Let A={1,2,3,....9} and R be relation in $A\times A$ defined by (a, b) R (c, d) if a + d = b + c for (a, b),(c, d) in $A\times A$. Prove that R is an equivalence relation. Also obtain the equivalence class [(2,5)].
Answer
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Hint: For solving this problem, first we have to prove that the relation is reflexive, symmetric and transitive by using appropriate conditions. Once these three relations are established, then R becomes an equivalence relation on A. After that we can use the given relation to obtain the equivalence class.
Complete step-by-step answer:
The conditions for a set to be reflexive, transitive and symmetric are:
1) For a relation to be reflexive, $\left( a,a \right)\in R$.
2) For a relation to be symmetric, $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$.
3) For a relation to be transitive, $\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
4) For a relation to be equivalence, when it is reflexive, symmetric and transitive.
Given: A = {1, 2, 3, ……9}
We have to prove R belongs to the equivalence class.
Now, (a, b) R (c, d) if $\left( a,\text{ }b \right)\text{ }\left( c,\text{ }d \right)\in A$
$a+d=b+c$
Consider, (a, b) R (a, b)
$\therefore a+b=b+a$
Hence, R is reflexive.
Consider (a, b) R (c, d) given by (a, b) (c, d)$\in A\times A$
Now, by using a + d = b + c => c + b = d + a, we get
$\Rightarrow \left( c,d \right)R\left( a,b \right)=\left( a,b \right)R\left( c,d \right)$
Hence, R is symmetric.
Let (a, b), (c, d), (e, f),$\in A\times A$
For (a, b) R (c, d) and (c, d) R (e, f), we have
a + d = b + c and c + f = d + e
$\begin{align}
& \Rightarrow a-c=b-d...\left( 1 \right) \\
& \Rightarrow c+f=d+e...\left( 2 \right) \\
\end{align}$
Adding equation 1 and 2 we get:
a – c + c + f = b - d + d + e
a + f = b + e
(a, b) R (e, f)
Hence, R is transitive.
Since, R is reflexive, transitive and symmetric. Therefore, R is an equivalence relation.
For, equivalence class of (2, 5): (2, 5) R (c, d)
a and b such that 2 + d = 5 + c
so, d = c + 3
consider (1, 4) to make possible pairs of (c, d)
(2, 5) R (1, 4) $\Rightarrow 2+4=5+1$
$\left[ \left( 2,5 \right)=\left( 1,4 \right),\left( 2,5 \right),\left( 3,6 \right),\left( 4,7 \right),\left( 5,8 \right),\left( 6,9 \right) \right]$ is the equivalent class under relation R.
Note: The knowledge of equivalence of a relation is must for solving this problem. Students must remember all the necessary conditions for proving a set a reflexive, symmetric and transitive. All the possible pairs from the given number must be obtained for equivalence class.
Complete step-by-step answer:
The conditions for a set to be reflexive, transitive and symmetric are:
1) For a relation to be reflexive, $\left( a,a \right)\in R$.
2) For a relation to be symmetric, $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$.
3) For a relation to be transitive, $\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
4) For a relation to be equivalence, when it is reflexive, symmetric and transitive.
Given: A = {1, 2, 3, ……9}
We have to prove R belongs to the equivalence class.
Now, (a, b) R (c, d) if $\left( a,\text{ }b \right)\text{ }\left( c,\text{ }d \right)\in A$
$a+d=b+c$
Consider, (a, b) R (a, b)
$\therefore a+b=b+a$
Hence, R is reflexive.
Consider (a, b) R (c, d) given by (a, b) (c, d)$\in A\times A$
Now, by using a + d = b + c => c + b = d + a, we get
$\Rightarrow \left( c,d \right)R\left( a,b \right)=\left( a,b \right)R\left( c,d \right)$
Hence, R is symmetric.
Let (a, b), (c, d), (e, f),$\in A\times A$
For (a, b) R (c, d) and (c, d) R (e, f), we have
a + d = b + c and c + f = d + e
$\begin{align}
& \Rightarrow a-c=b-d...\left( 1 \right) \\
& \Rightarrow c+f=d+e...\left( 2 \right) \\
\end{align}$
Adding equation 1 and 2 we get:
a – c + c + f = b - d + d + e
a + f = b + e
(a, b) R (e, f)
Hence, R is transitive.
Since, R is reflexive, transitive and symmetric. Therefore, R is an equivalence relation.
For, equivalence class of (2, 5): (2, 5) R (c, d)
a and b such that 2 + d = 5 + c
so, d = c + 3
consider (1, 4) to make possible pairs of (c, d)
(2, 5) R (1, 4) $\Rightarrow 2+4=5+1$
$\left[ \left( 2,5 \right)=\left( 1,4 \right),\left( 2,5 \right),\left( 3,6 \right),\left( 4,7 \right),\left( 5,8 \right),\left( 6,9 \right) \right]$ is the equivalent class under relation R.
Note: The knowledge of equivalence of a relation is must for solving this problem. Students must remember all the necessary conditions for proving a set a reflexive, symmetric and transitive. All the possible pairs from the given number must be obtained for equivalence class.
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