It's NOT a homework. I'm self-studying Rudin's Analysis and I came to this part that I can't follow the argument after a certain point:(adsbygoogle = window.adsbygoogle || []).push({});

statement: if x,y [itex]\in[/itex] R, and x<y, then there exists a p [itex]\in[/itex] Q such that x<p<y.

Proof: since x<y, we have y-x>0, and the Archemedean property furnishes a positive integer n such that:

n(y-x)>1

Apply the Archemedes' 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}.

[Here is where I get confused].

Hence there is an integer m (with -m_{2}<=m<=m_{1}) such that: *

m-1 <= nx < m. *

If we combine these inequalities, we obtain

nx < m <= 1 + nx < ny. *

Since n>0, it follows that

x < m/n <y.

I don't understand the places that I've marked with *. I fail to follow the argument after the place that I've mentioned it. Can anyone clarify up these statements for me please?

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# A simple question from Rudin's principles of mathematical analysis

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