Question

Let R be the real line consider the following subsets of the plane $R\times R$$S=\left\{ \left( x,y \right)/y=x+1\ \text{and}\ 0A. T is an equivalence relation on R but S is not.
B. Neither S nor T is an equivalence relation on R.
C. Both S and T are equivalence relations on R.
D. S is an equivalence relation on R but T is not.

Answer 1

Hint: We will be using the concepts of functions and relations to solve the problem. We will be using the definitions of reflexive relation, symmetric relations and transitive relations to verify whether the relation T and S are equivalence or not by checking each relation for reflexive, symmetric and transitive.

Complete step-by-step answer:

Now, we have been given a relation and we have to find whether the relation is reflexive, symmetric, transitive or a combination of these.
Now, we know that reflexive relations are those in which every element is mapped to itself i.e. $\left( a,a \right)\in R$ while symmetric relations are those for which if a R b then b R a. Also, holds and transitive are those relations in which if a R b and b R c then a R c must be held.
Now, we know different types of relations, we will check the given relation for these.
Now, we have two relation on R defined as,
S=\left\\{ \left( x,y \right)/y=x+1\ \text{and}\ 0< x <2 \right\\},\ T=\left\\{ \left( x,y \right)/x-y\ \text{is an integer} \right\\}
Now, we know that if a relation is reflexive, symmetric and transitive then it is an equivalence relation.
Now, we will first check T for equivalence.
$\left( x,x \right)\in T\ as\ \left( x-x \right)=0$ is always an integer.
Now, for symmetric, we have
$\left( x,y \right)\in T\Rightarrow \left( x-y=I \right)$ an integer.
Now, $\left( y-x \right)=-I$ is also an integer.
Therefore, $\left( x,y \right)\in T\Rightarrow \left( y,x \right)\in T$.
Hence, T is a symmetric relation also.
Now, for transitive we have,
$\begin{aligned} & \left( x,y \right)\in T\Rightarrow x-y={{I}_{1}}\ \ \left( \text{an integer} \right) \\\ & \left( y,z \right)\in T\Rightarrow y-z={{I}_{2}}\ \ \left( \text{an integer} \right) \\\ \end{aligned}$
Now, if we add both of them we have,

& x-y+y-z={{I}_{1}}+{{I}_{2}} \\\ & x-z={{I}_{1}}+{{I}_{2}}\ \ \ \left( \text{an integer} \right) \\\ \end{aligned}$$ Since, the sum of two integers $${{I}_{1}}\ and\ {{I}_{2}}$$ is an integer. Therefore, we have if, $\left( x,y \right)\in T\ and\ \left( y,z \right)\in T\Rightarrow \left( x,z \right)\in T$ Therefore, relation T is equivalence relation. Now, for S we will check first that the relation is reflexive or not. So, $\left( x,x \right)\in T\Rightarrow x=x+1$ $x=x+1$ is not possible for 0 < x < 2\. Hence, S is not reflexive therefore, neither an equivalence relation. Hence, the correct option is (A). Note: To solve these types of questions it is important to note that a R b means that a is related to b by a relation R. It is important to note that we have used a fact that sum of two integers is an integer in relation T also for relation S we have checked first whether the relation is reflexive or not which comes out to be false therefore the relation can’t be reflexive as it is not reflexive and hence we haven’t checked the relation further.