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Question: The relation R in N x N such that \(\left( {a,b} \right)R\left( {c,d} \right) \Leftrightarrow a + d ...

The relation R in N x N such that (a,b)R(c,d)a+d=b+c\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

Explanation

Solution

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(a,b)R(c,d)a+d=b+c\left( {a,b} \right)R\left( {c,d} \right) \Leftrightarrow a + d = b + c

Above equation can be written as
c+b=d+a(c,d)R(a,b)c + b = d + a \Rightarrow (c,d)R(a,b)
Therefore R is symmetric.

And a+a=a+a(a,a)R(a,a)a + a = a + a \Rightarrow (a,a)R(a,a)
Therefore R is reflexive.

Now let (a,b)R(c,d)\left( {a,b} \right)R\left( {c,d} \right)and (c,d)R(e,f)\left( {c,d} \right)R\left( {e,f} \right)
a+d=b+c and c+f=d+e\Rightarrow a + d = b + c{\text{ and }}c + f = d + e
Add these two equations
a+d+c+f=b+c+d+e a+f=b+e(a,b)R(e,f)  \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.