2. For any subsets A and B of universal set U, prove that:

Question

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:

• \( A \Delta B = B \Delta A \)
• \( A \Delta B = (A \cup B) \setminus (A \cap B) \)