Question Number 8846 by Rasheed Soomro last updated on 31/Oct/16
$$\mathrm{Let}\:\mathrm{by}\:\left(\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,…\mathrm{a}_{\mathrm{n}} \right)\:\mathrm{we}\:\mathrm{mean}\:\mathrm{LCM} \\ $$$$\mathrm{of}\:\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,…\mathrm{a}_{\mathrm{n}} \:,\mathrm{where}\:\mathrm{a}_{\mathrm{i}} \in\mathbb{N}. \\ $$$$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{that}\:\left(\:\left(\mathrm{a},\mathrm{b}\right),\left(\mathrm{b},\mathrm{c}\right)\:\:\right)=\left(\mathrm{a},\mathrm{b},\mathrm{c}\right). \\ $$
Answered by 123456 last updated on 01/Nov/16
$$\mathrm{lets}\:\mathrm{say} \\ $$$${x}=\left({a},{b}\right) \\ $$$${y}=\left({b},{c}\right) \\ $$$${z}=\left({a},{b},{c}\right) \\ $$$$\mathrm{by}\:\mathrm{definition} \\ $$$${x}=\left({a},{b}\right)\Rightarrow{a}\mid{x}\wedge{b}\mid{x} \\ $$$${y}=\left({b},{c}\right)\Rightarrow{b}\mid{y}\wedge{c}\mid{y} \\ $$$${z}=\left({a},{b},{c}\right)\Rightarrow{a}\mid{z}\wedge{b}\mid{z}\wedge{c}\mid{z} \\ $$$$\mathrm{call}\:{u}=\left({x},{y}\right) \\ $$$${u}=\left({x},{y}\right)\Rightarrow{x}\mid{u}\wedge{y}\mid{u} \\ $$$$\mathrm{since}\:{a}\mid{x}\:\mathrm{and}\:{x}\mid{u},\:{a}\mid{u} \\ $$$$\mathrm{by}\:\mathrm{same}\:\mathrm{reason} \\ $$$${b}\mid{x}\wedge{x}\mid{u}\Rightarrow{b}\mid{u} \\ $$$${b}\mid{y}\wedge{y}\mid{u}\Rightarrow{b}\mid{u} \\ $$$${c}\mid{y}\wedge{y}\mid{u}\Rightarrow{c}\mid{u} \\ $$$$\mathrm{so}\:{a}\mid{u}\wedge{b}\mid{u}\wedge{c}\mid{u} \\ $$$$\mathrm{the}\:\mathrm{min}\:\mathrm{value}\:\mathrm{that}\:\mathrm{hold}\:\mathrm{this}\:\mathrm{is}\:{z} \\ $$$$\mathrm{so}\:{z}={u}\:\mathrm{or} \\ $$$$\left({a},{b},{c}\right)=\left(\left({a},{b}\right),\left({b},{c}\right)\right) \\ $$
Commented by Rasheed Soomro last updated on 01/Nov/16
$$\mathcal{N}{ice}! \\ $$