TITLE: Proof of greatest integer theorem: floor function is well-defined
QUESTION [6 upvotes]: I have to prove that

$$\forall x \in \mathbb{R},\exists !\,n \in \mathbb{Z} \text{ s.t. }n \leq x < n+1\;.$$

where $\exists !\,n $ means there exists a unique (exactly one) $n$.
I'm done with proving that there are at least one integers for the solution.
I couldn't prove the "uniqueness" of the solution, and so I looked up the internet, and here's what I found:

Let $\hspace{2mm}n,m \in \mathbb{Z} \text{ s.t. }n \leq x < n+1$ and $m \leq x < m+1$.
Since $n \leq x \text{ and } -(m+1) < -x$, by adding both, $n + (-m-1) < (x-x) = 0$. And (some steps here) likewise, $(x-x) < n+m+1$.

Now, can I add up inequalities like that, even when the book is about real analysis (and in which assumptions are supposed to be really minimal)?
Or should I also prove those addition of inequalities?
Thank you :D

REPLY [8 votes]: As to your "adding inequalities," I do not see any issue. The basic properties of $<$ are used frequently in proofs, ordinarily without explicit mention, except at the very beginning, when ordered field has just been defined. 
Multiplication by $-1$, followed by addition, is cute, but has a somewhat magical character. I think it is more natural to argue as follows. Suppose that $m$ and $n$ are integers such that 
$$m\le x<m+1\qquad\text{and} \qquad n \le x<n+1.$$
Then 
$$m \le x <n+1 \qquad\text{and}\qquad n \le x <m+1.$$
So we have $m<n+1$ and $n<m+1$.  From $m<n+1$ we conclude that $m-n<1$. From $n<m+1$ we conclude that $m-n>-1$.  Thus $m-n=0$.