A relation on the set A = {x : |x| < 3, x $$\in$$ Z}, where Z is the set of integers is defined by R = {(x, y) : y = |x|, x $$\neq$$ $$-1$$}. Then the number of elements in the power set of R is:

First, we need to understand the set A. The set A is defined as A = {x : |x| < 3, x ∈ Z}, where Z is the set of integers. This means A consists of all integers x such that the absolute value of x is less than 3. The condition |x| < 3 translates to -3 < x < 3. Since x must be an integer, we list all integers satisfying this inequality: x = -2, because | -2 | = 2 < 3; x = -1, because | -1 | = 1 < 3; x = 0, because |0| = 0 < 3; x = 1, because |1| = 1 < 3; and x = 2, because |2| = 2 < 3. Note that x = -3 is not included because | -3 | = 3 is not less than 3, and similarly x = 3 is not included. Therefore, A = {-2, -1, 0, 1, 2}.

Next, we define the relation R on the set A. The relation is given as R = {(x, y) : y = |x|, x ≠ -1}. This means R consists of ordered pairs (x, y) where y is the absolute value of x, and x is not equal to -1. Since R is a relation on A, both x and y must be elements of A.

We now list all possible pairs (x, y) that satisfy the condition. The elements of A are x = -2, -1, 0, 1, 2. However, the condition x ≠ -1 excludes x = -1. So we consider the remaining x values: -2, 0, 1, 2.

For each x, we compute y = |x|:

• When x = -2, y = | -2 | = 2, so the pair is (-2, 2).
• When x = 0, y = |0| = 0, so the pair is (0, 0).
• When x = 1, y = |1| = 1, so the pair is (1, 1).
• When x = 2, y = |2| = 2, so the pair is (2, 2).

We must ensure that each y is in A. Checking: y = 2 is in A, y = 0 is in A, y = 1 is in A, and y = 2 is in A. Also, there is no pair for x = -1 because it is excluded. Therefore, the relation R consists of the ordered pairs: (-2, 2), (0, 0), (1, 1), and (2, 2). Thus, R has 4 elements.

Now, we need to find the power set of R, denoted P(R). The power set of a set is the set of all its subsets. If a set has n elements, the number of elements in its power set is $$2^n$$. Here, R has 4 elements, so the number of elements in P(R) is $$2^4$$.

Calculating: $$2^4 = 2 \times 2 \times 2 \times 2 = 16$$. Therefore, the power set of R has 16 elements.

Looking at the options: A. 32, B. 16, C. 8, D. 64. The value 16 corresponds to option B. Hence, the correct answer is Option B.