
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 3rd, 2015, 12:12 PM  #1 
Newbie Joined: Nov 2015 From: haifa Posts: 2 Thanks: 0  about discrete math
hey ive got a problem in Discrete Math i need some help if some one can help me. the exercise : Will be B, X, Y groups have shown: X U B = Y U B IFF X  B = Y  B Thanks. 
November 3rd, 2015, 12:32 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond 
Why did you post this in the calculus subforum?

November 3rd, 2015, 12:41 PM  #3 
Newbie Joined: Nov 2015 From: haifa Posts: 2 Thanks: 0  
November 4th, 2015, 06:13 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
I don't see that X, Y, and B being groups has anything to do with this problem. It is true for any sets, X, Y, and B that "$\displaystyle X\cup B= Y\cup B$ if and only if X B= Y B". Since that is "if and only if" it has to be proved both ways: 1) If X B= Y B then $\displaystyle X\cup B= Y\cup B$. To prove, for sets, "P= Q" we must prove both $\displaystyle Q\subset Q$ and $\displaystyle Q\subset P$. And to prove "$\displaystyle P\subset Q$" we start "If $\displaystyle x\in P$" and use the definitions and properties of P and Q to conclude "therefore $\displaystyle x\in Q$". So if $\displaystyle x\in X\cup B$ then either (1) $\displaystyle x\in B$ or (2) $\displaystyle x\in X$. (a) If $\displaystyle x\in B$ then $\displaystyle x\in Y\cup B$ and we are done. (b) if x is not in B then it is in X and so in $\displaystyle X B= Y B$ so $\displaystyle x\in Y$ and therefore $\displaystyle x\in Y B$ and so $\displaystyle Y B\subset X B$ (a) If $\displaystyle x\in Y\cup B$ we can do the same thing, replacing X above with Y to conclude $\displaystyle X B\subset Y B$. Therefore X B= Y B (b) if X B= Y B then $\displaystyle X\cup B= Y\cup B$. (2) If $\displaystyle X\cup B= Y\cup B$ then X B= Y B. (a) $\displaystyle X B\subset Y B$ if $\displaystyle x\in Y B$ then x is in Y but not in B. So x is in $\displaystyle X\cup B= Y\cup B$ means that x is in X but not in B. Therefore $\displaystyle x\in X B$. (b) $\displaystyle Y B\subset X B$ Again the same proof, swapping X and Y shows that if $\displaystyle x\in X B$ implies $\displaystyle x\in Y B$ so $\displaystyle X B= Y B$. 
November 21st, 2015, 02:58 PM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  

Tags 
discrete, math 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Discrete Math Help?  pjlloyd100  Applied Math  11  April 18th, 2013 11:03 AM 
Please help  Discrete math  OriaG  Applied Math  5  February 11th, 2013 10:27 PM 
discrete math....i need help  amyporter17  Applied Math  1  November 17th, 2010 07:23 AM 
Discrete Math  rainysomber  Applied Math  10  December 3rd, 2009 02:55 AM 
Plz help, no one seem to be able to ! Discrete Math  Arturo  Applied Math  6  April 3rd, 2008 08:06 PM 