Question

Let N denote the set of all natural numbers and R be the relation on $N\times N$ defined by $\left( a,b \right)R\left( c,d \right)$ if $ad\left( b+c \right)=bc\left( a+d \right)$ , then $R$ is:
(A). Symmetric only
(B). Reflexive only
(C). Transitive only
(D). An equivalence relation

Answer 1

Hint: First observe the relation given in question. Find all the properties related to the relation. Try to substitute values or variables to check their properties. Try to use definitions of Reflexive, transitive, symmetric, anti symmetry. First and the foremost thing to do is use the definition of relation and say that the given relation satisfies all the conditions needed for it to be a relation. Next see the range on which relation is defined. While checking properties, substitute a few elements only which satisfy the range given in the question.

Complete step-by-step solution -
Relation:- The relations in maths are also called as binary relation over se x,y is a subset of the Cartesian product of the set x,y that is the relation contains the elements of set XxY. XxY, is a set of ordered pairs consisting of element x in X and y in Y. It encodes the information, $\left( x,y \right)$ will be in relation set if and only if x is related y in the relation property given.
Reflexive relation:- A relation R is said to be reflexive, if a is an element in the range. If the pair $\left( a,a \right)$ belongs to relation then relation is reflexive. In other words, if ‘a’ is related to itself by a given relation then relation is reflexive.
Symmetric Relation:- A relation R is said to symmetric, if a,b are elements in the range. If the pair $\left( a,b \right)$ belongs to relation then $\left( b,a \right)$ must belong to relation. In other words if a is related to b, then b must also be related to a for a relation to be symmetric.
Transitive Relation:- If $\left( a,b \right)$ belongs to relation and $\left( b,c \right)$ also belongs to the relation, then $\left( a,c \right)$ must belong to relation to make it transitive.
Anti symmetric Relation:- If $\left( a,b \right)$ and $\left( b,a \right)$ are given belong to relation, then a must be equal to b to make the relation as anti symmetric. If a relation is reflexive, Transitive and Symmetric, then it is an equivalence relation. The set is the set of all natural numbers. The relation R between $\left( a,b \right);\left( c,d \right)$ is given in the question as $ad\left( b+c \right)=bc\left( a+d \right)$ .
Reflexive:- Use the points $\left( a,b \right);\left( a,b \right)$ substitute $c=a,d=b$ in the relation, By substituting the values we get,
$ab\left( b+a \right)=ab\left( a+b \right)$ which is true.
It is Reflexive
Symmetric:- Give $ad\left( b+c \right)=bc\left( a+d \right)$ Substitute $\left( c,d \right)$ in place of $\left( a,b \right)$ and vice versa $cb\left( d+a \right)=da\left( c+b \right)$ which is true. It is symmetric.
Transitive:- Assume $\left( a,b \right)R\left( c,d \right)$ and $\left( c,d \right)R\left( e,f \right)$
$ad\left( b+c \right)bc = \left( a+d \right)$ and $cf\left( d+e \right)=\left( c+f \right)$
By cross multiplying we get,
$\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{d}$ and $\dfrac{1}{d}+\dfrac{1}{e}=\dfrac{1}{c}+\dfrac{1}{f}$ .
By adding both equations, we get, $\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}=\dfrac{1}{a}+\dfrac{1}{d}+\dfrac{1}{c}+\dfrac{1}{f}$
$\left( a,b \right)R\left( e,f \right)$ So, it is transitive.
Therefore, it is equivalence and option (D) is correct.

Note: The idea of turning it into a sum of its reciprocals is crucial to prove that it is transitive. So, do it carefully. For solving this type of problem we need to remember the definitions of reflexive, symmetric and transitive relation.