Question

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.

Answer 1

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`