Question Number 191937 by Mastermind last updated on 04/May/23
$$\mathrm{Check}\:\mathrm{whether}\:\left(\mathrm{Q},\:\centerdot\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{or} \\ $$$$\mathrm{not} \\ $$$$ \\ $$$$\mathrm{Hello}\:\mathrm{bosses}! \\ $$
Answered by deleteduser1 last updated on 04/May/23
$${Let}\:{a},{b}\in{Q}\left({Q}=\frac{{x}}{{y}}\right)\:{where}\:{a}=\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} },{b}=\frac{{x}_{\mathrm{2}} }{{y}_{\mathrm{2}} } \\ $$$${a}\centerdot{e}={e}.{a}={a} \\ $$$$\Rightarrow{e}={a}\centerdot{a}^{−\mathrm{1}} =\mathrm{1} \\ $$$${where}\:{a}\:{unique}\:{a}^{−\mathrm{1}} \in{Q} \\ $$$${There}\:{does}\:{not}\:{exist}\:{a}\:{unique}\:{a}^{−\mathrm{1}} \in{Q}\:{for}\:\:{a}=\mathrm{0}\: \\ $$$${such}\:{that}\:{a}\centerdot{a}^{−\mathrm{1}} =\mathrm{1}.\:{Hence},\:{Q}\:{cannot}\:{be}\:{a}\:{group}. \\ $$
Commented by Mastermind last updated on 04/May/23
$$\mathrm{Thak}\:\mathrm{you}\:\mathrm{BOSS},\:\mathrm{i}\:\mathrm{do}\:\mathrm{really}\:\mathrm{appreciate} \\ $$
Commented by Mastermind last updated on 04/May/23
$$\mathrm{200l} \\ $$