My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news


Thanks Tree1Thanks
  • 1 Post By Maschke
Reply
 
LinkBack Thread Tools Display Modes
May 16th, 2017, 12:12 PM   #1
Adm
Newbie
 
Joined: May 2017
From: Neverland

Posts: 6
Thanks: 0

Unhappy [Set theory] - countering(oppositing)

K is a Group in relation to *, and a,b,c are different parameters that belong to K. and given this equation a*b = b*c.
I need to prove that if a counters itself, then c does too.
I can't really express how it works...
may I have some guidance here please?
Thanks.

Sorry for bad English.

Last edited by skipjack; May 17th, 2017 at 05:59 AM.
Adm is offline  
 
May 16th, 2017, 12:21 PM   #2
Senior Member
 
Joined: Aug 2012

Posts: 1,249
Thanks: 292

Quote:
Originally Posted by Adm View Post
K is a Group in relation to *, and a,b,c are different parameters that belong to K. and given this equation a*b = b*c.
I need to prove that if a counters itself, then c does too.
I can't really express how it works...
may I have some guidance here please?
Thanks.

Sorry for bad English.
No prob. By "counters itself" I assume you mean self-inverse. So the question is

$(K, *)$ is a group, $a, b, c \in K$. If $ab = bc$ then if $a = a^{-1}$ then $b = b^{-1}$. [Ignoring the * entirely. Don't need it].

Last edited by skipjack; May 17th, 2017 at 06:00 AM.
Maschke is offline  
May 16th, 2017, 12:27 PM   #3
Adm
Newbie
 
Joined: May 2017
From: Neverland

Posts: 6
Thanks: 0

Quote:
Originally Posted by Maschke View Post
No prob. By "counters itself" I assume you mean self-inverse. So the question is

$(K, *)$ is a group, $a, b, c \in K$. If $ab = bc$ then if $a = a^{-1}$ then $b = b^{-1}$. [Ignoring the * entirely. Don't need it].
Thank you for replying sir!
And yes I meant self-inverse,
but didn't get it how is b = b^{-1} ?
I've only started studying set theory 3 weeks ago, so I'm not that advanced yet...

Last edited by skipjack; May 17th, 2017 at 06:01 AM.
Adm is offline  
May 16th, 2017, 01:19 PM   #4
Senior Member
 
Joined: Aug 2012

Posts: 1,249
Thanks: 292

Quote:
Originally Posted by Adm View Post
Thank you for replying sir!
And yes I meant self-inverse,
but didn't get it how is b = b^{-1} ?
I've only started studying set theory 3 weeks ago, so I'm not that advanced yet...
Most of the time you just bang these out whether or not there's insight. Groups are very simple, one operation. There's so little you can do, that you might as well just try things.

We have $ab = bc$.

Ok, what can we do? I just tried the most mindless thing, I multiplied both sides on the left by $a$ to get

(1) $aab = abc$.

Under the assumption that $a = a^{-1}$ we have $a^2 = aa =1$ so

(2) $aab = abc \implies b = abc$.

Since $ab = bc$ (given) we can write

$(ab)c = (bc)c = bc^2$ using associativity (referring back to (2), so now we have

$b = bc^2$. (Combining (1) and (2)) And then hit it hard with $b^{-1}$ on the left to get

$b^{-1}b = b^{-1}bc^2 \implies 1 = c^2 \implies c = c^{-1}$.

I freely used associativity a few times without mentioning it but you should notice where I did it.

This is the general pattern for a lot of these kinds of problems. You just start multiplying both sides by this or that and shuffle things around till the result falls out. After a while you get better at it.

Don't worry about insight, just learn to crank the symbols. Insight comes much later when you need to use groups for something specific. Do be aware that group theory is everywhere in math. There is light at the end of the abstract algebra tunnel
Thanks from Adm

Last edited by skipjack; May 17th, 2017 at 06:03 AM.
Maschke is offline  
May 16th, 2017, 01:33 PM   #5
Adm
Newbie
 
Joined: May 2017
From: Neverland

Posts: 6
Thanks: 0

Smile

Quote:
Originally Posted by Maschke View Post
Most of the time you just bang these out whether or not there's insight. Groups are very simple, one operation. There's so little you can do, that you might as well just try things.

We have $ab = bc$.

Ok, what can we do? I just tried the most mindless thing, I multiplied both sides on the left by $a$ to get

$aab = abc$.

Under the assumption that $a = a^{-1}$ we have $a^2 = aa =1$ so

$aab = abc \implies b = abc$.

Since $ab = bc$ (given) we can write

$(ab)c = (bc)c = bc^2$ using associativity, so now we have

$b = bc^2$. And then hit it hard with $b^{-1}$ on the left to get

$b^{-1}b = b^{-1}bc^2 \implies 1 = c^2 \implies c = c^{-1}$.

I freely used associativity a few times without mentioning it but you should note those.

This is the general pattern for a lot of these kinds of problems. You just start multiplying both sides by this or that and shuffle things around till the result falls out. Don't worry about insight, just learn to crank the symbols, and don't worry about insight. That comes much later when you need to use groups for something specific.
My god I've spent 30 mins trying to figure out something and didn't come to just trying adding same elements to each side for some reason and given up...
I've got the picture sun bright, I'll treasure your advice!
Many thanks!

Last edited by skipjack; May 17th, 2017 at 06:14 AM.
Adm is offline  
Reply

  My Math Forum > Math Forums > Math

Tags
counteringoppositing, set, theory



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
The 9 theory and The Infinity Theory rohantomar10 Number Theory 5 October 22nd, 2015 11:05 AM
Kalphauer Theory Groundbreaking Theory? Yes or No? Need Help Completing Theory. flextera New Users 0 July 30th, 2014 12:12 PM
Set theory gaussrelatz Algebra 3 August 8th, 2012 09:35 AM
Graph theory and number theory proglote Number Theory 3 October 30th, 2011 04:20 PM
Category-theory (finite group theory) prove butabi Abstract Algebra 8 September 3rd, 2011 01:52 PM





Copyright © 2017 My Math Forum. All rights reserved.