let R be a relation on the set of integers given by aRb if a = 2^kb, for some integer k. show that R is an equivalence relation.

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$$\begin{gathered} \mathrm{for} \mathrm{some} \mathrm{k} \mathrm{means} \mathrm{that} \mathrm{we} \mathrm{are} \mathrm{free} \mathrm{to} \mathrm{choose} \mathrm{k} \mathrm{of} \mathrm{our} \mathrm{choice} \\ 1 \mathrm{Reflexive} \\ \mathrm{aRb}\Rightarrow \mathrm{a}=2^{\mathrm{k}}\mathrm{b} \\ \Rightarrow \mathrm{aRa}\Rightarrow \mathrm{a}=2^{\mathrm{k}}\mathrm{a} \\ \mathrm{This} \mathrm{will} \mathrm{be} \mathrm{true} \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} \mathrm{k}=0 \\ \mathrm{Hence} \mathrm{it} \mathrm{will} \mathrm{be} \mathrm{reflexive} \mathrm{when} \mathrm{k}=0 \\ 2 \mathrm{Symmetric} \\ \mathrm{aRb}\Rightarrow \mathrm{a}=2^{\mathrm{k}}\mathrm{b} \\ \Rightarrow \mathrm{bRa}\Rightarrow \mathrm{b}=2^{\mathrm{k}}\mathrm{a} \\ \Rightarrow \mathrm{bRa}\Rightarrow \mathrm{b}\frac{1}{2^{\mathrm{k}}}=\mathrm{a} \\ \Rightarrow \mathrm{bRa}\Rightarrow \mathrm{a}=2^{-\mathrm{k}}\mathrm{b} \\ \mathrm{this} \mathrm{will} \mathrm{be} \mathrm{true} \mathrm{for} -\mathrm{k} \mathrm{i}.\mathrm{e}. -\mathrm{k}\in \mathrm{Z} \\ \mathrm{Hence} \mathrm{R} \mathrm{is} \mathrm{symmetric} \\ 3 \mathrm{Transitive} \\ \mathrm{aRb}\Rightarrow \mathrm{a}=2^{\mathrm{k}}\mathrm{b} \left[\mathrm{which} \mathrm{gives} \mathrm{b}=\frac{\mathrm{a}}{2^{\mathrm{k}}}\right] \\ \mathrm{bRc}\Rightarrow \mathrm{b}=2^{\mathrm{m}}\mathrm{c}, \mathrm{m}\in \mathrm{Z} \\ \Rightarrow \frac{\mathrm{a}}{2^{\mathrm{k}}}=2^{\mathrm{m}}\mathrm{c} \\ \Rightarrow \mathrm{a}=2^{\mathrm{k}}{2}^{\mathrm{m}}\mathrm{c} \\ \Rightarrow \mathrm{a}=2^{\mathrm{k}+\mathrm{m}}\mathrm{c}, \mathrm{where} \mathrm{k}+\mathrm{m}\in \mathrm{Z} \\ \mathrm{Hence} \mathrm{R} \mathrm{is} \mathrm{transitive}. \\ \mathrm{Hence} \mathrm{R} \mathrm{is} \mathrm{an} \mathrm{equivalence} \mathrm{relation}. \end{gathered}$$

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