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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]
30 Jun 2013, 06:37

fozzzy wrote:

For statement 2

we get b^2=c^2 then b^2 - C^2 = 0 >> (b-c) (b+c) = 0

Why is this manipulation incorrect?

That is not incorrect.

You get (b-c) (b+c) = 0 so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0 Hence a(b-c)=0 => sufficient _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]
30 Jun 2013, 06:52

Zarrolou wrote:

That is not incorrect.

You get (b-c) (b+c) = 0 so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0 Hence a(b-c)=0 => sufficient

So basically B+C is just redundant information once we solve till that point? Hence that case is ignored and we focus on the second case? _________________

Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]
30 Jun 2013, 06:56

1

This post received KUDOS

fozzzy wrote:

Zarrolou wrote:

That is not incorrect.

You get (b-c) (b+c) = 0 so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0 Hence a(b-c)=0 => sufficient

So basically B+C is just redundant information once we solve till that point? Hence that case is ignored and we focus on the second case?

I do not know what you mean by "redundant information", but yes: we know that b+c cannot be zero, so the other term (b-c) must be zero.

All we get from b^2-c^2=0 in this problem is that b-c=0 _________________

It is beyond a doubt that all our knowledge that begins with experience.