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 Post subject: is that proof correctPosted: Thu, 10 Feb 2011 04:15:05 UTC
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In proving :if |x|<3 ,then ,the following proof is followed:

|x|<3 => -3<x<3

for => 0<x<3 => ...............................1

for x<0 => -3<x<0 => 6>-2x => 7>1-2x => 1-2x< 7.................................2

Hence from (1) and (2) we conclude :

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 Post subject: Posted: Thu, 10 Feb 2011 05:48:32 UTC
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 Post subject: Posted: Thu, 10 Feb 2011 12:35:42 UTC
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In other words you approve of the proof given by the OP

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 Post subject: Posted: Thu, 10 Feb 2011 12:43:51 UTC
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idi wrote:
In other words you approve of the proof given by the OP

No. The OP failed to prove anything relating to his/her desired conclusion.

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 Post subject: Posted: Thu, 10 Feb 2011 13:21:49 UTC
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outermeasure wrote:
idi wrote:
In other words you approve of the proof given by the OP

No. The OP failed to prove anything relating to his/her desired conclusion.

Where did the OP failed??

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 Post subject: Posted: Thu, 10 Feb 2011 14:18:56 UTC
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idi wrote:
outermeasure wrote:
idi wrote:
In other words you approve of the proof given by the OP

No. The OP failed to prove anything relating to his/her desired conclusion.

Where did the OP failed??

In that he did nothing to advance his cause, as outermeasure says. He has two results which aren't helpful in proving his result. It's like saying "prove on " and then using something completely unrelated like the classification of compact, connected surfaces.

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 Post subject: Posted: Fri, 11 Feb 2011 01:17:07 UTC
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. He has two results which aren't helpful in proving his result. .

Which are those two results and why are they not helpful in proving his result??

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 Post subject: Posted: Fri, 11 Feb 2011 01:37:37 UTC
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x²<9 and 1-2x<7. They are for two DIFFERENT cases. What is the point of adding together inequalities from completely different situations?

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 Post subject: Posted: Fri, 11 Feb 2011 01:50:24 UTC
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idi wrote:
. He has two results which aren't helpful in proving his result. .

Which are those two results and why are they not helpful in proving his result??

They are what you have marked as (1) and (2).

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 Post subject: Posted: Fri, 11 Feb 2011 11:14:01 UTC
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aswoods wrote:
x²<9 and 1-2x<7. They are for two DIFFERENT cases. What is the point of adding together inequalities from completely different situations?

Why can't you add inequalities ,coming from completely different situations??

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 Post subject: Posted: Fri, 11 Feb 2011 14:06:47 UTC
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idi wrote:
aswoods wrote:
x²<9 and 1-2x<7. They are for two DIFFERENT cases. What is the point of adding together inequalities from completely different situations?

Why can't you add inequalities ,coming from completely different situations??

Because they have nothing to do with one another. That's like saying "why can't you apply the Pythagorean theorem to things which aren't triangles" It's because the theorem is a statement about triangles, so you HAVE to have a right triangle to use it. Similarly, if 0<x<3 AND -3<x<0, then there is NO x for which this is true, so you're not proving anything at all, rather you're proving that for all nonexistent x your statement is true, and who cares about that, you're aiming to prove it for |x|<3, and those x actually exist.

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 Post subject: Posted: Sat, 12 Feb 2011 02:19:41 UTC
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idi wrote:
aswoods wrote:
x²<9 and 1-2x<7. They are for two DIFFERENT cases. What is the point of adding together inequalities from completely different situations?

Why can't you add inequalities ,coming from completely different situations??

Because they have nothing to do with one another. That's like saying "why can't you apply the Pythagorean theorem to things which aren't triangles" It's because the theorem is a statement about triangles, so you HAVE to have a right triangle to use it. Similarly, if 0<x<3 AND -3<x<0, then there is NO x for which this is true, so you're not proving anything at all, rather you're proving that for all nonexistent x your statement is true, and who cares about that, you're aiming to prove it for |x|<3, and those x actually exist.

So in my proof ,which law of logic ,theorem,definition or axiom is ,do you think is violated??

Also in my proof i did not add 0<x<3 and -3<x<0 ,but add and

1-2x<7 . Now there is an x satisfying those two inequalities

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 Post subject: Posted: Sat, 12 Feb 2011 02:34:09 UTC
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idi wrote:
idi wrote:
aswoods wrote:
x²<9 and 1-2x<7. They are for two DIFFERENT cases. What is the point of adding together inequalities from completely different situations?

Why can't you add inequalities ,coming from completely different situations??

Because they have nothing to do with one another. That's like saying "why can't you apply the Pythagorean theorem to things which aren't triangles" It's because the theorem is a statement about triangles, so you HAVE to have a right triangle to use it. Similarly, if 0<x<3 AND -3<x<0, then there is NO x for which this is true, so you're not proving anything at all, rather you're proving that for all nonexistent x your statement is true, and who cares about that, you're aiming to prove it for |x|<3, and those x actually exist.

So in my proof ,which law of logic ,theorem,definition or axiom is ,do you think is violated??

Also in my proof i did not add 0<x<3 and -3<x<0 ,but add and

1-2x<7 . Now there is an x satisfying those two inequalities

I can't argue with that, but how do you know it's one of your original x?

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 Post subject: Posted: Sat, 12 Feb 2011 02:59:08 UTC
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from:

and 1-2x<7 we get |x|<3 and x>-3 which lead us to |x|<3.

Isn't that our original ,x

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