S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Fri, 18 Apr 2014 21:43:06 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 5 posts ] 
Author Message
PostPosted: Wed, 29 Feb 2012 15:07:15 UTC 
Offline
S.O.S. Newbie

Joined: Wed, 29 Feb 2012 14:46:49 UTC
Posts: 2
I am curious about your sites solution to inequalities in the form

|ax + b| < cx + d

You apply the definition of absolute value to state

(ax + b > 0 and ax + b < cx + d) or (ax + b < 0 and -(ax + b) < cx + d)

After solving the conjunctions, the union of the results produces the answer.

I am looking for a case where this would omit any extraneous solutions produced by assuming

-(cx + d) < ax + b < cx + d (similar to your proposed solution to |ax + b| < c)

but have not been able to find one. Would this assumption be valid for linear expressions?


Top
 Profile  
 
PostPosted: Wed, 29 Feb 2012 15:10:35 UTC 
Online
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13550
Location: Austin, TX
mwetzel wrote:
I am curious about your sites solution to inequalities in the form

|ax + b| < cx + d

You apply the definition of absolute value to state

(ax + b > 0 and ax + b < cx + d) or (ax + b < 0 and -(ax + b) < cx + d)

After solving the conjunctions, the union of the results produces the answer.

I am looking for a case where this would omit any extraneous solutions produced by assuming

-(cx + d) < ax + b < cx + d (similar to your proposed solution to |ax + b| < c)

but have not been able to find one. Would this assumption be valid for linear expressions?


It's valid for all such. The two cases or the one case listed at the same time, you get the same answers.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
PostPosted: Thu, 1 Mar 2012 13:36:59 UTC 
Offline
S.O.S. Newbie

Joined: Wed, 29 Feb 2012 14:46:49 UTC
Posts: 2
Why would we need to go through the longer discussion if this is the case?


Top
 Profile  
 
PostPosted: Thu, 1 Mar 2012 16:05:02 UTC 
Online
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13550
Location: Austin, TX
mwetzel wrote:
Why would we need to go through the longer discussion if this is the case?


It's a matter of style, some people like cases, some people like doing them all at once, it's a matter of preference. Think of it as texting someone versus calling them. Both can convey information, but you make a choice as to which you want to use at the time, as doing both would certainly be wasteful.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
PostPosted: Thu, 1 Mar 2012 16:49:26 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6621
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Shadow wrote:
mwetzel wrote:
Why would we need to go through the longer discussion if this is the case?


It's a matter of style, some people like cases, some people like doing them all at once, it's a matter of preference. Think of it as texting someone versus calling them. Both can convey information, but you make a choice as to which you want to use at the time, as doing both would certainly be wasteful.


Another reason is that splitting into cases works for many other inequalities, such as \lvert p\rvert+\lvert q\rvert\leq r (or with \lvert r\rvert) for some functions p,q,r, where there isn't a way to do them all at once without introducing erroneous solutions that you will need to investigate later.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA