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Question Number 158976 by physicstutes last updated on 11/Nov/21

Answered by Rasheed.Sindhi last updated on 11/Nov/21

A={0,1,2,...,11}; R={(x,y)∈A×A:−x+2y=9} −x+2y=9⇒y=((9+x)/2) y∈W⇒x∈O (b) Dom(A)={1,3,5,7,9,11} Range(A) ={((9+1)/2),((9+3)/2),((9+5)/2),((9+7)/2),((9+9)/2),((9+11)/2)} ={5,6,7,8,9,10} Dom(A)∩Range(A)={5,7,9} (a)R={(1,5),(3,6),(5,7),(7,8),(9,9),(11,10)} (d) (i)R is not reflexive, ∵ (1,1)∈R (ii)R is not symmetric ∵(1,5)∈R but (5,1)∉R (iii)R is not transitive (1,5),(5,7)∈R but (1,7)∉R (c)

$${A}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{11}\right\};\: \\ $$$${R}=\left\{\left({x},{y}\right)\in{A}×{A}:−{x}+\mathrm{2}{y}=\mathrm{9}\right\} \\ $$$$−{x}+\mathrm{2}{y}=\mathrm{9}\Rightarrow{y}=\frac{\mathrm{9}+{x}}{\mathrm{2}} \\ $$$${y}\in\mathbb{W}\Rightarrow{x}\in\mathbb{O} \\ $$$$\left({b}\right) \\ $$$${Dom}\left({A}\right)=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9},\mathrm{11}\right\} \\ $$$${Range}\left({A}\right) \\ $$$$\:\:\:=\left\{\frac{\mathrm{9}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{3}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{5}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{7}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{9}}{\mathrm{2}},\frac{\mathrm{9}+\mathrm{11}}{\mathrm{2}}\right\} \\ $$$$\:\:\:=\left\{\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\} \\ $$$${Dom}\left({A}\right)\cap{Range}\left({A}\right)=\left\{\mathrm{5},\mathrm{7},\mathrm{9}\right\} \\ $$$$\left({a}\right){R}=\left\{\left(\mathrm{1},\mathrm{5}\right),\left(\mathrm{3},\mathrm{6}\right),\left(\mathrm{5},\mathrm{7}\right),\left(\mathrm{7},\mathrm{8}\right),\left(\mathrm{9},\mathrm{9}\right),\left(\mathrm{11},\mathrm{10}\right)\right\} \\ $$$$\left({d}\right)\:\left({i}\right){R}\:{is}\:{not}\:{reflexive}, \\ $$$$\:\:\:\:\:\:\:\:\because\:\left(\mathrm{1},\mathrm{1}\right)\in{R} \\ $$$$\:\:\:\:\:\:\left({ii}\right){R}\:{is}\:{not}\:{symmetric} \\ $$$$\:\:\:\:\:\because\left(\mathrm{1},\mathrm{5}\right)\in{R}\:{but}\:\left(\mathrm{5},\mathrm{1}\right)\notin{R} \\ $$$$\:\:\:\:\:\left({iii}\right){R}\:{is}\:{not}\:{transitive} \\ $$$$\:\:\:\:\left(\mathrm{1},\mathrm{5}\right),\left(\mathrm{5},\mathrm{7}\right)\in{R}\:{but}\:\left(\mathrm{1},\mathrm{7}\right)\notin{R} \\ $$$$\left({c}\right) \\ $$$$ \\ $$

Commented by physicstutes last updated on 14/Nov/21

thanks so much sir

$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$

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