Question

Which of the following is not an equivalence relation on Z?
(a)$\text{aRb }\Leftrightarrow \text{ a}+\text{b}$ is an even integer
(b)$\text{aRb }\Leftrightarrow \text{ a}-\text{b}$ is an even integer
(c)$\text{aRb }\Leftrightarrow \text{ a}<\text{b}$
(d)$\text{aRb }\Leftrightarrow \text{ a}=\text{b}$

Answer 1

Hint: For solving this problem, we consider all options individually. An equivalence set is defined as a set which is reflexive, transitive and symmetric simultaneously. If any of the options fails to satisfy the above condition, then it is called not equivalence relation on Z.

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, $\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
Considering option (a), the set is defined as $\text{aRb }\Leftrightarrow \text{ a}+\text{b}$ is an even integer. Putting (a, a) in the relation, we get
a + a = 2a which is an even integer because a is an even integer. So, it is reflexive.
Putting (a, b) and (b, a) in the relation, we get
a + b is an even integer and b + a is also an even integer. So, it is symmetric.
Putting (a, b) and (b, c) in the relation, we get
a + b is an even integer and b + c is also even an integer. The sum of two even integers is even integers if the number belongs to a family of integer, whole or natural number. So, a and c are integers which can be both even or both odd and thus a + c is also an even integer. So, it is transitive. Rejecting option (a) because it is equivalence.

Considering option (b), the set is defined as $\text{aRb }\Leftrightarrow \text{ a}-\text{b}$ is an even integer. Putting (a, a) in the relation, we get
a - a = 0 which is an even integer because a is an even integer. So, it is reflexive.
Putting (a, b) and (b, a) in the relation, we get
a - b is an even integer and b - a is also an even integer which is negative of the previous one. So, it is symmetric.
Putting (a, b) and (b, c) in the relation, we get
a - b is an even integer and b - c is also even an integer. The sum of two quantities yields a - c is also an even integer which is (a, c). So, it is transitive. Rejecting option (b) because it is equivalence.

Considering option (c), the set is defined as $\text{aRb }\Leftrightarrow \text{ a}<\text{b}$. Putting (a, a) in the relation, we get
a < a, which is not true as a and a must be equal. So, it is not reflexive and thus not equivalence.
Therefore, option (c) is correct.

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. By directly considering the option (c) due to inequality, we can tick mark it as inequality fails for symmetric and reflexive conditions.