So I was walking on some subset of a spherical metric, when I came across an idea that I haven't yet been able to disprove or shake. It goes as follows:

Let the rational numbers be countable an lie dense in the real numbers. Then the irrational numbers must also lay dense in the reals.

Proof: Suppose $x,y$ real numbers and $x\neq y$, assume without loss of generality that $x<y$. Then there exists a rational number $q$ such that $x+\pi<q<y+\pi$, so $x<q-\pi<y$, which is irrational. $\blacksquare$

  • Take any arbitrary 2 elements $x,y$ from the irrational numbers. Then because the denseness of the rationals, there lies a rational $c$ inbetween.
  • Take any arbitrary 2 elements $p,q$ from the rational numbers. Then because of the denseness of the irrationals, there lies a rational $c$ inbetween.

On a first glance, this would suggest that the cardinality of the irrationals is equal to the cardinality of the rationals. This is obviously not true. The question is, why not? For this I have been unable to find a rigorous argument that does not use the fact that the reals are uncountable.

PS: I'm aware of: Intuitive explanation for how could there be "more" irrational numbers than rational? . This question does however not have any satisfactory answers because they are all either built on intuition (This question specifically looks at rigorous ways to prove this.) or use the uncountability of the reals (I already know this proof. I'm looking for an alternative.)

