MCQ 11 Mark
If $R$ is the relation in the set $A=\{1,2,3,4,5\}$ given by $R=\{\{a, b):|a-b|$ is even $\}$,
Assertion (A): R is an equivalence relation.
Reason (R): All elements of $\{1,3,5\}$ are related to all elements of $\{2,4\}$.
Assertion (A): R is an equivalence relation.
Reason (R): All elements of $\{1,3,5\}$ are related to all elements of $\{2,4\}$.
- ABoth A and R are true and R is the correct explanation of A.
- BBoth A and R are true but R is not the correct explanation of A.
- ✓A is true but R is false.
- DA is false but R is true.
Answer
View full question & answer→Correct option: C.
A is true but R is false.
(c) A is true but R is false.
Explanation: Assertion: Given that, A = {1, 2, 3, 4, 5}
$R=\{(a, b):|a-b|$ is even $\}$
Let $a \in A \Rightarrow| a - a |=0$ (which is even), $\forall a$
So, R is reflexive.
Let $( a , b ) \in R \Rightarrow| a - b |$ is even.
$\Rightarrow| a - b |=|-( b - a )|=| b - a |$, therefore $| b - a |$ is also even.
$\Rightarrow(b, a) \in R$. So, R is symmetric.
Now, let $(a, b) \in R$ and $(b, c) \in R$.
$\Rightarrow|a-b|$ is even and $| b - c |$ is even.
$\Rightarrow(a-b)$ is even and $( b - c )$ is even.
$\Rightarrow(a-c)=(a-b)+(b-c)$ is even
$[\because$ sum of two even integers is even $]$
$\Rightarrow|a-c|$ is even $\Rightarrow(a, c) \in R$.
So, R is transitive.
Hence, R is an equivalence relation.
Reason: Also, no element of the $\{1,3,5\}$ can be related to any element of $\{2,4\}$, as all elements of $\{1,3,5\}$ are odd and all elements of $\{2,4\}$ are even.
So, the modulus of the difference between the two elements (from each of these two subsets) will not be even. Hence Reason is not correct.
Explanation: Assertion: Given that, A = {1, 2, 3, 4, 5}
$R=\{(a, b):|a-b|$ is even $\}$
Let $a \in A \Rightarrow| a - a |=0$ (which is even), $\forall a$
So, R is reflexive.
Let $( a , b ) \in R \Rightarrow| a - b |$ is even.
$\Rightarrow| a - b |=|-( b - a )|=| b - a |$, therefore $| b - a |$ is also even.
$\Rightarrow(b, a) \in R$. So, R is symmetric.
Now, let $(a, b) \in R$ and $(b, c) \in R$.
$\Rightarrow|a-b|$ is even and $| b - c |$ is even.
$\Rightarrow(a-b)$ is even and $( b - c )$ is even.
$\Rightarrow(a-c)=(a-b)+(b-c)$ is even
$[\because$ sum of two even integers is even $]$
$\Rightarrow|a-c|$ is even $\Rightarrow(a, c) \in R$.
So, R is transitive.
Hence, R is an equivalence relation.
Reason: Also, no element of the $\{1,3,5\}$ can be related to any element of $\{2,4\}$, as all elements of $\{1,3,5\}$ are odd and all elements of $\{2,4\}$ are even.
So, the modulus of the difference between the two elements (from each of these two subsets) will not be even. Hence Reason is not correct.