Solveeit Logo

Question

Question: A relation on the set \[A = [x:\left| x \right| < 3,x \in Z]\] where \[Z\], the set of integers, is ...

A relation on the set A=[x:x<3,xZ]A = [x:\left| x \right| < 3,x \in Z] where ZZ, the set of integers, is defined by R=[(x,y):y=x,x1]R = [(x,y):y = \left| x \right|,x \ne - 1]. Then the number of elements in the power set of R is:
A. 3232
B.1616
C.88
D.6464

Explanation

Solution

We find the elements of set A using the general form of elements given in the set and then define the relation R. We can write the elements of set R as the elements from A X A and exclude the value -1 from A while writing the relation R. In the end we calculate the number of elements of R which we substitute in the formula of number of elements in the power set to obtain our answer.

Complete step-by-step answer:
We have the set A=[x:x<3,xZ]A = [x:\left| x \right| < 3,x \in Z]
We see all elements of the set A are such that the modulus of the element is less than 3 and the elements are from the set of integers. We can write set A in form of its elements as
A=[2,1,0,1,2]A = [ - 2, - 1,0,1,2]
Now we have a relation R defined on the set A is given by R=[(x,y):y=x,x1]R = [(x,y):y = \left| x \right|,x \ne - 1]
Where the elements of R are an ordered pair (x,y)(x,y) where x belongs to set A and y is a function of x.
So, we can write RA×AR \subseteq A \times A
Now, from the condition in the relation R, x1x \ne - 1, so we remove the element x=1x = - 1 from the set A and then write A×AA \times A
Now the set becomes A=[2,0,1,2]A = [ - 2,0,1,2]
So, x belongs to the set [2,0,1,2][ - 2,0,1,2]
The elements of R are given by R=[(x,y):y=x,x1]R = [(x,y):y = \left| x \right|,x \ne - 1]
So, R=[(2,2),(0,0),(1,1),(2,2)]R = [( - 2,2),(0,0),(1,1),(2,2)]
Number of elements of R is 4,
R=4\left| R \right| = 4
We know that number of elements of power set is given by P(R)=2R\left| {P(R)} \right| = {2^{\left| R \right|}}, where P(R) is the power set of R.
Substitute the value of R=4\left| R \right| = 4 in the formula.

P(R)=24 P(R)=16  \left| {P(R)} \right| = {2^4} \\\ \left| {P(R)} \right| = 16 \\\

So, the correct answer is “Option B”.

Note: Students many times get confused while writing the relation R because they think the values inside the modulus are positive so we will take pairs of only positive values but that is wrong because for ordered pair (x,y)(x,y)we are taking the value x from integers in the set A.