Question

If $R$ is a relation on the set of integers given by $aRb$ such that $a=2^k\cdot b$ 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

The correct option is A

an equivalence relation

Explanation for the correct option:

Step 1: Check for reflexive relation.

A relation is reflexive if $aRa$ exists.

A relation is symmetric, if both $aRb$ and $bRa$ exists.

A relation is transitive, if $aRb$ and $bRc$ exists implies that $aRc$ also exists.

Given $aRb$ such that $a=2^k\cdot b$ for some integer $k$.

In a reflexive relation, $aRa$ exists, substitute $b=a$, which gives $a=2^k\cdot a$

On solving

$\begin{array}{rcl}1& =& 2^k\\ & \Rightarrow & k=0\end{array}$

Thus, a value of $k$ exists,

Therefore, the given relation is reflexive.

Step 2: Check for symmetric relation.

Assume $aRb$ exists which implies $a=2^k\cdot b$ and $bRa$ also exists which implies $b=2^k\cdot a$.

Let

$\begin{array}{rcl}a& =& 2^k\cdot b\\ & \Rightarrow & b=2^{-k}\cdot a\\ & \Rightarrow & b=2^n\cdot a\end{array}$

where, $n=-k$

here, $k$ is an integer, then $n=-k$ is also an integer.

Therefore, the relation is symmetric.

Step 3: Check for transitive relation.

Assume $aRb$ exists which implies and $bRc$ also exists which implies $b=2^k\cdot c$.

Let $a=2^k\cdot b$ and $b=2^n\cdot c$ exists where $k,n$ are integers.

Multiply both the equation

$\begin{array}{rcl}ab& =& 2^k\cdot b\times 2^n\cdot c\\ & \Rightarrow & a=2^{k+n}\cdot c\end{array}$

Here $k,n$ are integers, then its sum $k+n$ is also an integer.

Therefore, $\begin{array}{rcl}a& =& 2^{k+n}\cdot c\end{array}$ also exist that is $aRc$ also exists.

Hence, the relation is transitive.

If a relation is reflexive, symmetric, and transitive, it is said to be an equivalence relation.

Hence, the correct option is (A).