Rudin's proof of theorem 1.20b (Archimedean Principle)

Question

The theorem states:

(a) If $x\in \mathbb{R}, y\in \mathbb{R}$, and $x>0$, then there is a positive integer $n$ such that $$nx>y.$$ (b) If $x\in \mathbb{R}, y\in \mathbb{R}$, and $x<y$, then there exists a $p\in Q$ such that $x<p<y$.

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 am having a hard time understanding the reasoning behind the assertion "Hence there is an integer $m$ (with $-m_2\leq m\leq m_1$)" How is this inequality true? If $m$ = $-m_2$, then $-m_2>nx$ which is obviously a contradiction. I understand the finality of the proof is a result of the well ordering of integers. I just don't understand the justification of this particular inequality.

Answer 1

$-m_2 < nx < m_1$, then $nx$ must be inside one of these interval $$[-m_2, -m_2+1), [-m_2+1, -m_2+2), \ldots, [m_1-2, m_1-1), [m_1-1, m_1)$$

so $m$ can take one of the values from $-m_2+1$ up to $m_1$.

Answer 2

Note that $m_1,m_2$ are positive integers, so that in the interval $$[-m_2,m_1]$$ there will always be some integer -- at the very worst, the integer $0.$

Answer 3

A non-empty subset of $\Bbb Z^+$ has a least member.

The set $S=\{j\in \Bbb Z^+: (j-m_2)-nx>0\} \,$ is not empty because $m_1+m_2\in S.$ So let $j_0=\min S$ and let $m=j_0-m_2.$

Now $m>nx$ because $m-nx=(j_0-m_2)-nx>0.$

We have $m_1+m_2\ge j_0.$ So $m_1\ge j_0-m_2=m.$

And $j_0-1\not \in S.$ So $(m-1)-nx=((j_0-1)-m_2)-nx\le 0.\;$ Hence $m-1\le nx.$

And $j_0\ge 0$ (because $j_0\in \Bbb Z^+$) so $m=j_0-m_2\ge 1-m_2>-m_2.$