# 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, 2017 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. Jul 6, 2017 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, 2017 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? Jul 6, 2017 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. Jul 6, 2017 at 22:02

## 1 Answer

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:

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.