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
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)
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