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 \[a,b,c \in A\].