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Old 2007-10-10, 05:38   #1
wreck
 
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"Bo Chen"
Oct 2005
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Default Ask an Advanced Algebra question.

a,b,c is Euclid space's three vectors.
Then: |a-b||c|<=|b-c||a|+|c-a||b|

It takes me a long time, but I still dont know how to prove it.
I dont know I can ask this type question here,though this is a forum to discuss about mersenne prime.
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Old 2007-10-10, 12:49   #2
R.D. Silverman
 
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Quote:
Originally Posted by wreck View Post
a,b,c is Euclid space's three vectors.
Then: |a-b||c|<=|b-c||a|+|c-a||b|

It takes me a long time, but I still dont know how to prove it.
I dont know I can ask this type question here,though this is a forum to discuss about mersenne prime.
Hint 1: What do you know about |x-y| vs |x| and |y|?
Think "triangle inequality".

Hint 2: Divide by |a||b||c|
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Old 2007-10-11, 04:47   #3
wreck
 
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"Bo Chen"
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I know the following result
||x|-|y|| <= |x-y| <= |x|+|y|
(a-b) = (a-c)+(c-b)
(a-b,c) = (a-c,b)+(c-b,a)
but I still can't prove the conclusion.

I can't understand the meaning of "Hint 2: Divide by |a||b||c|".
|a-b||c|<=|b-c||a|+|c-a||b|
<=> |a-b|/(|a||b|)<=|b-c|/(|b||c|)+|c-a|/(|c||a|)
There seems no apparent connection among (a-b)/(|a||b|), (b-c)/(|b||c|) and (c-a)/(|c||a|).
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Old 2007-10-11, 07:08   #4
Patrick123
 
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Quote:

I can't understand the meaning of "Hint 2: Divide by |a||b||c|".
|a-b||c|<=|b-c||a|+|c-a||b|
<=> |a-b|/(|a||b|)<=|b-c|/(|b||c|)+|c-a|/(|c||a|)
There seems no apparent connection among (a-b)/(|a||b|), (b-c)/(|b||c|) and (c-a)/(|c||a|).
This stuff is way beyond my league, but using basic algebra, could this not then be expanded to:

(a-b)/(|a||b|) = (a)/(|a||b|) - (b)/(|a||b|)

doing the same for the RHS. Then simplifying the result.

Regards
Patrick
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Old 2007-10-13, 11:47   #5
davieddy
 
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Call me lazy, but I haven't nailed this one yet.
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Old 2007-10-14, 09:46   #6
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If c has the smallest norm of the three, we can write
|a-b||c| ≤ |b-c||c| + |c-a||c|, assume |c| != 0 (or the inequality trivially holds) and divide by |c|
|a-b| ≤ |b-c| + |c-a| ⇐
|a-b| ≤ |(b-c) + (c-a)| ⇔
|a-b| ≤ |b-a|

So this case works. If any vector is 0 or any two vectors are identical, it also works. The rest eludes me.

Alex
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Old 2007-10-15, 09:00   #7
davieddy
 
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A proof in one dimension might be a good start.
A,B,C all lie on the x axis.
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Old 2007-10-15, 10:19   #8
Patrick123
 
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Dividing by |a||b||c|

Code:
LHS:


|a-b|/(|a||b|) = |a/(|a||b|) - b/(|a||b|)|
                     = |1/|b| - 1/|a| |


RHS

|b-c|/(|b||c|)+|c-a|/(|c||a|) = |b/(|b||c|) - c/(|b||c|)| + 
                                             |c/(|c||a|) - a/(|c||a|)|

thus
|1/|b| - 1/|a| | <= |1/|c| -1/|b|| + |1/|a| - 1/|c|| 

|1/|b| - 1/|a| | <= |-1/|b| + 1/|a| |

|1/|b| - 1/|a| | <= |1/|a| - 1/|b| |
Is my logic correct?

Last fiddled with by Patrick123 on 2007-10-15 at 10:21
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Old 2007-10-16, 00:49   #9
wreck
 
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Quote:
Originally Posted by Patrick123 View Post
Dividing by |a||b||c|
|a-b|/(|a||b|) = |a/(|a||b|) - b/(|a||b|)|
= |1/|b| - 1/|a| |
...
Is my logic correct?
since a,b,c are vectors, this equation not always right.
for example,
a=(1,0),b=(0,1)
then a-b=(1,-1)
|a-b|/(|a||b|)=sqrt(2),
but |1/|b| - 1/|a| | = |1-1| = 0
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