Question

If R is a relation on the set of integers given by $aRb \Rightarrow a = {2^k}.b$ 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
E). Symmetric and transitive but not reflexive

Answer 1

Here we are asked to find what kind of relation is $R$ on the set of integers. For that, we will check the given relation whether it satisfies the conditions of symmetric, reflexive, and transitive. According to that we can find the answer from the given options. The following are the conditions for each relation:
Let $A$ be any set and $R$ be the relation on the set $A$ then the relation is said to be
Symmetric: if aRb \Rightarrow bRa$$$$\forall a,b \in A
Transitive: if $aRb$and bRc \Rightarrow aRc$$$$\forall a,b,c \in A
Reflexive: if (a,a) \in R$$$$\forall a \in A
Equivalence: if $R$ is symmetric, transitive, and reflexive

Complete step-by-step solution:
It is given that $R$ is a relation on the set of integers and defined as $aRb \Rightarrow a = {2^k}.b$ we aim to find whether what kind of relation is $R$.
To find what kind of relation is $R$ we first need to check whether $R$ satisfies the conditions of the relations symmetric, reflexive, and transitive. We will check it one by one.
Let $A$be a set of integers and $R$ be a relation on the set $A$ and it is defined as aRb \Rightarrow a = {2^k}.b$$$$\forall a,b \in A
Symmetricity:
Consider $a = {2^k}.b$ this implies$b = {2^{ - k}}a$. We can see that
$a = {2^k}.b \Rightarrow aRb$
$b = {2^{ - k}}a \Rightarrow bRa$ where $k, - k,k + m \in \mathbb{Z}$
Thus, $aRb \Rightarrow bRa$
Therefore, $R$ is a symmetric relation.
Transitivity:
If $a = {2^k}.b$ and $b = {2^m}c$
$a = {2^k}.b \Rightarrow b = \dfrac{a}{{{2^k}}}$………$(1)$
$b = {2^m}c \Rightarrow c = \dfrac{b}{{{2^m}}}$
Substituting the equation $(1)$ in the above we get
$c = \dfrac{b}{{{2^m}}} \Rightarrow c = \dfrac{a}{{{2^k}{{.2}^m}}} \Rightarrow \dfrac{a}{{{2^{k + m}}}}$
$\Rightarrow a = {2^{k + m}}c$ where $k,m,k + m \in A$
$\Rightarrow aRc$
Thus, we got that $aRb$ and bRc \Rightarrow aRc$$$$\forall a,b,c \in A
Therefore, $R$ is a transitive relation.
Reflexivity:
We have $R$ since $k \in \mathbb{Z}$ let $k = 0$ then we get
$a = {2^0}.b \Rightarrow a = b$
$\Rightarrow (a,a) \in R$
Thus, $R$ is a reflexive relation.
Therefore, we have got that $R$ is a symmetric, reflexive, and transitive relation which implies that $R$ is an equivalence relation.
Now let us see the options, option (a) an equivalence relation is the correct option as we got $R$ is an equivalence relation from the above solution.
Option (b) reflexive but not symmetric is an incorrect option as we got $R$ is an equivalence relation.
Option (c) reflexive and transitive but not symmetric is an incorrect option as we got $R$is an equivalence relation.
Option (d) reflexive and symmetric but not transitive is an incorrect option as we got $R$ is an equivalence relation.
Option (e) Symmetric and transitive but not reflexive is an incorrect option as we got $R$ is an equivalence relation.
Hence, option (a) an equivalence relation is the correct answer.

Note: In this problem it is necessary to check all the conditions since the condition for a relation to be equivalence is only if it is symmetric, transitive, and reflexive all at a time. There are few more relations are there in set theory other than the ones that are mentioned above they are empty relation, identity relation, inverse relation, and universal relation.