# Ask Uncle Colin: Consecutiveness

Dear Uncle Colin,

Can there be two or more consecutive irrational numbers?

- Between A Number And Consecutive… Huh?

Hi, BANACH, and thanks for your message!

We… have a problem here. When you’re dealing with integers, consecutive is really neatly defined: every number has a single successor, a number that’s exactly one greater.

When you get into the domain of non-integers, that relationship breaks down: we can’t really talk about the “next” number when we’re looking at real numbers. ((With complex numbers, it’s even worse: they’re not even well ordered!))

For example, you might ask what the next number after 4 is. In the integers, boom, it’s 5. Done. In the rationals, you might suggest “4 and a billionth!” - to which I’d respond, “but 4 and a billion-and-first is smaller than that, why is *that* not the next one? For any rational greater than four that you suggest, I can always find a rational number closer to 4 than yours. The same goes for irrational numbers - whatever number you pick, I can find an irrational number closer to four than yours.

### Going beyond that

In fact, if you pick any two real numbers you like, I can show that there are countably many rational numbers between them, and uncountably many irrationals.

To begin with, I’ll assume you’re happy with the idea that there are countably many rational numbers between 0 and 1, but uncountably many irrationals. I’m not going down that rabbit hole here.

Think about the decimal expansions of your two different numbers. At some point, these decimals begin to differ - and at that point, it’s easy to find two rational numbers - call them $a$ and $b$ - between the two numbers by rounding ((Details. Interested reader. Exercise)).

Then the function $f(x) = a + (b-a)x$ maps every rational number between 0 and 1 to a rational number between $a$ and $b$; every irrational number between 0 and 1 is mapped to another irrational in the correct domain.

Since there are countably many rationals in $[0,1]$ and uncountably many irrationals, the same is true for $[a,b]$ - so my proposition holds.

I hope that helps!

- Uncle Colin