Question

If A = \left\\{ {1,2,3,..........,9} \right\\} and $R$ be the relation in $A \times A$ defined by $\left( {a,b} \right)R\left( {c,d} \right)$. If $a + d = b + c$ for $\left( {a,b} \right),\left( {c,d} \right)$ in $A \times A$. Prove that $R$ is an equivalence relation. Also, obtain the equivalence class $\left[ {\left( {2,5} \right)} \right]$.

Answer 1

Hint: To prove $R$ is an equivalence relation, first we have to prove $R$ is Reflexive, Symmetric and Transitive, then the relation $R$ is said to be an Equivalence relation. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given A = \left\\{ {1,2,3,..........,9} \right\\}
And $R$ is the relation in $A \times A$
Case 1: To prove $R$ is Reflexive
Given $\left( {a,b} \right)R\left( {c,d} \right)$ if $\left( {a,b} \right)\left( {c,d} \right) \in A \in A$
$a + d = b + c$
Consider, $\left( {a,b} \right)R\left( {a,b} \right)$ if $\left( {a,b} \right) \in A \in A$
$a + b = b + a$
Hence, $R$ is Reflexive
Case 2: To prove $R$ is symmetric
Consider $\left( {a,b} \right)R\left( {c,d} \right)$ given by $\left( {a,b} \right)\left( {c,d} \right) \in A \times A$
$a + d = b + c \Rightarrow c + b = a + d$
$\therefore \left( {c,a} \right)R\left( {a,b} \right)$
Hence, $R$ is symmetric
Case 3: To prove $R$ is transitive
Let $\left( {a,b} \right)R\left( {c,d} \right)$ and $\left( {c,d} \right)R\left( {e,f} \right)$
$\left( {a,b} \right),\left( {c,d} \right),\left( {c,d} \right) \in A \times A$
$a + b = b + c$ and $c + f = d + e$
$a + b = b + c$

$$\Rightarrow a - c = b - d..........................\left( 1 \right) \\\ \Rightarrow c + f = d + e..........................\left( 2 \right) \\\$$

Adding equation (1) and (2) we get
$\left( {a,b} \right)R\left( {e,f} \right)$
Hence, $R$ is transitive.
As $R$ is reflexive, symmetric and transitive we can say that $R$ is an equivalence relation.
Now, we select $a$ and $b$ from set A = \left\\{ {1,2,3,..........,9} \right\\} such that $2 + b = 5 + a$
So, $b = a + 3$
Let us consider $\left( {1,4} \right)$
If $\left( {2,5} \right)R\left( {1,4} \right)$ then $2 + 4 = 5 + 1$
Thus, $\left[ {\left( {2,5} \right) = \left( {1,4} \right)\left( {2,5} \right),\left( {3,6} \right)\left( {4,7} \right),\left( {5,8} \right)\left( {6,9} \right)} \right]$ is the equivalent class under relation $R$.

Note: A relation $R$ on a set $A$ is called reflexive if $\left( {a,a} \right) \in R$ for every element $a \in A$. A relation $R$ on a set $A$ is called symmetric if $\left( {b,a} \right) \in R$ whenever $\left( {a,b} \right) \in R$, for all $a,b \in A$. A relation $R$ on a set $A$ is called transitive if whenever $\left( {a,b} \right) \in R$ and $\left( {b,c} \right) \in R$, then $\left( {a,c} \right) \in R$, for all $a,b,c \in A$.