Question Number 93950 by oustmuchiya@gmail.com last updated on 16/May/20
$${Let}\:\ast\:{be}\:{the}\:{binary}\:{operation}\:{on}\:\mathrm{N} \\ $$$${given}\:{by}\:\mathrm{a}\ast\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{\mathrm{b}}.\:\mathrm{F}{ind} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{5}\ast\mathrm{7}\:,\:\:\mathrm{20}\ast\mathrm{16}\:\:\:\left(\mathrm{ii}\right)\:\mathrm{is}\:\ast\:{communitative}? \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{is}\:\ast\:{associative}? \\ $$$$\left(\mathrm{iv}\right)\mathrm{F}{ind}\:{the}\:{identity}\:{of}\:\ast\:{in}\:\boldsymbol{\mathrm{N}} \\ $$$$\left(\mathrm{v}\right)\:\mathrm{which}\:\mathrm{elements}\:\mathrm{of}\:\boldsymbol{\mathrm{N}}\:{are}\:{invertible} \\ $$$$\:\:\:\:\:\:\:{for}\:{the}\:{operation}\:\ast? \\ $$
Answered by som(math1967) last updated on 16/May/20
![i) 5∗7=L.C.M of 5,7=35 20∗16=L.C.M of 20,16=80 ii) L.C.M of a,b=L.C.M of b,a ∴∗ is commutative [for all a,b∈N] iii)( L.C.M of a,b)∗c =L.C.M of a,b,c = a∗(L.C.M of b,c) =L.C.M of a,b,c ∴ ∗ is associative](https://www.tinkutara.com/question/Q93997.png)
$$\left.\mathrm{i}\right)\:\mathrm{5}\ast\mathrm{7}=\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\:\mathrm{5},\mathrm{7}=\mathrm{35} \\ $$$$\mathrm{20}\ast\mathrm{16}=\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\mathrm{20},\mathrm{16}=\mathrm{80} \\ $$$$\left.\mathrm{ii}\right)\:\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\mathrm{a},\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of} \\ $$$$\mathrm{b},\mathrm{a} \\ $$$$\therefore\ast\:\mathrm{is}\:\mathrm{commutative}\:\left[\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b}\in\mathrm{N}\right] \\ $$$$\left.\mathrm{iii}\right)\left(\:\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\mathrm{a},\mathrm{b}\right)\ast\mathrm{c} \\ $$$$=\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\mathrm{a},\mathrm{b},\mathrm{c} \\ $$$$=\:\mathrm{a}\ast\left(\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\mathrm{b},\mathrm{c}\right) \\ $$$$=\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\mathrm{a},\mathrm{b},\mathrm{c} \\ $$$$\therefore\:\ast\:\mathrm{is}\:\mathrm{associative} \\ $$
Answered by som(math1967) last updated on 16/May/20
![iv) Let e∈N be the identity element in N ∴a∗e=a [a∈N] L.C.M of a,e=a but a∈N ∴e=1 is identity eliment in N ans](https://www.tinkutara.com/question/Q94007.png)
$$\left.\mathrm{iv}\right)\:\mathrm{Let}\:\mathrm{e}\in\mathrm{N}\:\mathrm{be}\:\mathrm{the}\:\mathrm{identity} \\ $$$$\mathrm{element}\:\mathrm{in}\:\mathrm{N} \\ $$$$\therefore\mathrm{a}\ast\mathrm{e}=\mathrm{a}\:\left[\mathrm{a}\in\mathrm{N}\right] \\ $$$$\mathrm{L}.\mathrm{C}.\mathrm{M}\:\mathrm{of}\:\mathrm{a},\mathrm{e}=\mathrm{a} \\ $$$$\mathrm{but}\:\mathrm{a}\in\mathrm{N}\:\therefore\mathrm{e}=\mathrm{1}\:\mathrm{is}\:\mathrm{identity}\:\mathrm{eliment} \\ $$$$\mathrm{in}\:\mathrm{N}\:\mathrm{ans} \\ $$
Commented by oustmuchiya@gmail.com last updated on 16/May/20
$${Wow}!\:{A}\:{million}\:{thanks}\:{indeed}\:{Papa}. \\ $$
Commented by oustmuchiya@gmail.com last updated on 16/May/20
$${How}\:{about}\:{this}\:{one}\:{please}. \\ $$$${Let}\:\ast'\:{be}\:{the}\:{binary}\:{opetation}\:{on}\:{the} \\ $$$${set}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\}\:{defined}\:{by}\:\boldsymbol{\mathrm{a}}\ast'\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{H}}.\boldsymbol{{C}}.\boldsymbol{\mathrm{F}}. \\ $$$${of}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}.\:{Is}\:{the}\:{operation}\:\ast'\:{same} \\ $$$${as}\:{the}\:{operation}\:\ast.\:{justify}\:{your} \\ $$$${answer}. \\ $$
Commented by som(math1967) last updated on 16/May/20
![a∗b=HCF of a,b [a,b ∈{1,2,3,4,5}] =b∗a=HCF of b,a so a∗b commutative ....](https://www.tinkutara.com/question/Q94043.png)
$$\mathrm{a}\ast\mathrm{b}=\mathrm{HCF}\:\mathrm{of}\:\mathrm{a},\mathrm{b}\:\left[\mathrm{a},\mathrm{b}\:\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\}\right] \\ $$$$=\mathrm{b}\ast\mathrm{a}=\mathrm{HCF}\:\mathrm{of}\:\mathrm{b},\mathrm{a} \\ $$$$\mathrm{so}\:\mathrm{a}\ast\mathrm{b}\:\mathrm{commutative}\:…. \\ $$