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Given $a,b,c,d\mathbb\in{R}$ and $a>b$ and $d>c$ how do I prove that $a-c>b-d$?

I started the proof like this: given: $a>b$ I can add to both sides of the ineqality -c therefore I get: $a-c > b-c$

How can I show now that $a-c>b-d$?

Thanks a lot for your time and help.

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if $d > c$, then $-c > -d$ and therefore $a-c>b-c>b-d$ – Kirthi Raman Mar 30 '12 at 14:59
up vote 2 down vote accepted

if $d > c$, then $-c > -d$ and therefore $a-c>b-c>b-d$

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Sweet, thanks :) – Anonymous Mar 30 '12 at 15:09

You could verify that $(a-c) - (b-d)>0$. Towards this end, use the fact that both $a-b>0$ and $d-c>0$.

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