Question

2. For any subsets A and B of universal set U, prove that:
a) A∆B =B∆A
b) A∆B =(AUB)-(AnB)

Answer 1

The symmetric difference of two sets \( A \) and \( B \), denoted \( A \Delta B \), is defined as the set of elements that are in either \( A \) or \( B \) but not in their intersection. Mathematically, the symmetric difference can be expressed as:

\[
A \Delta B = (A \setminus B) \cup (B \setminus A)
\]

Now, let's prove the two parts of the problem:

### Part (a): Prove that \( A \Delta B = B \Delta A \)

To prove that \( A \Delta B = B \Delta A \), we will use the definition of the symmetric difference:

1. **Left Side**:
\[
A \Delta B = (A \setminus B) \cup (B \setminus A)
\]

2. **Right Side**:
\[
B \Delta A = (B \setminus A) \cup (A \setminus B)
\]

3. We can see that both expressions represent the same union of two sets. Specifically:
- The first term of the left side, \( A \setminus B \), becomes the second term of the right side when we swap \( A \) and \( B \).
- The second term of the left side, \( B \setminus A \), becomes the first term of the right side.

Thus:
\[
A \Delta B = (A \setminus B) \cup (B \setminus A) = (B \setminus A) \cup (A \setminus B) = B \Delta A
\]

Therefore, we conclude that:
\[
A \Delta B = B \Delta A
\]

### Part (b): Prove that \( A \Delta B = (A \cup B) \setminus (A \cap B) \)

To prove this identity, we will again use the definition of the symmetric difference and set operations.

1. **Start with the left side**:
\[
A \Delta B = (A \setminus B) \cup (B \setminus A)
\]

According to the set definitions:
- \( A \setminus B \) consists of elements that are in \( A \) but not in \( B \).
- \( B \setminus A \) consists of elements that are in \( B \) but not in \( A \).

Therefore, \( A \Delta B \) contains elements that are either in \( A \) or in \( B \), but excludes those that are in both \( A \) and \( B \).

2. **Now consider the right side**:
\[
(A \cup B) \setminus (A \cap B)
\]
- \( A \cup B \) is the set of all elements that are in \( A \), in \( B \), or in both.
- \( A \cap B \) is the set of all elements that are in both \( A \) and \( B \).

Therefore, \((A \cup B) \setminus (A \cap B)\) includes all elements in \( A \) and \( B \) except those which are common to both, which is exactly what the symmetric difference \( A \Delta B \) represents.

3. From our definitions and worked through steps:
- The elements in \( A \Delta B \) (i.e., from the union of the differences) exactly match the elements in \((A \cup B) \setminus (A \cap B)\).

Thus, we conclude that:
\[
A \Delta B = (A \cup B) \setminus (A \cap B)
\]

In summary, we have proven both properties:
- \( A \Delta B = B \Delta A \)
- \( A \Delta B = (A \cup B) \setminus (A \cap B) \)