Suppose we have positive integers $n, m_1, m_2$ obeying $$-m_2 < nx < m_1$$
According to my textbook (Baby Rudin), this means there exists an $m \in \mathbb Z$ with $-m_2 \leq m \leq m_1$ such that $$m-1 \leq nx < m$$
How is this acquired? More specifically, (a) how do we derive this consequence, and (b) why the change from $<$ to $\leq$ on only one side of the equation?
The first method that comes to my mind is to use to the floor of $nx$.
Let $\beta=\lfloor nx \rfloor=\sup \{z \in \mathbb{Z} | z \leq nx \}$. Since $m_1 \in \mathbb{Z}$ and $nx<m_1$, the set $\{z \in \mathbb{Z} | nx \leq z \}$ is bounded above by $m_1$. Consequently we know that $\beta$ exists by the least upper bound property of $\mathbb{Z}$. We have, $$\beta \leq nx.$$ Now we need to show that $nx<\beta+1$. Suppose for the sake of contradiction that $nx \geq \beta+1$. Then $\beta+1 \in \{z \in \mathbb{Z} | nx \leq z \}$, but this contradicts $\beta$ bounding $\{z \in \mathbb{Z} | nx \leq z \}$ from above since $\beta<\beta+1$. Therefore it must be that $nx < \beta+1$. So we have, $$\beta \leq nx < \beta+1.$$ If you're so inclined you can substitute $m=\beta+1$, so that, $$m-1 \leq nx \leq m$$