# Homework Help: Symmetric difference in sets

1. Jul 13, 2011

### Uiiop

1. The problem statement, all variables and given/known data

There is a symmetric difference in sets X & Y, X Y is defined to be the sets of elements that are either X or Y but not both
Prove that for any sets X,Y & Z that
(X$\oplus$Y)$\oplus$(Y$\oplus$Z) = X$\oplus$Z

2. Relevant equations

$\oplus$ = symmetic difference

3. The attempt at a solution
i can see it in the venn diagams, but im not good at converting what i see into set statements this is my attempt in words

the symmetric difference of A and C is contained in the union of the symmetric difference of A and B and that of B and C because the symmetric difference of two repeated symmetric differences is the repeated symmetric difference of the join of the two multisets, where for each double set both can be removed.

i dont have a clue how to show that properly in subsets etc that why i need some help

2. Jul 13, 2011

### Uiiop

3. Jul 13, 2011

### micromass

How did you define $X\oplus Y$ in symbols??

You can split the proof up in this parts:
- Associativity: show that $X\oplus (Y\oplus Z)=(X\oplus Y)\oplus Z$
- Show that $Y\oplus Y=\emptyset$
- Show that $Y\oplus \emptyset=Y$

(this actually implies that $\oplus$ forms a group operation). These three things together imply what you want to show, do you see that?

4. Jul 14, 2011

### Uiiop

ok here is my revised attempt at the answer with your help:

In order to prove that (X⊕Y)⊕(Y⊕Z) = X⊕Z you must imply that ⊕ forms a group operation this proof is split up into three parts

The symmetric difference is associative which means that
(X⊕Y)⊕(Y⊕Z)= (Y⊕Y)⊕(X⊕Z) --I think that is correct?--
or X⊕(Y⊕Z ) = (X⊕Y) ⊕Z

The symmetric difference of the same set yields an empty set, Y⊕Y= ∅

The symmetric difference of a set and empty set yields a Y⊕∅= Y

so (Y⊕Y) = ∅
∅ ⊕(X⊕Z)= X⊕Z
hence for any sets X,Y & Z ,(X⊕Y)⊕(Y⊕Z) = X⊕Z

THAT ALL CORRECT?

5. Jul 15, 2011

### micromass

Well, you're using commutativity here (which is fine, but you must first show that you can do that). IF you don't want to do that, then

$$(X\oplus Y)\oplus (Y\oplus Z)=X\oplus (Y\oplus Y)\oplus Z=X\oplus \emptyset \oplus Z=X\oplus Z$$

is also fine...