Question Number 33369 by Rasheed.Sindhi last updated on 15/Apr/18
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\mathrm{gcd}\left(\:\:\mathrm{gcd}\left(\mathrm{A},\mathrm{B}\right),\mathrm{gcd}\left(\mathrm{B},\mathrm{C}\right),\mathrm{gcd}\left(\mathrm{C},\mathrm{A}\right)\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{gcd}\left(\mathrm{A},\mathrm{B},\mathrm{C}\right) \\ $$
Answered by MJS last updated on 15/Apr/18
$$\mathrm{set}\:\mathrm{theory} \\ $$$${U}\cap{V}={V}\cap{U} \\ $$$${U}\cap{U}={U} \\ $$$${U}\cap\left({V}\cap{W}\right)=\left({U}\cap{V}\right)\cap{W}={U}\cap{V}\cap{W} \\ $$$$ \\ $$$${P}_{{A}} ,\:{P}_{{B}} ,\:{P}_{{C}} :\:\mathrm{sets}\:\mathrm{of}\:\mathrm{prime}\:\mathrm{factors} \\ $$$$\mathrm{of}\:{A},\:{B},\:{C} \\ $$$${P}_{{gcd}\left({A},\:{B}\right)} ={P}_{{A}} \cap{P}_{{B}} \\ $$$$ \\ $$$$\left({P}_{{A}} \cap{P}_{{B}} \right)\cap\left({P}_{{B}} \cap{P}_{{C}} \right)\cap\left({P}_{{A}} \cap{P}_{{C}} \right)= \\ $$$$={P}_{{A}} \cap{P}_{{B}} \cap{P}_{{B}} \cap{P}_{{C}} \cap{P}_{{A}} \cap{P}_{{C}} = \\ $$$$=\left({P}_{{A}} \cap{P}_{{A}} \right)\cap\left({P}_{{B}} \cap{P}_{{B}} \right)\cap\left({P}_{{C}} \cap{P}_{{C}} \right)= \\ $$$$={P}_{{A}} \cap{P}_{{B}} \cap{P}_{{C}} ={P}_{{gcd}\left({A},\:{B},\:{C}\right)} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Apr/18
$${e}^{{x}} {cellent}\:\mathcal{S}{ir}!\:\mathrm{Th}\alpha\mathrm{n}\Bbbk\mathrm{s}! \\ $$