is that proof correct

idi · **Joined:** Fri, 28 May 2010 02:57:26 UTC **Posts:** 53

In proving :if |x|<3 ,then (x-1)^2<16

,the following proof is followed:

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

for $x\geq 0$ => 0<x<3 => x^2<9

...............................1

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

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

(x-1)^2<16

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9633

idi · **Joined:** Fri, 28 May 2010 02:57:26 UTC **Posts:** 53

In other words you approve of the proof given by the OP

outermeasure · **Posted:** Thu, 10 Feb 2011 12:43:51 UTC

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.

idi · **Joined:** Fri, 28 May 2010 02:57:26 UTC **Posts:** 53

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??

Shadow · **Posted:** Thu, 10 Feb 2011 14:18:56 UTC

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 $x\le x^2$ on [0,1]

" and then using something completely unrelated like the classification of compact, connected surfaces.

idi · **Joined:** Fri, 28 May 2010 02:57:26 UTC **Posts:** 53

Shadow 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??

aswoods · **Posted:** Fri, 11 Feb 2011 01:37:37 UTC

x²<9 and 1-2x<7. They are for two DIFFERENT cases. What is the point of adding together inequalities from completely different situations?

Shadow · **Posted:** Fri, 11 Feb 2011 01:50:24 UTC

idi wrote:

Shadow 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).

idi · **Joined:** Fri, 28 May 2010 02:57:26 UTC **Posts:** 53

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??

Shadow · **Posted:** Fri, 11 Feb 2011 14:06:47 UTC

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.

idi · **Joined:** Fri, 28 May 2010 02:57:26 UTC **Posts:** 53

Shadow 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 x^2<9

and

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

Shadow · **Posted:** Sat, 12 Feb 2011 02:34:09 UTC

idi wrote:

Shadow 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 x^2<9

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?

idi · **Joined:** Fri, 28 May 2010 02:57:26 UTC **Posts:** 53

from:

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

Isn't that our original ,x

S.O.S. Mathematics CyberBoard

is that proof correct

Who is online