If $r$ is irrational, there is a unique integer $n$ with $|r- n|<1/2$

I'm working on proving the following statement:

Show that if r is an irrational number, there is a unique integer n such that the distance between r and n is less than $\frac{1}{2}$.

How do I show uniqueness using a proof by contradiction? How do I first establish that such a case even exists?

I have no idea where to start.

• Suppose there were two integers $n,m$ with a distance to $r$ of less than $\frac12$. What would that imply about the distance between $m$ and $n$? – Daniel Fischer Oct 23 '13 at 20:53
• Uniqueness should be easy: If $r$ is at distance $a$ from $n$, what distance is it from $n+1$? From $n-1$? From $k>n+1$ or from $l<n-1$? – Andrés E. Caicedo Oct 23 '13 at 20:54
• Existence should also be easy, depending on what you can assume: Do you know that there is an integer $k$ such that $k<r<k+1$? – Andrés E. Caicedo Oct 23 '13 at 20:55
• @DanielFischer I'm not sure. – Bob Shannon Oct 23 '13 at 20:58
• @Bob Actually the first question to ask you is: What is an irrational number? – Hagen von Eitzen Oct 23 '13 at 20:59

Suppose there were two distinct integers $n$ and $m$ such that $|r - n| < 1/2$ and $|r - m| < 1/2$. Then $|n - m| \le |r - n| + |r - m| < 1/2 + 1/2 = 1$, i.e., $|n - m| < 1$. But as $n$ and $m$ are two distinct integers we have a contradiction since $|n - m| \ge 1$.
For existence consider $\lfloor r+\dfrac12 \rfloor$. For uniqueness note that if $m,n \in \mathbb{Z}$ are two such numbers, then $|m-n| = \left|(m-r)+(r-n)\right| \le |m-r| + |n-r|$.