Basic Abstact Algebra Proof

• Dec 13th 2008, 06:45 PM
TM51
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!
• Dec 13th 2008, 09:46 PM
JD-Styles
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 \$\displaystyle (Z,\bullet )\$, where Z is the set of integers and \$\displaystyle \bullet\$ is the regular addition but remove one from the sum, ie \$\displaystyle a \bullet b=a+b-1\$ for all a,b in Z. In this case, 0 is not the identity for the group, because \$\displaystyle a \bullet 0=a+0-1=a-1\$. The identity is 1, since \$\displaystyle 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...