Between any two rationals, there is an irrational. Between any two irrationals, there is a rational. It would be nice to use this to compare, but it's tricky, because between any two rationals, there are more rationals as well.
So, to really compare things fairly, we need to come up with some new device to talk about the numbers "between" rationals in a way that there are no extra rationals getting in the way.
To start, suppose we have two rationals $a < b$. Well, there is a rational $a_1$ between them in the way, so we can refine our view from $(a,b)$ to $(a_1,b)$. Or, we could have refined to $(a_0,a_1)$. Let's pick the first. Then there's a rational $a_2$ in $(a_1, b)$, so let's restrict our attention again, to $(a_1, a_2)$. Then....
Well, this is a bit of a mess, how can we organize it? Well, how about the following? Let's split all of the rational numbers into two (non-empty) sets: the "left" set and the "right" set, so that every number in the left set is smaller than every number in the right set.
Now, we can ask what's between the left set and the right set, and there are no rational numbers left to get in the way! This is called a "Dedekind cut".
It turns out there are exactly three sorts of Dedekind cuts:
- There is a rational number $q$ so that $L = (-\infty, q] \cap \mathbb{Q}$ and $R = (q, \infty) \cap \mathbb{Q}$
- There is a rational number $q$ so that $L = (-\infty, q) \cap \mathbb{Q}$ and $R = [q, \infty) \cap \mathbb{Q}$
- There is an irrational number $r$ so that $L = (-\infty, r) \cap \mathbb{Q}$ and $R = (r, \infty) \cap \mathbb{Q}$
So we've finally gotten to our goal of finding a device to talk about numbers that are between rationals, without other rational numbers getting in the middle. And, as we had hoped, there cannot be more than one irrational in-between.
But, surprise! There are uncountably many Dedekind cuts, so there really are more ways to talk about things "between" rationals than there are rationals themselves.
Fortunately, if we repeat the above for irrational numbers, we find that the number of Dedekind cuts of irrationals is the same as the number of Dedekind cuts for rationals, so at least we get those numbers the same.