>Unless I'm very wrong, there are more whole numbers than natural >numbers. An induction should prove that there are twice as many.
We're probably having a language and/or terminology collision. By natural numbers, I mean the positive integers. By whole numbers, I mean the natural numbers plus the number zero. Since both sets have infinite members, each has just as many members as the other has. It just *looks* like the whole numbers have "one more". But they don't, you know, because Inf+1 == Inf, as IEEE shows us in their seminal treatise on "How to Lie With Computers under IEEE Floating Point". It's not really relevant to figuring out how to evaluate equality testing on unbounded lists in Perl, but I think that your inductive proof would lead you to conclude the opposite of what you're thinking. You can pick a first member of both sets. Then you can pick a second member of both sets. Then a third, then a fourth, and so and so forth for all cardinal numbers. Even though your list of pairings one from each set itself stretches to infinity (not that that means it actually stops somewhere, of course, as though infinity were a place; I mean it just stretches ever upwards without bound), then I think induction will convince you that in the resulting pair-list, there are no missed members from either set. So we are comfortable saying that there are "just as many" of one as the other; well, *I* am comfortable saying that, at least, and I hope you are, too. :-) It's initially a bit disturbing, though, when you realize that this necessarily leads to saying there are "just as many" multiple of two as there are of, oh, eight. Maybe that's why Cantor died mad. :-) --tom
