Let I be the set of positve integers. R is a | vedclass

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Question

$\mathrm{a} \mathrm{R} \mathrm{b} \Leftrightarrow \mathrm{a}=2^{\mathrm{k}} \cdot \mathrm{a}$ it is true for $\mathrm{k}=0$

$	herefore$ reflexive

Vedclass · Accepted Answer

reflexive, transitive but not symmetric

Vedclass · Answer

$\mathrm{a} \mathrm{R} \mathrm{b} \Leftrightarrow \mathrm{a}=2^{\mathrm{k}} \cdot \mathrm{a}$ it is true for $\mathrm{k}=0$

$	herefore$ reflexive

$(2,1) \in \mathrm{R}$ but $(1,2) 
otin \mathrm{R} \Rightarrow$ it is not symmetric

if $a=2^{k_{1}} b$ and $b=2^{k_{2}} c,$ then $a=2^{k_{1}+k_{2}} c$

$\Rightarrow$ it is transistive.

Let $I$ be the set of positve integers. $R$ is a relation on the set $I$ given by $R =\left\{ {\left( {a,b} \right) \in I \times I\,|\,\,{{\log }_2}\left( {\frac{a}{b}} \right)} \right.$ is a non-negative integer$\}$, then $R$ is

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