Let A =`( 1,2,3,4,……250)` and R be the relation " is cube of " in A. Find R as subset of `A xx A ` Also find the domain and range of R.

Answer

To solve the problem, we need to find the relation \( R \) defined as "is cube of" within the set \( A = \{1, 2, 3, \ldots, 250\} \). We will identify the pairs \( (x, y) \) such that \( x \) is the cube of \( y \) and both \( x \) and \( y \) are elements of \( A \).

### Step-by-Step Solution:

1. **Identify Perfect Cubes in Set A**:
We need to find all perfect cubes \( x \) such that \( x \leq 250 \).
- The perfect cubes are calculated as follows:
- \( 1^3 = 1 \)
- \( 2^3 = 8 \)
- \( 3^3 = 27 \)
- \( 4^3 = 64 \)
- \( 5^3 = 125 \)
- \( 6^3 = 216 \)
- The next cube, \( 7^3 = 343 \), exceeds 250, so we stop here.

2. **List the Perfect Cubes**:
The perfect cubes in \( A \) are:
- \( 1, 8, 27, 64, 125, 216 \)

3. **Form the Relation R**:
Now we can form the relation \( R \) as a subset of \( A \times A \):
- The pairs will be:
- \( (1, 1) \) since \( 1 \) is \( 1^3 \)
- \( (8, 2) \) since \( 8 \) is \( 2^3 \)
- \( (27, 3) \) since \( 27 \) is \( 3^3 \)
- \( (64, 4) \) since \( 64 \) is \( 4^3 \)
- \( (125, 5) \) since \( 125 \) is \( 5^3 \)
- \( (216, 6) \) since \( 216 \) is \( 6^3 \)

Therefore, the relation \( R \) can be expressed as:
\[ R = \{(1, 1), (8, 2), (27, 3), (64, 4), (125, 5), (216, 6)\} \]

4. **Determine the Domain of R**:
The domain of \( R \) consists of the first elements of each ordered pair:
\[ \text{Domain} = \{1, 8, 27, 64, 125, 216\} \]

5. **Determine the Range of R**:
The range of \( R \) consists of the second elements of each ordered pair:
\[ \text{Range} = \{1, 2, 3, 4, 5, 6\} \]

### Final Results:
- The relation \( R \) as a subset of \( A \times A \) is:
\[ R = \{(1, 1), (8, 2), (27, 3), (64, 4), (125, 5), (216, 6)\} \]
- The domain of \( R \) is:
\[ \{1, 8, 27, 64, 125, 216\} \]
- The range of \( R \) is:
\[ \{1, 2, 3, 4, 5, 6\} \]