Existence of whole number between two real numbers $x$ and $x +1$? How to prove that there is a whole number, integer, between two real numbers $x$ and $x+1$ (in case $x$ is not whole)? 
I need this for an exercise solution in my Topology class, so I can, probably, use more than just axioms from set theory.
Any ideas? 
 A: There are two facts about the integers that we'll use, here:

(1) They are unbounded above and below in the reals. (Archimedean property of the reals)
(2) Any non-empty set of integers with a lower bound in the reals has a minimum. This is because any set of integers without a minimum has no lower bound in the integers, and so no lower bound in the reals by (1).

Take the least integer $n$ such that $n+1\ge x+1$. Then $x\le n$, and by our choice of $n$ we have $n=(n-1)+1<x+1$.
A: Consider $x > 0$.
Look at the set $\{ n ∈ ℕ;\; n > x\}$.
By the archimedian property of $ℝ$, this set is not empty.
Every nonempty subset of the natural numbers has a least element, say $N$ in this case.
Now, $N-1$ cannot be in the set, so $N - 1 ≤ x$.
Therefore $N ≤ x + 1$, so $x < N ≤ x + 1$.
The result for $x ≤ 0$ follows by reflecting.
A: This can be proved using the decimal expansion of $x$. If $x$ has the decimal expansion $n_0.n_1n_2n_3\cdots$, where $n_0$ is some whole number, then $x+1$ has the decimal expanion $(n_0+1).n_1n_2n_3\cdots$.
Then it is clear that $x < n_0+1 < x+1$ if $x$ is not a whole number itself.
A: Under the usual Lebesgue measure on $\Bbb R$, the quotient $\Bbb R/\Bbb Z$ has volume 1.
Also the interval $(x,x+1)$ has measure 1 and it is connected.
Therefore the image of $(x,x+1)$ under the quotient map $\pi$ is connected and has measure 1, so it is $\Bbb R/\Bbb Z\setminus\{\bar x\}$. Since $\{\bar x\}\neq\{\bar 0\}$, there must be a point $n\in(x,x+1)$ such that $\pi(n)=\bar 0$. Thus $n\in\Bbb Z$ by definition.
