What is the closure property of addition in Boolean groups?

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Homework Help Overview

The discussion revolves around the closure property of addition in Boolean groups, specifically focusing on proving that the sum of two subsets A and B of a set X remains a subset of X. The original poster expresses confusion regarding the definitions and requirements for demonstrating this property.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore how to show that the operation defined as A+B = (A-B) ∪ (B-A) results in a set that is a subset of X. The original poster questions the necessity of proving this, suggesting it seems obvious that the union of two sets is a set.

Discussion Status

There is an ongoing exploration of the definitions and implications of the closure property. Some participants provide guidance on how to approach the proof, while others express uncertainty about the need for formal proof of what seems intuitive. The discussion highlights the need to verify additional properties such as associativity and the existence of identities.

Contextual Notes

The conversation indicates that the definitions of operations in this context may differ from earlier learned concepts, which could contribute to the confusion. There is also mention of additional requirements for the exercise that have not yet been fully addressed.

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Homework Statement



I attached the question...

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

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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
 

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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]
 
rasmhop said:
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]

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?
 
Artusartos said:
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?

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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