Let R be a relation defined as `R={(x,y) : y=sqrt((x-1)^2) , xepsilon Z` and `-3<=x<=3}` Relation R is equals to :

Explanation

To solve the problem, we need to find the relation \( R \) defined as \( R = \{(x, y) : y = \sqrt{(x-1)^2}, x \in \mathbb{Z} \text{ and } -3 \leq x \leq 3\} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The equation \( y = \sqrt{(x-1)^2} \) simplifies to \( y = |x - 1| \). This means \( y \) is the absolute value of \( x - 1 \).

2. **Finding Integer Values of x**:
We need to evaluate \( y \) for all integer values of \( x \) in the range from -3 to 3, inclusive. The integers in this range are: -3, -2, -1, 0, 1, 2, 3.

3. **Calculating y for Each x**:
- For \( x = -3 \):
\[ y = |-3 - 1| = |-4| = 4 \]
- For \( x = -2 \):
\[ y = |-2 - 1| = |-3| = 3 \]
- For \( x = -1 \):
\[ y = |-1 - 1| = |-2| = 2 \]
- For \( x = 0 \):
\[ y = |0 - 1| = |-1| = 1 \]
- For \( x = 1 \):
\[ y = |1 - 1| = |0| = 0 \]
- For \( x = 2 \):
\[ y = |2 - 1| = |1| = 1 \]
- For \( x = 3 \):
\[ y = |3 - 1| = |2| = 2 \]

4. **Forming the Relation R**:
Now we can compile the pairs \( (x, y) \) based on our calculations:
- From \( x = -3 \), we have \( (-3, 4) \)
- From \( x = -2 \), we have \( (-2, 3) \)
- From \( x = -1 \), we have \( (-1, 2) \)
- From \( x = 0 \), we have \( (0, 1) \)
- From \( x = 1 \), we have \( (1, 0) \)
- From \( x = 2 \), we have \( (2, 1) \)
- From \( x = 3 \), we have \( (3, 2) \)

Thus, the relation \( R \) is:
\[ R = \{(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)\} \]