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 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 sub-forum?
November 3rd, 2015, 12:41 PM   #3
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Quote:
 Originally Posted by greg1313 Why did you post this in the calculus sub-forum?
sorry i didnt know where i need to post this question

 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
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Quote:
 Originally Posted by Country Boy To prove, for sets, "P= Q" we must prove both $\displaystyle Q\subset Q$ and $\displaystyle Q\subset P$
Sorry, this should have been "P= Q" we must prove both $\displaystyle P\subset Q$ and $\displaystyle Q\subset P$

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