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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: 7,958 Thanks: 1146 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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 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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 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 $\displaystyle Q\subset P$ Tags discrete, math Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pjlloyd100 Applied Math 11 April 18th, 2013 11:03 AM OriaG Applied Math 5 February 11th, 2013 10:27 PM amyporter17 Applied Math 1 November 17th, 2010 07:23 AM rainysomber Applied Math 10 December 3rd, 2009 02:55 AM Arturo Applied Math 6 April 3rd, 2008 08:06 PM

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