Question Number 119684 by help last updated on 26/Oct/20
Commented by help last updated on 26/Oct/20
$${is}\:{the}\:{operation}\:{closed} \\ $$
Commented by help last updated on 26/Oct/20
$${is}\:{it}\:{commutative} \\ $$$${is}\:{it}\:{associative} \\ $$$${identity}\:{of}\:{the}\:{elements} \\ $$$${inverse}\:{of}\:{i} \\ $$$${inverse}\:{of}\:−\mathrm{1} \\ $$
Answered by $@y@m last updated on 26/Oct/20
$${Let}\:{S}=\left\{−\mathrm{1},\mathrm{1},{i},−{i}\right\} \\ $$$${From}\:{operation}\:{table}, \\ $$$$\left({i}\right)\:{a}\ast{b}\in{S}\:\forall{a},{b}\in{S} \\ $$$$\therefore\:\ast\:{is}\:{closed}. \\ $$$$\left({ii}\right)\:{a}\ast{b}\in{b}\ast{a}\:\forall{a},{b}\in{S} \\ $$$$\therefore\:\ast\:{is}\:{commutative}. \\ $$$$\left({iii}\right)\:{Try}\:{yourself}. \\ $$$$\left({iv}\right)\:{a}\ast\mathrm{1}={a}\:\forall{a}\in{S} \\ $$$$\therefore\:\mathrm{1}\:{is}\:{the}\:{unity}\:{element}. \\ $$$$\left({iv}\right)\:{i}\ast\left(−{i}\right)=\mathrm{1} \\ $$$$\therefore\:−{i}\in{S}\:{is}\:{inverse}\:{of}\:{i} \\ $$$$\left({v}\right)\:{Try}\:{yourself} \\ $$