Question

Let P= $x\ne y$$\left\{ \left( x,y \right)|{{x}^{2}}+{{y}^{2}}=1,x,y\in R \right\}$then P is
(a)Reflexive
(b)Symmetric
( c)Transitive
(d)Antisymmetric

Answer 1

Hint: This question is based on relations. A relation refers to a collection of inputs and outputs that are associated with one another in some way. To solve this question, we put the conditions of reflexive, symmetric, transitive and antisymmetric. Then we check which condition satisfied the given function to get our desirable answer.

Complete step by step solution: 
We have given the question ${{x}^{2}}+{{y}^{2}}=1$
We have to check whether the given equation is reflexive, symmetric , transitive and antisymmetric.
We recall all the types of relations:-
R is reflexive if for all (x,x) $\in $R for every $x\in X$.
 R is symmetric if for all $(x,y)\in R$then $(y,x)\in R$
R is transitive if for all $(x,y)\in R$& $(y,x)\in R$then $(x,z)\in R$
R is antisymmetric if if $(x,y)\notin R$ or $(y,x)\notin R$whenever $x\ne y$
First we check whether P is reflexive.
$x\in R$
$\left( x,x \right)\in R$
Given question is ${{x}^{2}}+{{y}^{2}}=1$
That is ${{x}^{2}}+{{x}^{2}}=1$
$2{{x}^{2}}=1$
Hence, P is not reflexive.
Now we check whether P is symmetric.
As $(x,y)\in R$, $(y,x)\in R$
Given question is ${{x}^{2}}+{{y}^{2}}=1$
Then ${{y}^{2}}+{{x}^{2}}=1$
Hence, P is symmetric
Now we check whether P is transitive.
As $(x,y)\in R$, $(y,x)\in R$
Then $(x,z)\in R$
Given question is ${{x}^{2}}+{{y}^{2}}=1$
Then ${{y}^{2}}+{{z}^{2}}=1$
Hence, P is not transitive.
Now we check whether P is antitransitive.
For this given $(x,y)\in R$
But $(y,x)\notin R$unless x = y
Hence, P is not antisymmetric.
Option (B) is correct.

Note: To solve this question, students must know the conditions to check whether functions are reflexive, symmetric, transitive or antitransitive. If they don’t know the conditions, they will not be able to solve the question. Students must have the proper knowledge of the types of relations.