I have to prove that
$\forall x \in \mathbb{R},\exists\text{ exactly ONE }n \in \mathbb{Z} \text{ s.t. }n \leq x < n+1\;.$
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