1. ## symmetric difference

2. Hello,
Originally Posted by 1234567

$A \Delta B=(A \cup B) \backslash (A \cap B)$, this is according to the definition you're given.

It is pretty obvious that it is commutative, by commutativity of $\cap$ and $\cup$.
As for associativity, you have to prove that $(A \Delta B) \Delta C=A \Delta (B \Delta C)$. Just use the definition.
What is the identity, that is to say what is E such that $A \Delta E=A$ ? $A \Delta E=(A \cup E) \backslash (A \cap E)$. If $A \cap E=\emptyset$ and $A \cup E=A$, you're done. The set E verifying this condition is $E=\emptyset$

As for the inverse, you're looking for F such that $A \Delta F=\emptyset$. $(A \cup F) \backslash (A \cap F)=\emptyset$
If $A \cup F=A \cap F$, then you're done.
So $F=A$

For the last question, just rewrite the definition :
$(A \cup B) \backslash (A \cap B)=(A \cup B) \cap (A \cap B)^c$ (the complement)

3. Originally Posted by Moo
Hello,

$A \Delta B=(A \cup B) \backslash (A \cap B)$, this is according to the definition you're given.

It is pretty obvious that it is commutative, by commutativity of $\cap$ and $\cup$.
As for associativity, you have to prove that $(A \Delta B) \Delta C=A \Delta (B \Delta C)$. Just use the definition.
What is the identity, that is to say what is E such that $A \Delta E=A$ ? $A \Delta E=(A \cup E) \backslash (A \cap E)$. If $A \cap E=\emptyset$ and $A \cup E=A$, you're done. The set E verifying this condition is $E=\emptyset$

As for the inverse, you're looking for F such that $A \Delta F=\emptyset$. $(A \cup F) \backslash (A \cap F)=\emptyset$
If $A \cup F=A \cap F$, then you're done.
So $F=A$

For the last question, just rewrite the definition :
$(A \cup B) \backslash (A \cap B)=(A \cup B) \cap (A \cap B)^c$ (the complement)
THANKS FOR THE HELP, I NOW UNDERSTAND HOW TO DO THE
identity, and inverse laws BUT CAN NOT STILL DO THE