Please Explain Baby Rudin Theorem 1.20 (b) By Using Statements on Baby Rudin

Question

I have a question in Baby Rudin Theorem 1.20 (b).
I have checked other Q and A's of this theorem in mathstackexchange (and I can understand this theorem).
But, those answers did not explain statements in Baby Rudin (such as Baby Rudin Theorem 1.20 (b) Proof).

That theorem states that "If $x\in\mathbb{R}$, $y\in\mathbb{R}$, and$x<y$, then there exists a $p\in\mathbb{Q}$ such that $x<p<y$".
The proof of this theorem is below.

Since $x<y$, we have $y-x>0$, and Archimedian property furnishes a positive integer $n$ such that
$n(y-x)>1$.
Apply Archimedian property again, to obtain positive integers $m_1$ and $m_2$ such that $m_1>nx$, $m_2>-nx$. Then
$-m_2<nx<m_1$.

Hence there is an integer m(with $-m_2\leq m\leq m_1$) such that

$m-1\leq nx<m$.

If we conbine these inequalities, we obtain
$nx<m\leq 1+nx<ny$.
Since n>0, it follows that
$x<\frac{m}{n}<y$.
This proves this theorem, with $p=m/n$.

I want to know how Rudin gets a consequence that "Hence there is an integer m(with $-m_2\leq m\leq m_1$) such that $m-1\leq nx<m$".

Answer 1

You have positive integers $m_1$ and $m_2$ such that $-m_2 \lt nx \lt m_1$. You can split this range of length $m_1+m_2$ into $m_1+m_2$ intervals of length $1$ and $nx$ will be in one of them

So $nx$ falls in one of the intervals: $[-m_2,-m_2+1), [-m_2+1,-m_2+2), \ldots, [-1,0), [0,1), \ldots, [m_1-2,m_1-1), [m_1-1, m_1)$. If you prefer, one of the following statements is true: $-m_2 \le nx \lt -m_2+1, \ldots, -1 \le nx \lt 0, \ldots, m_1-1 \le nx \lt m_1$

So there is an integer $m$ with $-m_2 \le m \le m_1$ such that $nx$ is in $[m-1, m)$, i.e. such that $m-1 \le nx \lt m$