modify Cantor function so that all it's growth points are irrational. Then you have a mapping from an interval of irrational numbers to an interval of reals. Then extend to the whole range of numbers.
consider a sequence i[n] of irrationals that converges to a rational number. Then if for each member you take the inf of reals that map into it, and consider the limit L of those (it's a bounded sequence so there's a limit), then f(L) is irrational and greater than f(i[n]) for all n. It cannot be rational, so there would be a gap, but since it's an onto mapping there can't be any gaps.
if such embedding existed, irrationals and reals would be \infty-equivalent (think infinitely long Ehrenfeucht–Fraïssé game). But they are not. I think. (well, it's been shown that such embedding is actually possible, so my assumption was incorrect)
Grue
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