Question

If $R$ is a relation defined on the set of natural numbers $N$ such that $(a,b)R(c,d)$ if and only if $a+d=b+c$, then $R$ is

• A. Symmetric and transitive but not reflexive
• B. Reflexive and transitive but not symmetric
• C. Reflexive and symmetric but not transitive
• D. An equivalence relation

Answer 1

The correct option is D

An equivalence relation

Explanation of the correct Option:

The correct option is D : An equivalence relation

To show the equivalence relation we need to show that the relation is reflexive , symmetric and transitive

Reflexive: Let $(a,b)\in N\times N$. Then
$\Rightarrow a+b=b+a$ (Commutative law of Addition)
$\Rightarrow (a,b)R(a,b)$
$\Rightarrow R$is reflexive.

Symmetric: Let $(a,b),(c,d)\in N\times N$ such that
$(a,b)R(c,d)$. Then
$(a,b)R(c,d)$

$\Rightarrow a+d=b+c$

$\Rightarrow b+c=a+d$
$\Rightarrow c+b=a+d$ (By commutativity of addition on $N$)
$\Rightarrow (c,d)R(a,b)$
$\therefore R$ is symmetric.

Transitive : Let $(a,b),(c,d),(e,f)\in N\times N$ such that
$(a,b)R(c,d)$ and $(c,d)R(e,f)$. Then,
$\left.\begin{array}{r}(a,b)R(c,d)\Rightarrow a+d=b+c\\ (c,d)R(e,f)\Rightarrow c+f=d+e\end{array}\right\}$
$\Rightarrow (a+d)+(c+f)=(b+c)+(d+e)$
$\Rightarrow a+f=b+e$
$\Rightarrow (a,b)R(e,f)$
∴ $(a,b)R(c,d)$ and $(c,d)R(e,f)$ $\Rightarrow (a,b)R(e,f)$ on $N\times N$ so $R$ is transitive.

Hence $R$ is an equivalence relation on $N\times N$.

Therefore ,correct option is (D) An Equivalence Relation