# Thread: Basic Abstact Algebra Proof

1. ## 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!

2. 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...