I was trying to prove the following statement: $$\sqrt{a}\text{ is always irrational, }\forall a\in\left\{\mathbb{R}^+ : 1<a\neq b^2\right\}.\tag{$b\in\mathbb{Z}$}$$ I know there is at least one other post of this, but I have not looked at it because I already know how to prove this (using the Fundamental Theorm of Algebra, or FTOA). However, I wanted to come up with my own proof, instead of using a well-known proof. Below is what I came up with. I have used proof by contradiction (or in latin, reductio ad absurdum). May somebody verify whether or not it is true?
Supposed Proof:
Proof: Assume the statement otherwise, then $$\exists p, q\in\left\{\mathbb{R}^+ : \sqrt{a}=\frac{p}{q}\Leftrightarrow aq^2=p^2\right\}.\tag{$q\neq 1$}$$ Since $a\neq 1$ then $p\neq q$, and either $p>q$ or $p<q$. Consider the former, i.e. $p > q$. Subtract $q^2$ from both sides. $$\begin{align} aq^2 - q^2 &= p^2 - q^2 \\ \Leftrightarrow q^2(a-1) &= (p+q)(p-q) \\ \Leftrightarrow a - 1&{\ \mid} \ \ (p+q)(p-q) \\ \Leftrightarrow a &\equiv 1\pmod {p\pm q}.\end{align}$$ For the last statement, it is unknown whether $a-1\mid p+q$ and $a-1\nmid p-q$ or vice versa, but what is known, is that $a-1$ divides one of those factors.
Since there are infinitely many primes $s$, then there are infinitely many numbers $s_0$ such that $s-1=s_0$. By Fermat's Little Theorem, raising both sides of the congruence to the power of $s-1$ yields that the modulo is prime, i.e. $$a^{s-1}\equiv 1^{s-1}= 1\pmod {p\pm q},$$ thus $p+q$ and/or $p-q$ is prime. If we were to assume that the fraction $p/q$ was irreducible, then $\gcd(p,q)=1$, making it perfectly allowable for one of the conjugates to be prime. However, $$\begin{align} q^2(a-1)&=(p+q)(p-q) \\ \Leftrightarrow q^2&{ \ \mid} \ \ (p+q)(p-q) \\ \Leftrightarrow q&{ \ \mid} \ \ p\pm q.\end{align}$$ This is of course a contradiction, because now $q\mid p$, therefore $\gcd(p,q)\neq 1$; therefore $p/q$ can be simplified; and therefore $p\pm q$ is not prime, unless $p=0$. But that means $\sqrt{a} = 0$ $\Leftrightarrow a = 0 < 1$, even though $1 < a$. Contradiction.
Because of this contradiction, the first case cannot be true, i.e. $p\not > q$. Therefore, the only case left is that $p<q$. But, using the same steps, since $p < q$, then $p - q < 0$. $$\therefore \exists t\in\mathbb{R}^+ : q^2(a-1) = -t(p+q)(p-q).\tag{$p,q>0$}$$ But $q^2\nmid -t$ because otherwise $\sqrt{-1}\mid q$ or $-1\mid q$, and thus either way, $q\notin\mathbb{R}^+$.
And $a-1\nmid -t$ because if $1 < a$, then $a - 1 > 0 > -t$. And, trivially, $t\neq 0$, for otherwise $q^2 = 0$ $\Leftrightarrow q = 0$ and $ a = 1$. Contradiction.
Thus our second case cannot be true either, and since both of our two cases lead to a contradiction, then it is false to assume the statement otherwise. This completes the proof. $\qquad\qquad\qquad\quad\,\,\,\Box$
Edit: As noted in the comments, $a\neq 2$, but I can prove seperately that $\sqrt{2}$ is irrational, so $a\neq 2$.
If my proof is correct, how can I prove the same statement but with the condition that $a\in (0,1)$? I want to prove that in the same fashion. This post is entirely derived from my curiosity.
Thank you in advance.