Proof that the Irrationals are Countable

Question

Proof: Between any two irrationals lies a rational, by the Density of the rationals in the real number system. There are only countably many rationals; therefore, there are only countably many pairs of irrationals. Therefore the number of irrationals is countable since the cardinality of $2\mathbf{N}$ is $\mathbf{N}$.

I don't know why I came across this logic since I know the irrationals are uncountably infinite, but I don't see the hole in my logic.

Answer 1

The claim that between two rationals there is only one irrational is false. In fact between two rationals there are many irrationals, so you will have to map a lot of irrationals to the same pair (for most pairs too).

Therefore your proof does not constitute of a bijection, or even a well-defined function. This is a common mishap with infinite objects, though. They tend to get very confusing!

Answer 2

To make this argument work, you'd need a one-to-one correspondence between rational and irrational numbers. Part of the challenge of finding such a correspondence lies in the fact that between any to irrationals there is not just one rational number, but infinitely many.