Prove: $B=\{x\in \Bbb R\mid x^2\in\Bbb Q\}$ is countable

Question

Prove: The set $B=\{x\in \Bbb R\mid x^2\in\Bbb Q\}$ is countable

I have this idea but I'm not sure if it's correct:

We know $\Bbb Q$ is countable so we can list $\Bbb Q$ as $\Bbb Q= \{ q_1,q_2,q_3,\ldots,q_n \mid n \in \Bbb N\}$ On the other hand $x^2=a$ has finite solutions ($0$, $1$, or $2$)

Let $A_i= \{x\mid x^2= q_i, q_i \in \Bbb Q, i \in \Bbb N\}$ following the order of $\Bbb Q$.

Let $S= \bigcup_{i=1}^\infty A_i $ And S is a countable union of countable sets. Hence S is countable

Answer 1

Enumerate the members of $\mathbb Q$, thus: $q_1,q_2,q_3,\ldots\,{}$.

List the solutions of $x^2 = q_1$ (either $0$, $1$, or $2$ of them according as $q_1$ is negative, $0$, or positive), then those of $x^2=q_2$, then those of $x^2=q_3$, and so on. That gives you an enumeration of $\{x\in\mathbb R\mid x^2\in\mathbb Q\}$.

Answer 2

Use the fact that there only countable number of equations $qx^2-p=0$ and each has only finite number of solutions. So total number of solutions is no more than countable union of finite which is countable.

Another explanation is that $B$ is the subset of all algebraic number. Since all algebraic number is countable, $B$ is countable.

Answer 3

We can, indeed, use the fact that $\Bbb Q$ is countable. Consider the function $f:\Bbb Q\to\Bbb R$ given by $$f(x)=\begin{cases}\sqrt x & x\ge0\\-\sqrt{-x} & x<0.\end{cases}$$ This can be shown to be one-to-one without too much difficulty, and its range is precisely $B,$ so we're done.