
The relation R in N x N such that $\left( {a,b} \right)R\left( {c,d} \right) \Leftrightarrow a + d = b + c$ is
A. Reflexive but not symmetric
B. Reflexive and transitive but not symmetric
C. An equivalence relation
D. None of these
Answer
232.8k+ views
Hint: Here we will verify whether the given relation is reflexive or symmetric or transitive using their definitions.
Complete step-by-step answer:
It is given that$\left( {a,b} \right)R\left( {c,d} \right) \Leftrightarrow a + d = b + c$
Above equation can be written as
$c + b = d + a \Rightarrow (c,d)R(a,b)$
Therefore R is symmetric.
And $a + a = a + a \Rightarrow (a,a)R(a,a)$
Therefore R is reflexive.
Now let $\left( {a,b} \right)R\left( {c,d} \right)$and $\left( {c,d} \right)R\left( {e,f} \right)$
$ \Rightarrow a + d = b + c{\text{ and }}c + f = d + e$
Add these two equations
$
\Rightarrow a + d + c + f = b + c + d + e \\
\Rightarrow a + f = b + e \Rightarrow \left( {a,b} \right)R\left( {e,f} \right) \\
$
Therefore R is transitive.
Hence R is an equivalence relation.
So, option c is correct.
Note: If R satisfies all three conditions i.e symmetric, reflexive and transitive relation, then it is called an equivalence relation.
Complete step-by-step answer:
It is given that$\left( {a,b} \right)R\left( {c,d} \right) \Leftrightarrow a + d = b + c$
Above equation can be written as
$c + b = d + a \Rightarrow (c,d)R(a,b)$
Therefore R is symmetric.
And $a + a = a + a \Rightarrow (a,a)R(a,a)$
Therefore R is reflexive.
Now let $\left( {a,b} \right)R\left( {c,d} \right)$and $\left( {c,d} \right)R\left( {e,f} \right)$
$ \Rightarrow a + d = b + c{\text{ and }}c + f = d + e$
Add these two equations
$
\Rightarrow a + d + c + f = b + c + d + e \\
\Rightarrow a + f = b + e \Rightarrow \left( {a,b} \right)R\left( {e,f} \right) \\
$
Therefore R is transitive.
Hence R is an equivalence relation.
So, option c is correct.
Note: If R satisfies all three conditions i.e symmetric, reflexive and transitive relation, then it is called an equivalence relation.
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