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Math Help - Basic Abstact Algebra Proof

  1. #1
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    Basic Abstact Algebra Proof

    Hey guys, I am working on a problem in which the difficult part is finished, but I cannot wrap my head around how to prove the simple part.

    I want to be able to use zero and the null set interchangably, but how in the world do I prove that they are equal??

    To be more specific, I am trying to prove the following Axiom:
    There is 0 in Z such that a+0 = a for each a in Z.

    However, the values I am using for a are sets (such as {a,b,c}).

    I just need to show that 0 = null set.

    Thanks in advance for any help!
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  2. #2
    Junior Member
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    Nov 2008
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    You don't really need to prove anything. All you need to do is show that the null set fits the properties of 0. For example, let's say I define a group (Z,\bullet ), where Z is the set of integers and \bullet is the regular addition but remove one from the sum, ie a \bullet b=a+b-1 for all a,b in Z. In this case, 0 is not the identity for the group, because a \bullet 0=a+0-1=a-1. The identity is 1, since a \bullet 1=a+1-1=a. And that's all I need to do to show that 1 takes the place of 0 in the axiom.

    So for you all you'd need to do is say that a+null set = a for any set a. Therefore, the null set can be used interchangeably with 0 when talking about sets.

    I hope my explanation makes sense. It's these simple parts that are usually the toughest to explain well...
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