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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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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. –  Michael Hardy Jul 4 '12 at 22:32
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It is true that between any two irrational lies a rational, but the same rational can separate many pairs of irrationals! –  dtldarek Jul 4 '12 at 22:33
    
It might help if you try to turn this heuristic into a formal proof and see where it goes wrong. If you are trying to construct a map from the irrationals to the rationals, you should have a hard time ensuring that the mapping is one-to-one. –  guy Jul 4 '12 at 22:48
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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. –  Benedict Eastaugh Jul 8 '12 at 12:09
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up vote 12 down vote accepted

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! –  Asaf Karagila Jul 8 '12 at 12:05
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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.

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