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Homework Help: A question about rings

  1. Dec 25, 2012 #1
    1. The problem statement, all variables and given/known data

    I attached the question...

    I'm having trouble proving the closure property of the addition operation.

    2. Relevant equations

    3. The attempt at a solution

    Since the textbook states "The Boolean group B(X) is the family of all the subsets of X"

    I need to show that if A is in B(X) and B is in B(X), then A+B is in B(X). But I'm not sure how...since both A and B are subsets of X, I need to show that A+B is a subset...but X is just arbitrary. I'm a bit confused about this...

    Thanks in advance

    Attached Files:

    Last edited: Dec 25, 2012
  2. jcsd
  3. Dec 25, 2012 #2
    You know that A and B are subsets of X. From this can you figure out how to show that:
    [tex](A-B) \cup (B-A)[/tex]
    is a subset of X? This is all you need to prove the closure of addition.

    Maybe you are confused because earlier you learned:
    [tex]A+B = \{a+b|a\in A,\,b \in B\}[/tex]
    But note that the + operation is redefined in this exercise to mean:
    [tex]A+B=(A-B) \cup (B-A)[/tex]
  4. Dec 25, 2012 #3
    Thanks, but the reason that I'm confused is...isn't it already obvious that [tex](A-B) \cup (B-A)[/tex] is a set? So how can I "prove" it?
  5. Dec 25, 2012 #4
    Yes it is, but you also need to prove that it is a subset of X. That too is fairly obvious. Just mention that A is a subset of X so A-B is a subset of X, similarily B-A is a subset of X. Therefore (A-B) union (B-A) is a subset of X. If this is a slightly more advanced course you may even just say "Obviously [itex](B-A)\cup (A-B)[/itex] is a subset of X".

    However you need to check that the operations are associative, that multiplication distributes over addition, and you must find additive and multiplicative identities. Furthermore you must show that every subset of X has a negation. These parts of the exercise are non-trivial (and there are still a few parts left I didn't mention such as commutativity).
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