Can b always be greater than 0?

[Alpharup]

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?

 

[Samy_A]

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.

a=20, b=-5
a>0
a+b=15>0

But b<0.

So where is the mistake in your proof?

 

[Svein]

Let a>0. It is also true that a+b>0. Can we prove that b>0 always?

No.

 

[Alpharup]

Yeah...I get it. b>-a...and b. an be between 0 and -a

 

[Alpharup]

Yup, I did not consider between 0 and -a