This question is based on a previously asked one that I will quote in full here:
I have a question about Rudin's proof of Theorem 1.20 (b) in his book Principles of Mathematical Analysis. Theorem 1.20 is stated as follows:
(a) If $x\in\Bbb{R}, y\in\Bbb{R}$, and $x>0$, then there is a positive integer $n$ such that $$nx>y.$$ (b) If $x\in\Bbb{R}, y\in\Bbb{R}$, and $x<y$, then there exists a $p\in Q$ such that $x<p<y$.
I understand Rudin's proof of (a). The beginning of Rudin's proof of (b) is given below:
Since $x<y$, we have $y-x>0$, and (a) furnishes a positive integer $n$ such that $$n(y-x)>1.$$ Apply (a) 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 $$m-1\leq nx<m.$$
I don't understand the justification for this last sentence beginning "Hence...." How is $m$ found, and why are $m_1$ and $m_2$ needed to find $m$?
I want to know if my justification for why such an $m$ exists is correct. We are given that $-m_2<nx<m_1$, and we want to show that $$ \exists m \in[-m_2,m_1](m-1\le nx < m) \, . $$ We can proceed by contradiction. Suppose that $$ \forall m \in[-m_2,m_1](nx\not\in[m-1,m)) \, . $$ This means that \begin{align} x &\not\in [-m_2-1,-m_2) \\ x &\not\in [-m_2,-m_2+1) \\ x &\not\in [-m_2+1,-m_2+2) \\ \vdots \\ x &\not\in[m_1-2,m_1-1) \\ x &\not\in[m_1-1,m_1) \\ x &\not\in[m_1,m_1+1) \, .\\ \end{align} Putting these statements together, we get that $x\not\in[-m_2-1,m_1+1)$, and so it is not true that $-m_2<nx<m_1$. This is a contradiction.
