If `A=(x : x=4n+1,nle5,n in N}` and `B={3n : n le 8, n in N}`, then find `(A-(A-B))`.

Answer

To solve the problem, we need to find the sets A and B based on the given definitions and then compute \( A - (A - B) \).

### Step 1: Define Set A
Set A is defined as:
\[ A = \{ x : x = 4n + 1, n \leq 5, n \in \mathbb{N} \} \]

Here, \( n \) is a natural number (N = {1, 2, 3, 4, 5...}). We will substitute values of \( n \) from 1 to 5 into the expression \( 4n + 1 \).

- For \( n = 1 \): \( x = 4(1) + 1 = 5 \)
- For \( n = 2 \): \( x = 4(2) + 1 = 9 \)
- For \( n = 3 \): \( x = 4(3) + 1 = 13 \)
- For \( n = 4 \): \( x = 4(4) + 1 = 17 \)
- For \( n = 5 \): \( x = 4(5) + 1 = 21 \)

Thus, the set A is:
\[ A = \{ 5, 9, 13, 17, 21 \} \]

### Step 2: Define Set B
Set B is defined as:
\[ B = \{ 3n : n \leq 8, n \in \mathbb{N} \} \]

We will substitute values of \( n \) from 1 to 8 into the expression \( 3n \).

- For \( n = 1 \): \( 3(1) = 3 \)
- For \( n = 2 \): \( 3(2) = 6 \)
- For \( n = 3 \): \( 3(3) = 9 \)
- For \( n = 4 \): \( 3(4) = 12 \)
- For \( n = 5 \): \( 3(5) = 15 \)
- For \( n = 6 \): \( 3(6) = 18 \)
- For \( n = 7 \): \( 3(7) = 21 \)
- For \( n = 8 \): \( 3(8) = 24 \)

Thus, the set B is:
\[ B = \{ 3, 6, 9, 12, 15, 18, 21, 24 \} \]

### Step 3: Find \( A - B \)
Now we will find the difference \( A - B \). This means we will remove the elements of B from A.

- A = {5, 9, 13, 17, 21}
- B = {3, 6, 9, 12, 15, 18, 21, 24}

The common elements between A and B are 9 and 21. Therefore, we remove these from A:
\[ A - B = \{ 5, 13, 17 \} \]

### Step 4: Find \( A - (A - B) \)
Now we need to find \( A - (A - B) \). This means we will remove the elements of \( A - B \) from A.

- A = {5, 9, 13, 17, 21}
- \( A - B = \{ 5, 13, 17 \} \)

The common elements between A and \( A - B \) are 5, 13, and 17. Therefore, we remove these from A:
\[ A - (A - B) = \{ 9, 21 \} \]

### Final Answer
Thus, the final answer is:
\[ A - (A - B) = \{ 9, 21 \} \]