Given $r$ is a real number, show that there exists a unique integer $n$ with $n-1≤r

Question

The question is related to my real analysis course.

How do I use the Archimidean property of $\mathbb R$ to show that for every real number $r$ there is a unique integer $n\in \mathbb Z$ such that $n-1\leq r \lt n$?

This is supposed to be a simple corollary, which I need to know for my exam, but I can't see how to prove it.

A suggestion in terms of style: Do not write "any x belongs to real number". Simply say "any real x". — Andrés E. Caicedo, May 03 '14 at 16:49
I have rewritten this question, so I hope it is as intended. Please do take care how you ask, it will make a big difference to how or whether people answer. I appreciate you may need help preparing for an exam (my daughter is in that zone, so I have some sympathy) - if you ask carefully, you will find you can get a great deal of help. Very often questions very similar to your own will have been posted before, and will have a range of informative answers and comments, so do also look before asking. — Mark Bennet, May 03 '14 at 17:32
Possible duplicate of Proof of greatest integer theorem: floor function is well-defined — Caleb Stanford, Jul 17 '16 at 16:57

score 2 · Answer 1 · answered May 03 '14 at 17:45

The Archimedean property is that for any $r \in R$ then there exists $n \in N$ such that $r < n$. There is therefore also $m \in N$ such that $-r < m$ and this implies that $-m < r$.

If for the first n where $r < n$ also, $n -1 \le r$ then you have found your n, i.e. $n-1 \le r < n$, otherwise, $r < n-1$. You can repeat the checking if $(n -1 ) - 1 \le r$ but within a finite number of times $n-1, -1, ...$ becomes equal to $-m$ so we are bound to find some integer between $-m$ and $n$ that satisfies the inequality.

Given $r$ is a real number, show that there exists a unique integer $n$ with $n-1≤r

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