Prove the Cardinality Of the Reals is less than or equal to the Reals minus the Rationals?

Question

I need a proof that #(R) <= #(R\Q), where R is the reals and Q is the rationals. I don't think you can create an explict mapping from the Reals to the Irrationals as the function would have to output only irrationals. Therfore it would seem an indirect proof is in order, but if I knew what that proof was I wouldn't have asked the question.

Answer 1

Here's an explicit mapping, although another kind of proof might be more elegant.

By "e", I mean the famous irrational number e, but any irrational number will do.

For any real number x, compute f(x) in the following way:

If x can be written in the form x=ke+r, where k is a nonnegative integer and r is rational, then set f(x)=x+e. Otherwise, f(x)=x.

First, we show that this function sends R into R\Q:
Suppose that f(x) is rational for some x. If x can be written as x=ke+r, then f(x)=ke+r+e and is rational, from which we can show that e is rational (note that k is nonnegative). Contradiction. On the other hand, if x is not of the form ke+r, then f(x)=x; this implies that x is rational, so x=0e+r. Contradiction again.

Thus, f(x) is not rational.

Next, let's show that f is one-to-one.
Suppose that f(x)=f(y) for some x,y in R. If f(x) and f(y) are evaluated by the same rule, it's easy to show that x=y. So, let's suppose x=ek+r, while y cannot be written in that form. Then f(x)=ek+r+e, while f(y)=y; we now have y=f(y)=f(x)=(k+1)e+r, which contradicts our assumption that y cannot be written in that form.

So far, we've shown that f is an injection from R into R\Q. Shall we go for bijection? Why not.

Let z be a member of R\Q. If z-ne is rational for some nonnegative integer n, then n>0 (for otherwise z is in Q). We have z=ne+r (where r is rational), and z=f(z-e). If, on the other hand, z-ne is *not* rational for any nonnegative integer n, then f(z)=z.

So, the given function is a bijection between R and R\Q.