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a-b-c-R-Find-triple-of-positive-real-numbers-a-b-c-that-satisfy-a-b-5-b-c-5-c-a-12-




Question Number 58700 by naka3546 last updated on 28/Apr/19
a, b, c  ∈  R^+   Find  triple  of  positive  real  numbers (a, b, c)  that  satisfy             a⌊b⌋  =  5             b⌊c⌋  =  5             c⌊a⌋  =  12
$${a},\:{b},\:{c}\:\:\in\:\:\mathbb{R}^{+} \\ $$$${Find}\:\:{triple}\:\:{of}\:\:{positive}\:\:{real}\:\:{numbers}\:\left({a},\:{b},\:{c}\right)\:\:{that}\:\:{satisfy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}\lfloor{b}\rfloor\:\:=\:\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{b}\lfloor{c}\rfloor\:\:=\:\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{c}\lfloor{a}\rfloor\:\:=\:\:\mathrm{12} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Apr/19
In  a⌊b⌋=5 ,  ⌊b⌋ is an integer  Hence   a is also an integer  real×integer=integer⇒real=integer.  So a,b,c ∈ Z     ⌊a⌋=a ,⌊b⌋=b , ⌊c⌋=c        ab=5        bc=5       ca=12     a^2 b^2 c^2 =5×5×12=300=10(√3) ????  But,   a,b,c∈Z⇒a^2 ,b^2 ,c^2  ∈Z⇒a^2 b^2 c^2  ∈ Z  This contradiction leads that there  aren′t any such a,b,c.
$${In}\:\:{a}\lfloor{b}\rfloor=\mathrm{5}\:,\:\:\lfloor{b}\rfloor\:{is}\:{an}\:{integer} \\ $$$${Hence}\:\:\:{a}\:{is}\:{also}\:{an}\:{integer} \\ $$$${real}×{integer}={integer}\Rightarrow{real}={integer}. \\ $$$${So}\:{a},{b},{c}\:\in\:\mathbb{Z} \\ $$$$\:\:\:\lfloor{a}\rfloor={a}\:,\lfloor{b}\rfloor={b}\:,\:\lfloor{c}\rfloor={c} \\ $$$$\:\:\:\:\:\:{ab}=\mathrm{5} \\ $$$$\:\:\:\:\:\:{bc}=\mathrm{5} \\ $$$$\:\:\:\:\:{ca}=\mathrm{12} \\ $$$$\:\:\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{5}×\mathrm{5}×\mathrm{12}=\mathrm{300}=\mathrm{10}\sqrt{\mathrm{3}}\:???? \\ $$$${But}, \\ $$$$\:{a},{b},{c}\in\mathbb{Z}\Rightarrow{a}^{\mathrm{2}} ,{b}^{\mathrm{2}} ,{c}^{\mathrm{2}} \:\in\mathbb{Z}\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \:\in\:\mathbb{Z} \\ $$$${This}\:{contradiction}\:{leads}\:{that}\:{there} \\ $$$${aren}'{t}\:{any}\:{such}\:{a},{b},{c}. \\ $$
Commented by mr W last updated on 28/Apr/19
dear sir, a×⌊b⌋=integer doesn′t mean  that a is integer. it means only that  a is rational. e.g. a=(5/3), b=(7/2), ⌊b⌋=3,  ⇒a×⌊b⌋=5.
$${dear}\:{sir},\:{a}×\lfloor{b}\rfloor={integer}\:{doesn}'{t}\:{mean} \\ $$$${that}\:{a}\:{is}\:{integer}.\:{it}\:{means}\:{only}\:{that} \\ $$$${a}\:{is}\:{rational}.\:{e}.{g}.\:{a}=\frac{\mathrm{5}}{\mathrm{3}},\:{b}=\frac{\mathrm{7}}{\mathrm{2}},\:\lfloor{b}\rfloor=\mathrm{3}, \\ $$$$\Rightarrow{a}×\lfloor{b}\rfloor=\mathrm{5}. \\ $$
Commented by Rasheed.Sindhi last updated on 28/Apr/19
Sorry for deffective logic!   You′re very right sir! Thank you  to correct me.Actually I′m here to learn  from you!
$${Sorry}\:{for}\:{deffective}\:{logic}! \\ $$$$\:{You}'{re}\:{very}\:{right}\:{sir}!\:{Thank}\:{you} \\ $$$${to}\:{correct}\:{me}.{Actually}\:{I}'{m}\:{here}\:{to}\:{learn} \\ $$$${from}\:{you}! \\ $$
Commented by mr W last updated on 28/Apr/19
You′re welcome sir. In fact I have  learnt and still learn alot from you sir.  Many posts of you have become classic  of the forum. It′s a pleasure to study  them.
$${You}'{re}\:{welcome}\:{sir}.\:{In}\:{fact}\:{I}\:{have} \\ $$$${learnt}\:{and}\:{still}\:{learn}\:{alot}\:{from}\:{you}\:{sir}. \\ $$$${Many}\:{posts}\:{of}\:{you}\:{have}\:{become}\:{classic} \\ $$$${of}\:{the}\:{forum}.\:{It}'{s}\:{a}\:{pleasure}\:{to}\:{study} \\ $$$${them}. \\ $$
Commented by Rasheed.Sindhi last updated on 29/Apr/19
THαnks to enhearten me in such an  exciting way sir!!!
$$\mathcal{TH}\alpha{nks}\:{to}\:{enhearten}\:{me}\:{in}\:{such}\:{an} \\ $$$${exciting}\:{way}\:{sir}!!! \\ $$

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