Question

Show that, given any 3 integers, at least 2 of them must have the property that their difference is even.

Answer 1

Ans

It is certainly nonempty, so we would like to show that for no n ∈ Z+ is there a bijection ι : [n] → Z+. This seems obvious. Unfortunately, sometimes in mathematics we must struggle to show that the obvious is true (and sometimes what seems obvious is not true!). Here we face the additional problem of not hav- ing formally axiomatized things, so it’s not completely clear what’s “fair game” to use in a proof. But consider the following: does Z+ have one element? Absolutely not: for any function ι : [1] = {1} → Z+, ι is not surjective because it does not hit ι(1) + 1. Does Z+ have two elements? Still, no: if ι is not injective, the same argument as before works; if ι is injective, its image is a 2 element subset of Z+. Since Z+ is totally ordered (indeed well-ordered), one of the two elements in the image is larger than the other, and then that element plus one is not in the image of our map. We could prove it for 3 as well, which makes us think we should probably work by induction on n. How to set it up properly? Let us try to show that for allnandallι:[n]→Z+,thereexistsN =N(ι)suchthatι([n])⊂[N]. Ifwe can do this, then since [N] is clearly a proper subset of Z+ (it does not contain N + 1, and so on) we will have shown that for no n is there a surjection [n] → Z+ (which is in fact stronger than what we claimed). But carrying throughout by induction.