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Any help at all would be great. Thank you very much.

For all $m,n,p \in \mathbb{Z}$, If $p<0$ and $mp<np$ then $n<m$

user3182252
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    Duplicate of this? – 2012ssohn Feb 21 '14 at 01:08
  • @2012ssohn They're a bit different. – user3182252 Feb 21 '14 at 01:10
  • @user3182252 It's extremely similar to what you're asking, with a change of variables. See my answer below. –  Feb 21 '14 at 01:11
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1 Answers1

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To proof: For all $m,n,p\in\mathbb{Z}$, if $p<0$ and $mp<np$ then $n<m$.

If $p<0$, then $\exists$ some $y$ such that $y=-p$. Then, $$mp<np$$Becomes $$-my<-ny$$Dividing by $y$ gives $$-m<-n$$Dividing by $-1$ changes the inequality sign, so $n<m$.

Ragnar
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