| Version 10 |
Version 9 |
| The \emph{symmetric difference} between two sets $A$ and $B$, written |
The \emph{symmetric difference} between two sets $A$ and $B$, written |
| $A \symd\ B$, is the set of all $x$ such that either $x \in A$ or $x \in B$ but not both. |
$A \symd\ B$, is the set of all $x$ such that either $x \in A$ or $x \in B$ but not both. |
|
|
| \subsubsection*{Properties} |
\subsubsection*{Properties} |
| Suppose that $A$, $B$, and $C$ are sets. |
Suppose that $A$, $B$, and $C$ are sets. |
| \begin{itemize} |
\begin{itemize} |
| \item $A\ \symd\ B=(A\setminus B) \cup (B\setminus A)=(A \cup B)\setminus(A \cap B)$. |
\item $A\ \symd\ B=(A\setminus B) \cup (B\setminus A)=(A \cup B)\setminus(A \cap B)$. |
|
\item Note that for any set $A$, the symmetric difference satisfies $A\ \Delta \ A=\emptyset$ and $A\ \Delta\ \emptyset=A$.
|
\item Note that for any set $A$, we symmetric difference satisfies $A\ \Delta \ A=\emptyset$ and $A\ \Delta\ \emptyset=A$.
|
| \item The symmetric difference operator is commutative since $A \symd\ B=(A\setminus B) \cup (B\setminus A) = (B\setminus A) \cup (A\setminus B) = B \symd\ A$. |
\item The symmetric difference operator is commutative since $A \symd\ B=(A\setminus B) \cup (B\setminus A) = (B\setminus A) \cup (A\setminus B) = B \symd\ A$. |
| \item The symmetric difference operation is associative: $(A \symd\ B)\ \symd C = A \symd (B \symd C)$. (See the \PMlinkname{proof}{ProofOfTheAssociativityOfTheSymmetricDifferenceOperator}) |
\item The symmetric difference operation is associative: $(A \symd\ B)\ \symd C = A \symd (B \symd C)$. (See the \PMlinkname{proof}{ProofOfTheAssociativityOfTheSymmetricDifferenceOperator}) |
| \item In general, an element will be in the symmetric difference of several sets iff it is in an odd number of the sets. |
\item In general, an element will be in the symmetric difference of several sets iff it is in an odd number of the sets. |
| \end{itemize} |
\end{itemize} |
|
|
| It is worth noting that these properties show that the symmetric difference operation can be used as a group law to define an abelian group on the power set of some \PMlinkescapetext{fixed} set. |
It is worth noting that these properties show that the symmetric difference operation can be used as a group law to define an abelian group on the power set of some \PMlinkescapetext{fixed} set. |
|
|
| Finally, we note that intersection distributes over the symmetric difference operator: $A\cap(B\Delta C)=(A\cap B)\Delta(A\cap C)$, giving us that the power set of a given fixed set can be made into a Boolean ring using symmetric difference as addition, and intersection as multiplication. |
Finally, we note that intersection distributes over the symmetric difference operator: $A\cap(B\Delta C)=(A\cap B)\Delta(A\cap C)$, giving us that the power set of a given fixed set can be made into a Boolean ring using symmetric difference as addition, and intersection as multiplication. |
