1. ## Help with proof

Prove that given a real number x, there exist unique numbers n and ε such that x=n+ε, where n is an integer, and 0≤ε<1.

I started like this:

x is either an integer or a noninteger. If x=n+ε, then ε=x-n. If x is an integer, then x-n is an integer. So, ε must be an integer. The only integer k such that 0≤k<1 is k=0, so ε=0. This means that n=x-ε=x-0=x.

What do I conclude about the case where x is not an integer?

2. Let A be the set of all integers larger than x. That set has x as a lower bound and so contains a smallest integer. Let n be that smallest integer minus 1.

3. Ok, I think I get the hang of this (sort of). So if we rounded up x to the nearest integer and called it n, then since ε=x-n, and n>x, then ε would be negative, but epsilon can't be negative because ε≥0. So we round x up to the nearest integer x', where x'=n-1. Thus, ε=x-x'+1. Since x-x' is a negative number with -1<x-x'<0, adding 1 to it will give us a positive number with 0≤x-x'+1<1, which is what ε is.