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Proof: Between any two irrationals lies a rational, by the Density of the rationals in the real number system. There are only countably many rationals; therefore, there are only countably many pairs of irrationals. Therefore the number of irrationals is countable since the cardinality of $2\mathbf{N}$ is $\mathbf{N}$.

I don't know why I came across this logic since I know the irrationals are uncountably infinite, but I don't see the hole in my logic.

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    The basic problem here is surely with your first statement, "Between any two irrationals lies a rational, by the Density of the rationals in the real number system." There is a big difference between "There exists an $x$ such that so-and-so holds of $x$" and "There exists a _unique_ $x$ such that so-and-so holds of $x$". Density allows you to conclude the first, but not the second.2012-07-08

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The claim that between two rationals there is only one irrational is false. In fact between two rationals there are many irrationals, so you will have to map a lot of irrationals to the same pair (for most pairs too).

Therefore your proof does not constitute of a bijection, or even a well-defined function. This is a common mishap with infinite objects, though. They tend to get very confusing!

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    @lhf: Thanks for all the suggestions!2012-07-08
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To make this argument work, you'd need a one-to-one correspondence between rational and irrational numbers. Part of the challenge of finding such a correspondence lies in the fact that between any to irrationals there is not just one rational number, but infinitely many.