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The following statements are all true:

  • Between any two rational numbers, there is a real number (for example, their average).
  • Between any two real numbers, there is a rational number (see this proof of that fact, for example).
  • There are strictly more real numbers than rational numbers.

While I accept each of these as true, the third statement seems troubling in light of the first two. It seems like there should be some way to find a bijection between reals and rationals given the first two properties.

I understand that in-between each pair of rationals there are infinitely many reals (in fact, I think there's $2^{\aleph_0}$ of them), but given that this is true it seems like there should also be in turn a large number of rationals between all of those reals.

Is there a good conceptual or mathematical justification for why the third statement is tue given that the first two are as well?

Thanks! This has been bothering me for quite some time.

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    "It seems like there should be some way to find a bijection between reals and rationals given the first two properties." I highly encourage you to actually attempt to do this and see what goes wrong.2012-03-26
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    The answer, in a nutshell: real numbers correspond to *sets* of rationals rather to rationals, and there are *a lot* more sets than rationals.2012-03-26
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    Also, **very** related question here: http://math.stackexchange.com/questions/18969/ I'm not sure if it is a duplicate or not, but it's quite close.2012-03-26
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    @Asaf: (This addresses your *nutshell* comment.) Maybe, but not all these sets *at all* are required, so I am not sure I want to buy the argument as is.2012-03-26
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    @Didier: I have a bijection to guarantee that all sets are playing. Also, this argument is not for sale!2012-03-26
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    @Asaf: Not interested in buying it anyway... :-) My point is that the usual *coupures* construction does not require all the subsets of the rationals. Hence, to appeal to your (extremely clever, I am sure, but) unusual bijection to explain these things, especially *in a nutshell*, might be seen as an odd pedagogical choice.2012-03-26
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    @Didier: It's actually not that clever. It's just using $\omega^{<\omega}$ and its limits.2012-03-26
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    @Asaf: Precisely what I was alluding to.2012-03-26

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