Let A={1,2,3,4,5,6} and B = {2,4,6,8}. Find : A-B and B-A.
Show that `A-BneB-A`.

Answer

To solve the problem, we need to find the sets \( A - B \) and \( B - A \) using the definitions of set subtraction.

### Step 1: Identify the sets
We are given:
- \( A = \{1, 2, 3, 4, 5, 6\} \)
- \( B = \{2, 4, 6, 8\} \)

### Step 2: Calculate \( A - B \)
The set \( A - B \) contains all elements that are in \( A \) but not in \( B \).

- Check each element of \( A \):
- \( 1 \) is not in \( B \) → include \( 1 \)
- \( 2 \) is in \( B \) → do not include \( 2 \)
- \( 3 \) is not in \( B \) → include \( 3 \)
- \( 4 \) is in \( B \) → do not include \( 4 \)
- \( 5 \) is not in \( B \) → include \( 5 \)
- \( 6 \) is in \( B \) → do not include \( 6 \)

Thus, \( A - B = \{1, 3, 5\} \).

### Step 3: Calculate \( B - A \)
The set \( B - A \) contains all elements that are in \( B \) but not in \( A \).

- Check each element of \( B \):
- \( 2 \) is in \( A \) → do not include \( 2 \)
- \( 4 \) is in \( A \) → do not include \( 4 \)
- \( 6 \) is in \( A \) → do not include \( 6 \)
- \( 8 \) is not in \( A \) → include \( 8 \)

Thus, \( B - A = \{8\} \).

### Step 4: Show that \( A - B \neq B - A \)
Now we compare the two sets:
- \( A - B = \{1, 3, 5\} \)
- \( B - A = \{8\} \)

Since the elements of \( A - B \) (which are \( 1, 3, 5 \)) are different from the elements of \( B - A \) (which is \( 8 \)), we conclude that:
\[ A - B \neq B - A \]

### Final Result
Thus, we have:
- \( A - B = \{1, 3, 5\} \)
- \( B - A = \{8\} \)
- \( A - B \neq B - A \)

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