Question

Let R be a relation of the set of integers given by aRb such that $a = 2^k.b$ for some integer k. Then R is-
A. An equivalence relation
B. Reflexive but not symmetric
C. Reflexive and transitive but not symmetric
D. Reflexive and symmetric but not transitive

Answer 1

Hint: Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive.Using these definitions we will check the relation R.

Complete step-by-step answer:
It is given that aRb is defined such that $a = 2^k.b$. We will check if it is reflexive, symmetric or transitive or not.

For a reflexive relation, aRa should exist. That is, $a = 2^k.a$.
On solving this equation-
$2^k = 1$
k = 0
A value of k exists, hence the given relation is reflexive.

For a symmetric relation, both aRb and bRa should exist. That is if $a = 2^k.b$ exists then
$b = 2^k.a$ should also exist.
Let $a = 2^k.b$ exist then-
$b = 2^-k.a$
$b = 2^m.a$ where m = -k.
If k is an integer, then -k is also an integer. Hence, the given relation is symmetric.

For a transitive relation, if aRb and bRc exist, aRc should also exist. That is, if $a = 2^k.b$ exists and $b = 2^k.c$ also exists, then $a = 2^k.c$ should exist.
Let $a = 2^k.b$ and $b = 2^m.c$ exist where k, m are integers. On multiplying the two equations we get-
$ab = 2^kb.2^mc$
$a = 2^{(k+m)}.c$
If k and m are integers, then their sum is also an integer. Hence, $a = 2^{(k+m)}.c$ also exists in R. The given relation is transitive.

The relation R is symmetric, reflexive and transitive. Hence, it is an equivalence relation. The correct option is A.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified.