# Mark the tick against the correct answer in the following:

Question:

Mark the tick against the correct answer in the following:

Let $R$ be a relation on $N \times N$, defined by $(a, b) R(c, d) \Leftrightarrow a+d=b+c .$ Then, $R$ is

A. reflexive and symmetric but not transitive

B. reflexive and transitive but not symmetric

C. symmetric and transitive but not reflexive

D. an equivalence relation

Solution:

According to the question ,

$R=\{(a, b),(c, d): a+d=b+c\}$

Formula

For a relation $R$ in set $A$

Reflexive

The relation is reflexive if $(a, a) \in R$ for every $a \in A$

Symmetric

The relation is Symmetric if $(a, b) \in R$, then $(b, a) \in R$

Transitive

Relation is Transitive if $(a, b) \in R \&(b, c) \in R$, then $(a, c) \in R$

Equivalence

If the relation is reflexive, symmetric and transitive, it is an equivalence relation.

Check for reflexive

Consider, $(a, b) R(a, b)$

$(a, b) R(a, b) \Leftrightarrow a+b=a+b$

which is always true .

Therefore , R is reflexive ……. (1)

Check for symmetric

$(a, b) R(c, d) \Leftrightarrow a+d=b+c$

(c, d) R $(a, b) \Leftrightarrow c+b=d+a$

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

$(a, b) R(c, d) \Leftrightarrow a+d=b+c$

$(c, d) R(e, f) \Leftrightarrow c+f=d+e$

On adding these both equations we get, $a+f=b+e$

Also,

$(a, b) R(e, f) \Leftrightarrow a+f=b+e$

$\therefore$ It will always be true

Therefore, $\mathrm{R}$ is transitive (3)

Now, according to the equations (1), (2), (3)

Correct option will be (D)