# Given that the rationals are countable and the denseness of $\mathbb{Q}$ in $\mathbb{R}$ how can $\mathbb{R}$ be uncountable?

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.)

• The same rational might be associated with many pairs of irrationals. For example, if $x,y$ are two irrationals with $x<0<y$ we are free to choose $0$ as your rational in between them. – lulu Jul 6 '17 at 21:41
• 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. – Omnomnomnom Jul 6 '17 at 21:47
• @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 6 '17 at 21:49
• 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 6 '17 at 21:53
• 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 6 '17 at 22:02

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.