Register FAQ Search Today's Posts Mark Forums Read

 2007-10-10, 05:38 #1 wreck     "Bo Chen" Oct 2005 Wuhan,China 101010102 Posts 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.
2007-10-10, 12:49   #2
R.D. Silverman

Nov 2003

746010 Posts

Quote:
 Originally Posted by wreck 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|

 2007-10-11, 04:47 #3 wreck     "Bo Chen" Oct 2005 Wuhan,China 2·5·17 Posts 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|).
2007-10-11, 07:08   #4
Patrick123

Jan 2006
JHB, South Africa

157 Posts

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

 2007-10-13, 11:47 #5 davieddy     "Lucan" Dec 2006 England 11001010010102 Posts Call me lazy, but I haven't nailed this one yet.
 2007-10-14, 09:46 #6 akruppa     "Nancy" Aug 2002 Alexandria 9A316 Posts 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
 2007-10-15, 09:00 #7 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts A proof in one dimension might be a good start. A,B,C all lie on the x axis.
 2007-10-15, 10:19 #8 Patrick123     Jan 2006 JHB, South Africa 157 Posts 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
2007-10-16, 00:49   #9
wreck

"Bo Chen"
Oct 2005
Wuhan,China

17010 Posts

Quote:
 Originally Posted by Patrick123 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

 Similar Threads Thread Thread Starter Forum Replies Last Post wingman99 Information & Answers 3 2017-02-18 05:47 dwarfvader Puzzles 2 2017-01-24 21:20 sashamkrt Msieve 5 2013-12-16 04:10 Unregistered Information & Answers 2 2008-02-23 20:23 moo LMH > 100M 2 2005-10-28 22:52

All times are UTC. The time now is 20:36.

Sat Jan 22 20:36:50 UTC 2022 up 183 days, 15:05, 0 users, load averages: 1.80, 1.99, 1.98