## Prove that {An with n as an element of the natural numbers) is a pairwise disjoint...

For each natural number n, let An = {x be an element of real number | n - 1 < x < n}. Prove that {An | n be an element of natural number} is a pairwise disjoint family of sets and that

U An = (R^+ - N)
n is an element of N

I know that They're pairwise disjoint because if x were in the intersection of A_i and A_j then i-1< x < i and j-1<x<j where j and i are naturals, and thus i = j. Clearly, n is not in A_i for any i, because then n would be in (i-1, i) for some natural i, but by definition i is the smallest natural after i-1. Clearly any x in R^+ / N is in some A_i, namely represent x as a decimal let i=floor(x), then x is in A_i+1.

But I'm not sure how to set this up in a proof.