Question

The relation defined on set A=\left\\{ x:\left| x \right|<3,x\in \mathbb{Z} \right\\} by R=\left\\{ \left( x,y \right):y=\left| x \right| \right\\} is
A. \left\\{ \left( -2,2 \right),\left( -1,1 \right),\left( 0,0 \right),\left( 1,1 \right),\left( 2,2 \right) \right\\}
B. \left\\{ \left( -2,2 \right),\left( -1,1 \right),\left( 0,0 \right),\left( 1,-2 \right),\left( 2,-1 \right),\left( 2,-2 \right) \right\\}
C. \left\\{ \left( 0,0 \right),\left( 1,1 \right),\left( 2,2 \right) \right\\}
D. None of these

Answer 1

We first need to define the notion of modulus function. Then we try to form the elements of the set A=\left\\{ x:\left| x \right|<3,x\in \mathbb{Z} \right\\}. From the given relation of R=\left\\{ \left( x,y \right):y=\left| x \right| \right\\}, we find the possible dual form of the relation that satisfies the result.

Complete step by step solution:
Modulus function $f\left( x \right)=\left| x \right|$ works as the distance of the number from 0. The number can be both positive and negative but the distance of that number will always be positive.
In mathematical notation we express it with modulus value. Let a number be x whose sign is not mentioned. The absolute value of that number will be $\left| x \right|$. We can say $\left| x \right|\ge 0$.
We can express the function $f\left( x \right)=\left| x \right|$ as f\left( x \right)=\left\\{ \begin{aligned} & x; \left( x\ge 0 \right) \\\ & -x; \left( x<0 \right) \\\ \end{aligned} \right..
Therefore, the set A=\left\\{ x:\left| x \right|<3,x\in \mathbb{Z} \right\\} becomes A=\left\\{ -2,-1,0,1,2 \right\\}.
We have been given relation R=\left\\{ \left( x,y \right):y=\left| x \right| \right\\} on set A=\left\\{ -2,-1,0,1,2 \right\\}.
We know that $f\left( x \right)=\left| x \right|=\pm x$ depending on the value of the number x.
For the relation R=\left\\{ \left( x,y \right):y=\left| x \right| \right\\}, we have $y=\pm x$.
Therefore, we can have the value of y as a positive number which are 0, 1, 2.
For these values of y, we can have values of x as $y=\pm x=0,\pm 1,\pm 2$.
With the given individual values of x and y, we can form the relation as \left\\{ \left( -2,2 \right),\left( -1,1 \right),\left( 0,0 \right),\left( 1,1 \right),\left( 2,2 \right) \right\\}.
Therefore, the correct option is A.

Note: We need to be careful about the condition $\left| x \right|<3$. The equality is not provided and that’s why we can’t use the boundary points. Also, we have to remember that modulus distance can never be negative.