Background Information
Hi there, so I've started doing a real analysis course, and am obviously learning about proofs. I'm a bit new to proof writing, and am just wondering if this proof is logically valid or if I have made some goofy non-proof.
Maybe Valid Proof
Claim: If $x\in \mathbb{R} \backslash \mathbb{Q}_{\square}$ then $\sqrt{x}\not\in\mathbb{Q}$
Where $\mathbb{Q}_{\square} = \{\frac{a^2}{b^2}\in\mathbb{Q}\text{ | }(a,b)\in\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})\}$
Proof:
Let $x\in \mathbb{R} \backslash \mathbb{Q}_{\square}$, and assume $\sqrt{x}\in\mathbb{Q}$ $$\implies \exists(a,b)\in \mathbb{Z}\times(\mathbb{Z}\backslash\{0\}) \text{ s.t. }\sqrt{x} = \frac{a}{b}$$ $$\implies x = \frac{a^2}{b^2}, (a,b)\in \mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$$ $$\implies x \in \mathbb{Q}_{\square}$$ Hence a contradiction.
Therefore $\sqrt{x} \not \in \mathbb{Q}$