Question

Prove that the relation R in the set of non zero integers I0 defined by aRb if a^b = b^a for every a,b element of I0 is reflexive, symmetric but not transitive.

Answer 1

$$\begin{gathered} \mathrm{For}\mathrm{reflexive}:- \\ \left(\mathrm{a},\mathrm{a}\right)\in \mathrm{R} \\ \Rightarrow {\mathrm{a}}^{\mathrm{a}}={\mathrm{a}}^{\mathrm{a}},\mathrm{true}. \\ \mathrm{Hence}\mathrm{relation}\mathrm{R}\mathrm{is}\mathrm{reflexive}. \\ \mathrm{For}\mathrm{symmetric}:- \\ \left(\mathrm{a},\mathrm{b}\right)\in \mathrm{R} \\ \Rightarrow {\mathrm{a}}^{\mathrm{b}}={\mathrm{b}}^{\mathrm{a}} \\ \Rightarrow {\mathrm{b}}^{\mathrm{a}}={\mathrm{a}}^{\mathrm{b}} \\ \Rightarrow \left(\mathrm{b},\mathrm{a}\right)\in \mathrm{R} \\ \mathrm{Hence}\mathrm{R}\mathrm{is}\mathrm{symmetric}. \\ \mathrm{For}\mathrm{transistive}:- \\ \mathrm{Let}\left(\mathrm{a},\mathrm{b}\right)\in \mathrm{R}\mathrm{and}\left(\mathrm{b},\mathrm{c}\right)\in \mathrm{R} \\ \Rightarrow {\mathrm{a}}^{\mathrm{b}}={\mathrm{b}}^{\mathrm{a}}\mathrm{and}{\mathrm{b}}^{\mathrm{c}}={\mathrm{c}}^{\mathrm{b}} \\ \mathrm{but}{\mathrm{a}}^{\mathrm{c}}\ne {\mathrm{c}}^{\mathrm{a}} \\ \mathrm{so},\left(\mathrm{a},\mathrm{c}\right)\notin \mathrm{R} \\ \mathrm{therefore},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{transistive}. \end{gathered}$$