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I've been trying to prove that $\mathbb{Q} + i \mathbb{Q}$ is countable, but before I can start, I need to be sure what kind of set I'm dealing with. Are elements inside the set of the type $(a, b)$ as they are in $\mathbb{C}$ or is $\mathbb{Q} + i \mathbb{Q}$ the union of the sets $\mathbb{Q}$ and $i \mathbb{Q}$. That is, should I be proving that $\mathbb{Q} \times i \mathbb{Q}$ is countable, or that $\mathbb{Q} \cup i \mathbb{Q}$ is countable?

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    $\begingroup$ $\mathbb Q \times i\mathbb Q$. $\endgroup$ – GEdgar Apr 20 '15 at 12:47
  • $\begingroup$ @GEdgar Can you provide some details, please. And maybe suggest how should I go about establishing the countablilty of the set if I've already proven that $\mathbb{Q}$ is countable, and that the union and product of two countable sets is countable. $\endgroup$ – wutangclan Apr 20 '15 at 12:49
  • $\begingroup$ In addition to what you say you already know, you will need to show that $i\mathbb Q$ is countable... can you do that? $\endgroup$ – GEdgar Apr 20 '15 at 12:51
  • $\begingroup$ @GEdgar: That's simple enough, I think. There exists a natural bijection $f : \mathbb{Q} \to i \mathbb{Q}$ mapping $\frac{p}{q}$ into $i\frac{p}{q}$. $\endgroup$ – wutangclan Apr 20 '15 at 12:54
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The set you are dealing with is a subset of the complex numbers, it's elements are of the form $a+ib$ with $a,b$ both rational.

Considering, real and imaginary part, there is a bijection to $\mathbb{Q} \times \mathbb{Q}$ (in principle also to $\mathbb{Q} \times i\mathbb{Q}$ but it is not very common to write it that way).

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