^2$ is rational, by squaring both sides of $(1)$. But we know that $\pi^2$ is irrational.
In fact, $\pi$ can never be written down as: $$\pi=\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\,\cdots \sqrt{\color{white}{\overline{\color{black}{\dfrac pq\,}}}}}}}}}$$ since $\pi^n$ is irrational for every $n\in\mathbb N$. Here's the proof:
If $\pi^{n}$ was rational, then $\pi$ would not be transcendental, as it would be the root of $ax^n−b=0$ for some integers $a$,$b$.
Two things are wrong here. First of all, the square root of $p/q$ is not necessarily irrational. For example, $\sqrt{4/9} = 2/3$.
Second, it is known that $\pi$ can't be the square root (or even the $nth$ root) of any rational number. $\pi$, on top of being irrational, is in fact transcendental. That is, $\pi$ it is not a root of any non-zero polynomial equation with rational coefficients.