Set EqualityDate: 10/12/98 at 07:28:32 From: lior Subject: Groups algebra (discrete math) I have to prove the following equation, using groups algebra only. By groups algebra, I mean: DeMorgan's rules, absorption rules, and so on). The equation is: (A-B)-C = (A-C)-(B-C) Thank you very much for your help. Date: 10/12/98 at 10:12:21 From: Doctor Anke Subject: Re: Groups algebra (discrete math) Hello Lior, I assume that by A, B, and C, you meant sets (it is not necessary for them to have any special property). So we will do this proof in two parts: first we will show that (A-B)-C is a subset of (A-C)-(B-C), and then we will show that (A-C)-(B-C) is a subset of (A-B)-C. Showing these two things is equivalent to showing equality between the two sides. 1. (A-B)-C subset of (A-C)-(B-C) -------------------------------- To show this, we choose an arbitrary x in (A-B)-C. Now we have to show that x is in (A-C)-(B-C). Since x is in (A-B)-C, it holds that x is in A, x is not in B, and x is not in C. Therefore, it is also true that x is in A-C and x is not in B-C, so x is in (A-C)-(B-C), as well. 2. (A-C)-(B-C) subset of (A-B)-C -------------------------------- We proceed as in the above case. Let x be an arbitrary element in (A-C)-(B-C). From this we can conclude that x is in A-C and x is not in B-C. Since x is not in C and x is not in B-C, x is also not in B. So we know: x is in A, x is not in B, and x is not in C. Putting this together gives us that x is in (A-B)-C. I hope this made things clearer. I didn't state each time which law I used, but it shouldn't be too difficult to find this out and write it down. If you have further questions, please feel free to write again. - Doctor Anke, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/