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This question already has an answer here:

How can one show that there is a bijection from $\mathbb N$ to $\mathbb Q \times \mathbb Q $?

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marked as duplicate by Austin Mohr, Thomas Andrews, Yiorgos S. Smyrlis, Vishal Gupta, user127.0.0.1 Feb 5 '14 at 5:49

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Yes, $\mathbb{Q} \times \mathbb{Q}$ is countable (denumerable). Since $\mathbb{Q}$ is countable (this follows from the fact that $\mathbb{N} \times \mathbb{N}$ is countable), taking the cartesian product of two countable sets gives you back a countable set. This link: http://www.physicsforums.com/showthread.php?t=487173 should be helpful.

The basic idea is to make a matrix out of the (infinite) list of rational numbers across each row and column so that you have pairs $(p/q,p'/q')$ of rational numbers. Then you can follow a diagonal path through the matrix to demonstrate countability.

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