Question Number 158976 by physicstutes last updated on 11/Nov/21
Answered by Rasheed.Sindhi last updated on 11/Nov/21
$${A}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{11}\right\};\: \\ $$$${R}=\left\{\left({x},{y}\right)\in{A}×{A}:−{x}+\mathrm{2}{y}=\mathrm{9}\right\} \\ $$$$−{x}+\mathrm{2}{y}=\mathrm{9}\Rightarrow{y}=\frac{\mathrm{9}+{x}}{\mathrm{2}} \\ $$$${y}\in\mathbb{W}\Rightarrow{x}\in\mathbb{O} \\ $$$$\left({b}\right) \\ $$$${Dom}\left({A}\right)=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9},\mathrm{11}\right\} \\ $$$${Range}\left({A}\right) \\ $$$$\:\:\:=\left\{\frac{\mathrm{9}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{3}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{5}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{7}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{9}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{11}}{\mathrm{2}}\right\} \\ $$$$\:\:\:=\left\{\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\} \\ $$$${Dom}\left({A}\right)\cap{Range}\left({A}\right)=\left\{\mathrm{5},\mathrm{7},\mathrm{9}\right\} \\ $$$$\left({a}\right){R}=\left\{\left(\mathrm{1},\mathrm{5}\right),\left(\mathrm{3},\mathrm{6}\right),\left(\mathrm{5},\mathrm{7}\right),\left(\mathrm{7},\mathrm{8}\right),\left(\mathrm{9},\mathrm{9}\right),\left(\mathrm{11},\mathrm{10}\right)\right\} \\ $$$$\left({d}\right)\:\left({i}\right){R}\:{is}\:{not}\:{reflexive}, \\ $$$$\:\:\:\:\:\:\:\:\because\:\left(\mathrm{1},\mathrm{1}\right)\in{R} \\ $$$$\:\:\:\:\:\:\left({ii}\right){R}\:{is}\:{not}\:{symmetric} \\ $$$$\:\:\:\:\:\because\left(\mathrm{1},\mathrm{5}\right)\in{R}\:{but}\:\left(\mathrm{5},\mathrm{1}\right)\notin{R} \\ $$$$\:\:\:\:\:\left({iii}\right){R}\:{is}\:{not}\:{transitive} \\ $$$$\:\:\:\:\left(\mathrm{1},\mathrm{5}\right),\left(\mathrm{5},\mathrm{7}\right)\in{R}\:{but}\:\left(\mathrm{1},\mathrm{7}\right)\notin{R} \\ $$$$\left({c}\right) \\ $$$$ \\ $$
Commented by physicstutes last updated on 14/Nov/21
$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$