Q:

Is my proof that the square root of all imperfect squares are irrational correct?

I was answering a Quora question about whether $\sqrt{13}$ is irrational or not (link if needed), and I tried to prove that, in fact, the square root of all imperfect squares are irrational.
This is the first proof I have ever attempted, not knowing anything about them before-hand, and I barely know the mathematical symbols, never-mind how to properly set out a proof. So, keeping in mind that I am a complete newbie, can you tell me whether my proof is in fact correct or not, and if it isn't, where I went wrong and how I could improve it next time.
Also, if I chose the wrong symbol, please point out the where the mistake was and what the correct symbol would have been.
Start of Proof
Let's suppose that $n \in \mathbb{N} = \mathbb{Z}^{+}$ is not a perfect square.
This is going to be a proof by contradiction, so we are going to start out by assuming that $\sqrt{n}$ is indeed a rational number, that can be expressed in the irreducible fraction $\frac{A}{B}$ where $A, B \in \mathbb{Z}^{+}$ and $B \neq 1 \because \iff B = 1, \sqrt{n} = A$ which means $n = A^{2}$ which means $n$ is a perfect square.
$\sqrt{n} = \frac{A}{B}$
We can then square both sides to get:
$n = \frac{A^{2}}{B^{2}}$
Since $\frac{A}{B}$ is an irreducible fraction, $A$ and $B$ must not share any factors. When we square a number, we merely repeat its factors, therefore $A^{2}$ and $B^{2}$ must also not share any factors except $1$, making the fraction $\frac{A^{2}}{B^{2}}$ also irreducible.
Bacause it is irreducible, this means $\frac{A^{2}}{B^{2}} \notin \mathbb{Z}^{+} \because B^{2} > B \forall B > 1$ and $ B \in \mathbb{Z}^{+}$ and $B \neq 1$.
Since $n =  \frac{A^{2}}{B^{2}}$, this means that $n \notin \mathbb{Z}^{+}$ also.
$\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, \sqrt{n} \neq \frac{A}{B}$
As we had previously defined $n$ to be a positive integer, this is a contradiction. Therefore, our assumption that $\sqrt{n}$ could be expressed as the ratio of two integers was incorrect. Hence $\sqrt{n}$ is irrational $\forall n \in \mathbb{N} = \mathbb{Z}^{+}$ where $n$ is not a perfect square.
$\mathbb{Q.E.D.}$
End of Proof
Thanks for taking the time to read my proof. I would appreciate any and all feedback. As said, I am completely new at this so please show me where I went wrong and how to improve if I did in fact go wrong.
~Edits~:

Changed the penultimate statement $\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, n \neq \frac{A}{B}$ by adding a radical to the last $n$ that was previously missing: $\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, \sqrt{n} \neq \frac{A}{B}$
Added a concise contradiction as opposed to ending the proof by simply stating that $\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, \sqrt{n} \neq \frac{A}{B}$ without looping back to the opening when we defined $n$ as an integer.
Further reinstated why $\frac{A}{B} \notin \mathbb{Z}^{+}$ by adding reasoning that $\because B^{2} > B \forall B > 1$ and $ B \in \mathbb{Z}^{+}$ and $B \neq 1$, along with the fact that $\frac{A^{2}}{B^{2}}$ is irreducible.

Credit to Mathew Daly for helping me improve the summary.

A:

I will mention one (easily corrected) logical error and one stylistic piece of advice that could make the proof more readable.  But the upshot is that this is a well-argued proof by any standard, and especially impressive for a first effort.
When you said that $A^2$ and $B^2$ share no factors aside from 1, that does not imply that $\frac{A^2}{B^2}$ is not an integer.  You merely showed that it is a simplified rational number.  That is an important step, but ultimately, you need to add that $B^2\neq 1$.  As I said earlier, this is nearly trivial to address, since $B^2>B$ for all $B>1$.  But it is worth addressing key points in proofs even if they are trivial.
Stylistically, I felt a bit let down when you hit the contradiction.  It's kind of a climax of the proof, so you should feel free to emphasize it.  More importantly, you want to specifically point out the contradiction and the original assumption that you now know to be false (as this is a proof that's long enough that we've likely forgotten the beginning by now).  If I had written this, I might have ended it like this:

As we had previously defined $n$ to be a positive integer, this is a contradiction.  Therefore, our assumption that $\sqrt n$ could be expressed as the ratio of two integers was incorrect.  Hence $\sqrt n$  is irrational.

