No.Alpharup said:Let a>0. It is also true that a+b>0. Can we prove that b>0 always?
My attempt
b>-a... but 0>-a.
therefore min (b,0)>-a
case 1: b <0.
If, b <-a, then a <-b
so a+b <b-b
so, a+b <0.
Contradiction
Hence b>0.
Is my proof right?
No.Alpharup said:Let a>0. It is also true that a+b>0. Can we prove that b>0 always?
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