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: 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 m1 and m2 such that m1>nx, m2>-nx. Then: -m2<nx<m1. [Here is where I get confused]. Hence there is an integer m (with -m2<=m<=m1) 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?