Could you show me some bijection between the RR−QQ and RR?

I was wondering if is there some "nice" bijective function that connects the irrationals and the reals. There must be some bijection, once that RR and RR-QQ has the "same size".

P.S: It doesn't need to be a continuous function.

1 Answer
Feb 13, 2018

Here's one...

Explanation:

I can explain it, though I don't have a formula just yet.

The rational numbers are countable, so we can define a sequence:

a_0 = 0, a_1, a_2, ...

enumerating all of QQ.

Then we can define a function f(x): RR -> RR "\" QQ as follows:

f(x) = { (a_(2n + 1)sqrt(2) " if " x = a_n " for " n >= 0), (a_(2n)sqrt(2) " if " x = a_n sqrt(2) " for " n >= 1), (x " otherwise") :}

This maps all rational numbers and all rational multiples of sqrt(2) to the set of all non-zero rational multiples of sqrt(2).

This f(x) is a bijection between RR and RR "\" QQ