Question

Which of the following are not equivalence relations on \[I\] ?
A.\[aRb\] if \[a + b\] is an even integer
B.\[aRb\] if \[a - b\] is an even integer
C.\[aRb\] if \[a < b\]
D.\[aRb\] if \[a = b\]

Answer 1

Hint: Here we will first state the definition of equivalence relation. Then we will check each option individually whether the relation is equivalence relation or not. The option which does not satisfy the given condition of equivalence condition, will be our required answer.

Complete step-by-step answer:
Here we have to check which one of the following is an equivalence relation.
We know that a relation on any set is said to be an equivalence relation if and only if the relation is symmetric, reflexive and transitive.
Now, we will consider all the options and check each option individually whether the relation is equivalence relation or not.
A.\[aRb\] if \[a + b\] is an even integer
Let \[aRa\] and \[a\] be an integer here.
Therefore, \[a + a = 2a\], we can see that it is an even integer.
It is satisfying the condition.
Therefore, we can say that the given relation is reflexive.
Now, we will check if the relation is symmetric or not.
Let \[aRb\] if \[a + b\] is an even integer
We can write \[a + b\] as \[b + a\] i.e.
\[ \Rightarrow a + b = b + a\]
Therefore, we get \[bRa\].
Hence, the given relation is symmetric.
Now, we will check if the relation is transitive or not.
Let \[aRb\] and \[bRc\] if \[a + b\] and \[b + c\] is an even integer.
Let the \[a + b = 2k\] and \[b + c = 2s\]
On adding both these, we get
\[ \Rightarrow a + b + b + c = 2k + 2s\]
On further simplification, we get
\[\begin{array}{l} \Rightarrow a + c = 2k + 2s - 2b\\ \Rightarrow a + c = 2\left( {k + s - b} \right)\end{array}\]
We can say that the \[bRc\].
Therefore, the given relation is also a transitive.
As the given relation is symmetric, reflexive and transitive, therefore, the given relation is an equivalence relation.

B.\[aRb\] if \[a - b\] is an even integer
Let \[aRa\] and \[a\] be an integer here.
Therefore, \[a - a = 0\], we can see that it is an even integer.
It is satisfying the condition, therefore, we can say that the given relation is reflexive.
Now, we will check if the relation is symmetric or not.
Let \[aRb\] if \[a - b\] is an even integer.
Let \[a - b = 2k\]
Then
\[\left( {b - a} \right) = - 2k\] is also an even integer.
Therefore, we get \[bRa\].
Hence, the given relation is symmetric.
Now, we will check if the relation is transitive or not.
Let \[aRb\] and \[bRc\] if \[a - b\] and \[b - c\] is an even integer.
Let the \[a - b = 2k\] and \[b - c = 2s\]
On adding both these, we get
\[ \Rightarrow a - b - b + c = 2k - 2s\]
On further simplification, we get
\[\begin{array}{l} \Rightarrow a - c = 2k + 2s\\ \Rightarrow a - c = 2\left( {k + s} \right)\end{array}\]
We can see that \[\left( {a - c} \right)\] is also an even integer.
We can say that the \[bRc\]
Therefore, the given relation is also a transitive.
As the given relation is symmetric, reflexive and transitive, therefore, the given relation is an equivalence relation.

C.\[aRb\] if \[a < b\]
Let \[aRa\] and \[a\] be an integer here.
Therefore, \[a < a\], we can see that it is not possible.
It is not satisfying the condition, therefore, we can say that the given relation is not reflexive.
Now, we will check if the relation is symmetric or not.
Let \[aRb\] if \[a < b\]
We cannot write \[a < b\] as \[b < a\] i.e.
\[ \Rightarrow a < b \ne b < a\]
Hence, the given relation is not symmetric.
Now, we will check if the relation is transitive or not.
Let \[aRb\] and \[bRc\] if \[a < b\] and \[b < c\] is an even integer.
On adding both these, we get
\[ \Rightarrow a + b < b + c\]
On further simplification, we get
\[ \Rightarrow a < c\]
We can say that the \[bRc\]
Therefore, the given relation is also a transitive.
As the given relation is only transitive, therefore, the given relation is not an equivalence relation.

D.\[aRb\] if \[a = b\]
Let \[aRa\] and \[a\] be an integer here.
Therefore, \[a = a\], we can see that it is satisfying the condition.
Therefore, we can say that the given relation is reflexive.
Now, we will check if the relation is symmetric or not.
Let \[aRb\] if \[a = b\]
We can write \[a = b\] as \[b = a\] i.e.
Therefore, \[bRa\]
Hence, the given relation is symmetric.
Now, we will check if the relation is transitive or not.
Let \[aRb\] and \[bRc\] if \[a = b\] and \[b = c\] is an even integer.
On adding both these, we get
\[ \Rightarrow a + b = b + c\]
On further simplification, we get
\[ \Rightarrow a = c\]
We can say that the \[bRc\]
Therefore, the given relation is transitive.
As the given relation is symmetric, reflexive and transitive, therefore, the given relation is an equivalence relation.
We can see that the given relation in option C is not an equivalence relation.
Hence, the correct option is option C.

Note: Here, we need to find the equivalence relation by checking the relation if it is reflexive, symmetric and transitive or not. A relation is said to be reflexive, when every element is mapped to itself. A relation is said to be symmetric, if \[\left( {a,b} \right) \in R\], then \[\left( {b,a} \right) \in R\]. A relation is said to be transitive, \[\left( {x,y} \right) \in R\] and \[\left( {y,z} \right) \in R\], then \[\left( {x,z} \right) \in R\]. Also, we need to find which is not an equivalence relation, so we might make a mistake by writing the option which satisfies equivalence relation as an answer.