elementary number theory - 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.