Question

Show that the relation R on $N\times N$, defined by $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$ is an equivalent relation.

Answer 1

Hint: In order to solve this question, we should know that equivalent relation is nothing but a relation which satisfies reflexive relation, symmetric relation and transitive relation property. So, to prove the given relation, we will prove that as reflexive, symmetric, and transitive relation. Here we will consider $\left( a,b \right)R\left( a,b \right)$ for reflexive relation.

Complete step-by-step answer:
In this question, we have been asked to prove that relation R on $N\times N$, defined by $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$ is an equivalent relation. So, to prove this, we will prove that the given relation R is reflexive, symmetric, and transitive relation because we know that any relation is equivalent only when it is reflexive, symmetric, and transitive relation.
So, let us consider the reflexive relation first. Reflexive relation is a relation which is defined for itself. So, for, $\left( a,b \right)R\left( a,b \right)$, we can say that a = a, b = b, c = a and d = b. So, according to the relation, a + d = b + c. e get the LHS and RHS as a +b and b + a respectively and by commutative property of addition, we can say that a + b = b + a.
Hence, we can say that (a, b) is defined for (a, b), that is, relation R is a reflexive relation.
Now, let us consider the symmetric relation. Symmetric relation is a relation which also satisfies the converse relation. For example, if $aRb$ is satisfied, then, $bRa$ should also be satisfied for symmetric relations. Here we have been given a relation R, that is $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$. So, if it is true, then we have to prove that $\left( c,d \right)R\left( a,b \right)$ is also true. Here we will consider a = c, b = d, c = a and d = b in the relation a + d = b + c. So, we get, c + b = d + a, which is the same as a + d = b + c.
Hence, we can say that relation R is symmetric.
Now, let us go with the transitive relation, which states that if $aRb$ and $bRc$ are there, then $aRc$. So, to check whether the relation is transitive or not, we will consider (a, b), (c, d) and (e, f). So, let us consider, $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$ and $\left( c,d \right)R\left( e,f \right)\Leftrightarrow c+f=d+e$ and now we will see whether relation $\left( a,b \right)R\left( e,f \right)$ is satisfied or not. We know that, a + d = b + c, so we can write as,
a - c = b - a ………… (i)
Also, we know that c + f = d + e, and we can write it as,
d - c = f - e ………… (ii)
From equation (i) and (ii), we can state that,
b - a = f - e
And we can write it as a + f = b + e, which states that $\left( a,b \right)R\left( e,f \right)$ is satisfied.
Hence, we can say that relation R is a transitive relation.
Therefore, we have proved that relation R is reflexive, symmetric, and transitive relation. Hence, relation R is an equivalent relation.

Note: In this question, we need to remember that for reflexive relation, we will consider $\left( a,b \right)R\left( a,b \right)$. At times students tend to write $\left( a,b \right)R\left( b,b \right)$ for reflexive relation which will give us correct answers for wrong terms. Also, we have to be very patient and careful while proving the relations.