Can b always be greater than 0?

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    Inequality
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Discussion Overview

The discussion revolves around whether the condition \( a + b > 0 \) (with \( a > 0 \)) implies that \( b \) must always be greater than 0. Participants explore various proofs and counterexamples related to this inequality.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to prove that \( b > 0 \) by showing that if \( b < 0 \), it leads to a contradiction with \( a + b > 0 \).
  • Another participant provides a counterexample where \( a = 20 \) and \( b = -5 \), demonstrating that \( a + b > 0 \) can still hold true while \( b < 0 \).
  • Some participants acknowledge that \( b \) can be between 0 and \(-a\), indicating that \( b \) does not have to be strictly greater than 0.

Areas of Agreement / Disagreement

Participants generally disagree on whether \( b \) must be greater than 0, with some asserting that it can be negative under certain conditions, while others initially proposed that it must be positive.

Contextual Notes

Participants have not fully resolved the implications of their proofs and counterexamples, leaving open questions about the conditions under which \( b \) can be negative while still satisfying \( a + b > 0 \.

Alpharup
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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?
 
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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.

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

But b<0.

So where is the mistake in your proof?
 
Alpharup said:
Let a>0. It is also true that a+b>0. Can we prove that b>0 always?
No.
 
Yeah...I get it. b>-a...and b. an be between 0 and -a
 
Yup, I did not consider between 0 and -a
 

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