Mitchell Faas
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    The same rational might be associated with many pairs of irrationals. For example, if $x,y$ are two irrationals with $x<0 – lulu Jul 06 '17 at 21:41
  • *Counter argument:* The same irrational might be associated with many pairs of rationals. For example, if $x,y$ are two rationals with $x<\pi – Mitchell Faas Jul 06 '17 at 21:44
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    My impression is that you have the intuition that the rational/irrational numbers "alternate" (much in the way that the even and odd numbers do in $\Bbb Z$), and so they must have the same cardinality. The flaw in this intuition is that it depends on the existence of a *successor* to each element of the set: in the real numbers, we can have no *next* rational/irrational number. – Ben Grossmann Jul 06 '17 at 21:47
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    @MitchellFaas True, but I don't see the point you want to make. All my argument (and yours) shows is that this type of "association" doesn't do a good job of counting. – lulu Jul 06 '17 at 21:49
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    As you say, your observation _suggests_ that the cardinality of the irrationals is equal to the cardinality of the rationals. But a suggestion is not a proof. How would you propose actually writing a proof? – Eric Wofsey Jul 06 '17 at 21:53
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    @EricWofsey I wouldn't. I'd use cantor's diagonal proof to show that $\mathbb{R}\setminus \mathbb{Q}$ is uncountable. But that is not really the question that I have, it's more to do with why such a proof wouldn't work. – Mitchell Faas Jul 06 '17 at 21:59
  • @Omnomnomnom This is one of the basis of the glance, yes. But not actually part of my personal intuition. In essence, I'm trying to find a proof of the uncountable nature of the irrationals without using the uncountable nature of the reals. – Mitchell Faas Jul 06 '17 at 22:02
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    You're two bulleted statements are symmetric, that is, they make no distinction between set of rationals and the set of irrationals, and so just from these two statements alone, you cannot conclude that one is countable while the other is uncountable. You are required to use more outside information about the rationals and irrationals, and which you use is up to you. One question to ask: how do you define the irrational numbers? You may find that it intimately involves the reals. – Bob Krueger Jul 06 '17 at 22:02
  • The title suggests that you are trying to count the irrationals. Your argument just shows the existence of some irrational (and rational), but doesn't really count such numbers. – ChocolateAndCheese Jul 06 '17 at 22:02
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    I think Eric's point is that it's hard to say in a rigorous way why a "suggestion" is wrong. By attempting to turn the suggestion into a rigorous proof, it might be illuminating (i.e., you will see where the logic breaks down) – Alex Zorn Jul 06 '17 at 22:03
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    With measure theory, we can use the helpful intuition that the rationals have measure zero. In other words: we can build an open set in the interval $[0,1]$, whose total "weight" is arbitrarily close to $1$, that contains only irrational numbers. This would not be possible if the irrationals were countable. – Ben Grossmann Jul 06 '17 at 22:05
  • @MitchellFaas I'm confused now. In one comment, you say *I'm trying to find a proof of the uncountable nature of the irrationals without using the uncountable nature of the reals*. In another, you say *I'd use cantor's diagonal proof to show that $\Bbb R \setminus \Bbb Q$ is uncountable. But that is not really the question that I have, it's more to do with why such a proof wouldn't work.* So: are you trying to prove that the irrationals are uncountable, or are you trying to do something else? – Ben Grossmann Jul 06 '17 at 22:10
  • Actually, it might be more intuitive to build a closed set containing only rationals, following a construction similar to that of the Cantor set. – Ben Grossmann Jul 06 '17 at 22:11
  • @Omnomnomnom Apologies for the confusion. Yes, I'm trying to prove that the irrationals are uncountable. The comment you refer to was in response to the question *"How would you write a proof of this"*. – Mitchell Faas Jul 06 '17 at 22:13
  • @Omnomnomnom: You can't have a nonempty _open_ subset of $\mathbb R$ that contains only irrationals. – hmakholm left over Monica Jul 06 '17 at 22:18
  • @HenningMakholm Hmmm, you're quite right... I was thinking of the "cantor-set-like" construction that I mentioned. – Ben Grossmann Jul 06 '17 at 22:20
  • @Omnomnomnom I'd ask you to show me that proof, but apparently this question has lived its life :') – Mitchell Faas Jul 06 '17 at 22:25
  • "On a first glance, this would suggest that the cardinality of the irrationals is equal to the cardinality of the rationals. " Why? How are the irrationals between a pair of rationals in one to one correspondence to with the rationals between a pair of rationals. To suggest a correspondence we must clarify which correspondence is being suggested. Maybe you are thinking there are only countably many pairs of rationals so there are only countably many irrations in the pairs. But each pair has uncountable many. Or maybe you are thinking... to be continued... – fleablood Jul 06 '17 at 22:33
  • @MitchellFaas see [this post](https://math.stackexchange.com/a/178635/81360) – Ben Grossmann Jul 06 '17 at 22:35
  • ... there are uncountably many pairs of irrationals so if we can find uncountably distinct pairs and there'll be uncountably many rationals in them. That fails because the intervals formed by these pairs must intersect and rationals will be repeated uncountably many times. Try clarifying where you think the correspondence is. And I guarentee you will not actually find one. – fleablood Jul 06 '17 at 22:38
  • @MitchellFaas See also [this post](https://math.stackexchange.com/a/1975180/81360) – Ben Grossmann Jul 06 '17 at 22:45
  • @Omnomnomnom I suppose that I'll have to wait for measure theory :P. Thanks for your help! PS: The first one still uses that $\mathbb{R}$ is uncountable. – Mitchell Faas Jul 06 '17 at 22:53
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    @MitchellFaas: Proving that $\mathbb R$ is uncountable and proving that $\mathbb R\setminus \mathbb Q$ is uncountable are so close to each other that it is difficult to understand exactly what it is you're expecting when you say you want to do one without also doing the other. (For example, the diagonal proof does either equally well). – hmakholm left over Monica Jul 06 '17 at 22:55
  • @HenningMakholm Well, suppose the question *"Are the reals uncountable?"* is an open question, and me, as being an aspiring mathematician have set my eyes on answering this question. Now I've decided to try a solution in which I *first* prove that $\mathbb{R}-\mathbb{Q}$ is uncountable, from which the uncountability of the reals will directly follow. I'm looking for that solution. – Mitchell Faas Jul 06 '17 at 23:02
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    @MitchellFaas: It is still very unclear which kind of proof you would accept. For example it seems like you're rejecting a diagonal argument simply because the same argument also proves $\mathbb R$ uncountable. Which _conceivable_ shape could a proof that some subset of $\mathbb R$ is uncountable have that would not also, in the same breath, prove that $\mathbb R$ itself is uncountable (and therefore be rejected by you as proving too much)? – hmakholm left over Monica Jul 06 '17 at 23:10
  • @HenningMakholm There wouldn't exist any, I'm looking for alternatives. I came across this paper though: https://www.researchgate.net/publication/230996583_The_Uncountability_of_the_Unit_Interval Which has satisfied all my desires, it's still all focused on the reals, but some can be extended. I find Levy's of particular interest. – Mitchell Faas Jul 06 '17 at 23:23
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    If $A$ is an infinite subset of $B$ and we want to prove that $A$ is uncountable, our basic options are to prove either (1) that no function $\mathbb N\to A$ can be surjective, or (2) no function $A\to\mathbb N$ can be injective. But _either_ of these strategies would also immediately imply the same property for $B$. So it is not meaningful to demand a proof of the uncountability of $\mathbb R\setminus\mathbb Q$ that is not also a direct proof of the uncountability of $\mathbb R$. – hmakholm left over Monica Jul 06 '17 at 23:23
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    I can't tell what your question is. Is your question, "Why can't my argument be used to prove the irrationals and rationals have the same cardinality?"? If so, this is impossible to answer _rigorously_ unless you explain how you intend to turn your argument into a proof. Or is your question, "How do you prove the irrational numbers are uncountable?"? If so, you should make this explicit since this is not what it sounds like you're asking. And as Henning Makholm has said, you need to clarify exactly what you would mean by not using the uncountability of the reals. – Eric Wofsey Jul 07 '17 at 01:46

1 Answers1


Here is how I think of this intuitively; this is not mathematically rigorous, and so might not be the best answer to your question, but I suppose it's worth trying. Assume we have a box that is half filled with sand:

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At each point on the boundary between the air and the sand, there are both particles of sand and particles of air. Can we use this information to measure the amount of sand compared to the amount of air in the box? This depends a lot on what you mean by the "amount." Both the air and the sand do equally well at covering the boundary. But what if weighed the air and the box, and then weighed the sand? The sand would be much heavier because it's physical density is higher. In this sense, the property that both materials cover the boundary is useless; we have to use completely different tools to measure the weight of the air and the weight of the sand.

In my head I think of the air as the rationals and the sand as the irrationals. Both numbers are equally useful in "approximating the boundary;" i.e. both sets of numbers are dense in $\mathbb{R}$. However, to measure the "weight," or countability, of the two sets, we have to use other methods.

If you are looking for direct proofs that the irrational numbers are uncountable, then those exist: https://math.stackexchange.com/a/768936/223701 but intuitively this is always what I think of when people bring this fact up.

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