# Proof that the Irrationals are Countable

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.

• 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. Commented Jul 4, 2012 at 22:32
• It is true that between any two irrational lies a rational, but the same rational can separate many pairs of irrationals! Commented Jul 4, 2012 at 22:33
• It might help if you try to turn this heuristic into a formal proof and see where it goes wrong. If you are trying to construct a map from the irrationals to the rationals, you should have a hard time ensuring that the mapping is one-to-one.
– guy
Commented Jul 4, 2012 at 22:48
• math.stackexchange.com/questions/18969/… Commented Jul 4, 2012 at 23:59
• The basic problem here is surely with your first statement, "Between any two irrationals lies a rational, by the Density of the rationals in the real number system." There is a big difference between "There exists an $x$ such that so-and-so holds of $x$" and "There exists a unique $x$ such that so-and-so holds of $x$". Density allows you to conclude the first, but not the second. Commented Jul 8, 2012 at 12:09