Question

Let A=\left\\{ 1,2,3,4,5,6,7,8,9 \right\\} and R be a relation in $A\times A$, defined by $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$ for all $\left( a,b \right)$ and $\left( c,d \right)\in A\times A$. Prove that R is an equivalence relation. Also obtain the equivalence class determined by (2, 5).

Answer 1

Hint: To solve this question, we should know that an equivalent relation is nothing but a relation which satisfies reflexive, symmetric, and transitive relation property. So, to prove the given relation, we will prove it to be reflexive, symmetric, and transitive relation. Also, we will find the equivalence of (2, 5) by considering all the possible cases.

Complete step-by-step answer:
In this we have been asked to prove that relation R on $A\times A$, defined by $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$ is an equivalence relation where A=\left\\{ 1,2,3,4,5,6,7,8,9 \right\\}. So, to prove this, we will prove that the given relation R is reflexive, symmetric, and transitive relation as we know that any relation is an equivalent relation only when it is reflexive, symmetric, and transitive relation.
Now, let us first go with reflexive relation. 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 given relation a + d = b + c, we will get the LHS and the RHS as a + b and b + a respectively and by commutative property 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 $aRb$ then it satisfies $bRa$ for a symmetric relation. Here we have been given the relation R, $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$. 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 have 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 a symmetric relation.
Now, we will go with the transitive relation, which states that if $aRb$ and $bRc$, then $aRc$. So, to check if 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 check 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,
d - c = b - a ………… (i)
We also know that if c + f = d + e, then 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 + c, 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 equivalent relation.
Now, we have also been asked to find the equivalence class determined by (2, 5). So, let us consider (a, b) as the equivalence of (2, 5). So, according to the relation $\left( a,b \right)R\left( c,d \right)\Leftrightarrow a+d=b+c$, we can write, $\left( 2,5 \right)R\left( a,b \right)\Leftrightarrow 2+b=5+a\Rightarrow b-a=5-2\Rightarrow b-a=3$. So, we have to find all possible values of a and b from A=\left\\{ 1,2,3,4,5,6,7,8,9 \right\\} such that there is a difference of 3. So, we can say that (a, b) can possibly be, \left\\{ \left( 1,4 \right),\left( 2,5 \right),\left( 3,6 \right),\left( 4,7 \right),\left( 5,8 \right),\left( 6,9 \right) \right\\}.
Hence, the equivalence class determined by (2, 5) is \left\\{ \left( 1,4 \right),\left( 2,5 \right),\left( 3,6 \right),\left( 4,7 \right),\left( 5,8 \right),\left( 6,9 \right) \right\\}.

Note: In this question, we need to remember that for a reflexive relation, we have to always consider (a, b)R(a, b). At times we consider it as (a, a)R(b, b) which will result in the correct answer for wrong terms. Also, we can think of proving this question by taking values of a, b, c and d, which will give a correct answer, but it will be lengthy and time confusing.