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Is the infinity of irrational numbers equal to the infinity of rational numbers? Or is one is greater than other? And what is the proof? I could not find out a rigorous proof about this.

P.S. I am interested in which set of numbers is more dense - rational or irrational. Other similar questions do not consider the density of the numbers.

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marked as duplicate by Noah Schweber, David K, user228113, Watson, mathreadler Feb 19 '16 at 16:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Related : math.stackexchange.com/questions/93251/… $\endgroup$ – Watson Feb 19 '16 at 14:42
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    $\begingroup$ 1. There are more irrationals than rationals. 2. They are both equally dense, as this word has a precise mathematical meaning that they share. $\endgroup$ – vadim123 Feb 19 '16 at 14:44
  • $\begingroup$ @vadim123 Does not (2) contradict (1)? And can you show me a rigorous proof? $\endgroup$ – SchrodingersCat Feb 19 '16 at 14:45
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    $\begingroup$ @SchrodingersCat : density is a topological notion, and has in general nothing to do with cardinality. $\endgroup$ – Watson Feb 19 '16 at 14:46
  • $\begingroup$ @SchrodingersCat Re: a rigorous proof, this is Cantor's diagonal argument, which can be found in a number of questions on this site, or by googling in general. $\endgroup$ – Noah Schweber Feb 19 '16 at 15:04
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There is no formal definition of "more dense", you would need to explain what you mean by that.

Anyhow, note that the rationals are countable while the irrationals are uncountable.

One way to "measure" how dense they are is the following: the rationals have Lebesgue measure 0, while the irrationals have full measure in every interval. Every set which has full measure in every interval is automatically dense, while sets of measure zero are rarely dense.

Also, if we remove subsets of the same cardinality for both sets, the rationals can lose the denseness property, while the irrationals will never do that [subsets of same cardinality imply countable subsets]...

But again, you need to explain what you mean by "more" dense.

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  • $\begingroup$ So would I be wrong if I said "between any 2 rational numbers, there are infinite irrational numbers and between any 2 irrational numbers, there are infinite rational numbers."? $\endgroup$ – SchrodingersCat Feb 19 '16 at 14:57
  • $\begingroup$ @SchrodingersCat No, that's correct. $\endgroup$ – Amitesh Datta Feb 19 '16 at 14:58
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    $\begingroup$ @SchrodingersCat However, it's important to note that, despite this, there are a different number of irrationals and rationals in the same interval. Between any two distinct reals, there are infinitely many rationals, but only countably many; whereas there are uncountably many irrationals. $\endgroup$ – Noah Schweber Feb 19 '16 at 15:03
  • $\begingroup$ @NoahSchweber Thanks for your valuable comment. It helped me a lot. $\endgroup$ – SchrodingersCat Feb 19 '16 at 15:14

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