If the relation R be defined on the set `A={1,2,3,4,5}` by `R={(a,b): |a^(2)-b^(2)|lt 8}.` Then, R is given by …….. .

Explanation

To solve the problem, we need to find the relation \( R \) defined on the set \( A = \{1, 2, 3, 4, 5\} \) such that \( R = \{(a, b) : |a^2 - b^2| < 8\} \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
The condition \( |a^2 - b^2| < 8 \) can be rewritten using the difference of squares:
\[ |(a - b)(a + b)| < 8 \]
This means we need to find pairs \( (a, b) \) such that the product \( (a - b)(a + b) \) is less than 8.

2. **Evaluating Possible Values of \( a \)**:
We will evaluate each value of \( a \) from the set \( A \) and determine the corresponding values of \( b \).

3. **Case 1: \( a = 1 \)**:
\[ |1^2 - b^2| < 8 \implies |1 - b^2| < 8 \]
This leads to two inequalities:
\[ -8 < 1 - b^2 < 8 \]
Solving these gives:
\[ -7 < -b^2 < 7 \implies -7 < -b^2 \quad \text{(always true)} \]
\[ b^2 < 9 \implies b < 3 \]
Thus, \( b \) can be \( 1 \) or \( 2 \). So, the pairs are \( (1, 1) \) and \( (1, 2) \).

4. **Case 2: \( a = 2 \)**:
\[ |2^2 - b^2| < 8 \implies |4 - b^2| < 8 \]
This leads to:
\[ -8 < 4 - b^2 < 8 \]
Solving gives:
\[ -4 < -b^2 < 4 \implies -4 < -b^2 \quad \text{(always true)} \]
\[ b^2 < 12 \implies b < \sqrt{12} \approx 3.46 \]
Thus, \( b \) can be \( 1, 2, \) or \( 3 \). So, the pairs are \( (2, 1), (2, 2), (2, 3) \).

5. **Case 3: \( a = 3 \)**:
\[ |3^2 - b^2| < 8 \implies |9 - b^2| < 8 \]
This leads to:
\[ -8 < 9 - b^2 < 8 \]
Solving gives:
\[ 1 < b^2 < 17 \implies 1 < b < \sqrt{17} \approx 4.12 \]
Thus, \( b \) can be \( 2, 3, \) or \( 4 \). So, the pairs are \( (3, 2), (3, 3), (3, 4) \).

6. **Case 4: \( a = 4 \)**:
\[ |4^2 - b^2| < 8 \implies |16 - b^2| < 8 \]
This leads to:
\[ -8 < 16 - b^2 < 8 \]
Solving gives:
\[ 8 < b^2 < 24 \implies \sqrt{8} \approx 2.83 < b < \sqrt{24} \approx 4.89 \]
Thus, \( b \) can be \( 3, 4 \). So, the pairs are \( (4, 3), (4, 4) \).

7. **Case 5: \( a = 5 \)**:
\[ |5^2 - b^2| < 8 \implies |25 - b^2| < 8 \]
This leads to:
\[ -8 < 25 - b^2 < 8 \]
Solving gives:
\[ 17 < b^2 < 33 \implies \sqrt{17} \approx 4.12 < b < \sqrt{33} \approx 5.74 \]
Thus, \( b \) can only be \( 5 \). So, the pair is \( (5, 5) \).

8. **Combining All Pairs**:
Now, we combine all the pairs we found:
- From \( a = 1 \): \( (1, 1), (1, 2) \)
- From \( a = 2 \): \( (2, 1), (2, 2), (2, 3) \)
- From \( a = 3 \): \( (3, 2), (3, 3), (3, 4) \)
- From \( a = 4 \): \( (4, 3), (4, 4) \)
- From \( a = 5 \): \( (5, 5) \)

Thus, the relation \( R \) is:
\[ R = \{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)\} \]