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Question: Let R be a relation on the set of integers given by \(aRb \Leftrightarrow a = {2^k}b\) for some inte...

Let R be a relation on the set of integers given by aRba=2kbaRb \Leftrightarrow a = {2^k}b for some integer k. Then R is
A) An equivalence relation
B) Reflexive and transitive but not symmetric
C) Reflexive and transitive but not transitive
D) Symmetric and transitive but not Reflexive

Explanation

Solution

A relation on set A is called an equivalence relation if it is reflexive, symmetric, and transitive.
Reflexive relation: A binary relation R over a set X is reflexive if it relates every element of X to itself. Formally this may be written xX:xRx\forall x \in X:xRx
Symmetric relation: It is a type of binary relation. A binary relation R over a set X is symmetric if
a,bX:(aRbbRa)\forall a,b \in X:\left( {aRb \Leftrightarrow bRa} \right)
Transitive relation: A homogeneous relation R on the set X is a transitive relation is, for all a,b,cXa,b,c \in X, if aRb and bRc, then aRc. Or in terms of first-order logic:
a,b,cX:(aRbbRc)aRc\forall a,b,c \in X:\left( {aRb \wedge bRc} \right) \Rightarrow aRc

Complete step-by-step solution:
First, let us check for reflexive relations.
Let us assume any integer a.
We have,
a=2ka\Rightarrow a = {2^k}a
Take the value of k is equal to 0.
Therefore,
a=20a\Rightarrow a = {2^0}a
As we know that the value of 20{2^0} is 1.
a=a\Rightarrow a = a
So, we can say that
(a,a)R\Rightarrow \left( {a,a} \right) \in R
Hence R is reflexive on Z.
Now, let us check for symmetric relations.
Let (a,b)R\left( {a,b} \right) \in R
Then, for some integer k
a=2kba = {2^k}b
Let us divide both sides by 2k{2^k}
So,
a2k=2kb2k\Rightarrow \dfrac{a}{{{2^k}}} = \dfrac{{{2^k}b}}{{{2^k}}}
That is equal to
b=2kab = {2^{ - k}}a
Hence R is symmetric on Z.
Now, let us check for transitive relations.
Let (a,b)R\left( {a,b} \right) \in R and (b,c)R\left( {b,c} \right) \in R.
a=2kba = {2^k}b and b=2mcb = {2^m}c for some integers k and m.
a=2k+mc\Rightarrow a = {2^{k + m}}c
(a,c)R\Rightarrow \left( {a,c} \right) \in R
Hence R is transitive on Z.
Therefore, R is an equivalence relation on Z.

Option A is the correct answer.

Note: The examples of reflexive relations include:
Is equal to
Is a subset of
Divides
Is greater than or equal to
Is less than or equal to
The examples of symmetric relations include:
Is equal to
Is comparable to
... and ... are odd
The examples of symmetric relations include:
Is a subset of
Divides
implies