$b \in \mathbb{R}$ such that for any $r>0$, there exists $n\in\mathbb{N}$ we have $n - r < -b < n + r$ Suppose I've got a real number $b \in \mathbb{R}$ which obeys the property:
For every $r>0$, there exists an $n \in \mathbb{N}$ such that $n - r < -b < n + r $.
(equivalent to: for every $r>0$, there exists an $n \in \mathbb{N}$ such that $|n+b|<r$, I think)
Is it true that $b=-n$? How can I formally prove this? This looks suspiciously like the definition of a limit, but I'm not entirely convinced.
EDIT: I was wrong, it forces $ (-b) \in \mathbb{N} $. But how can I prove this?

We can see that $b < 0$, as $r = \frac{1}{2}$ shows that $0 < n - \frac{1}{2} < -b$
Suppose $b = -k + \epsilon$ for $k \in \mathbb{N}$ and $0 < \epsilon < 1$, and take $0 < r < \min(1-\epsilon,\epsilon)$. 
Then, if $n - r < -b$, we have that $n < -b + r = k - (\epsilon - r) < k$. If $-b < n + r$, we have that $n > -b -r = k  - (r + \epsilon) > k - 1$. So $k - 1 < n < k$, where $k$ and $n$ are natural numbers, which is a contradiction. Therefore, we must have had $\epsilon = 0$, and $b = -k$ for some $k \in \mathbb{N}$.
