Question

A relation on the set A = {x : |x| < 3, x ∈ Z}, where Z is the set of integers is defined by R = {(x, y) : y = |x|, x ≠ – 1}. Then the number of elements in the power set of R is:
(a) 32
(b) 16
(c) 8
(d) 64

Answer 1

Consider set $A = \{x:\ |x|<3,\ x \in \mathbb{Z}\}$.

All elements of the set A, defined by x, are also contained in Z, the set of integers. It's also given that the absolute value of these elements are strictly less than 3.

$|x|<3\ \ \Longrightarrow\ \ |x| \in \{0,\ 1,\ 2\} \subset A$

Hence possibly this set in roster form will be,

$A = \{-2,\ -1,\ 0,\ 1,\ 2\}$

A relation R on the set A where  $R \subseteq (A \times A)$  is defined in set builder notation as  $R=\{(x,\ y):y=|x|,\ x\neq -1\}$.

Thus $xRy$ is possible if y is the absolute value of x and x is not equal to -1.

And also,  $(x,\ y) \in (A \times A)$

$|x|=y\in \{0,\ 1,\ 2\}\\ \\ \therefore\ x\in \{-2,\ 0,\ 1,\ 2\}\ \ \ but\ \ \ x\neq -1$

Hence possibly R in roster form will be,

$R = \{(-2,\ 2),\ (0,\ 0),\ (1,\ 1),\ (2,\ 2)\}$

Here we get that  $\mid R \mid\ =4$.

Thus,

$\mid P(R) \mid =2^{\mid R \mid}=2^4=\bold{16}$

Hence the answer is (b) 16.

Answer 2

please see the attachment