Menu Close

If-a-1-b-b-1-c-c-1-a-then-prove-that-abc-1-a-b-c-

Tinku Tara 0 Comments
Pin




Question Number 181438 by Agnibhoo98 last updated on 26/Nov/22
If a + (1/b) = b + (1/c) = c + (1/a) then prove that abc = ±1. a ≠ b ≠ c
$$\mathrm{If}\:{a}\:+\:\frac{\mathrm{1}}{{b}}\:=\:{b}\:+\:\frac{\mathrm{1}}{{c}}\:=\:{c}\:+\:\frac{\mathrm{1}}{{a}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${abc}\:=\:\pm\mathrm{1}.\:\:\:{a}\:\neq\:{b}\:\neq\:{c} \\ $$
Commented by mr W last updated on 26/Nov/22
it′s not true. e.g. you can take a=b=c=2, and get abc=8.
$${it}'{s}\:{not}\:{true}.\:{e}.{g}.\:{you}\:{can}\:{take}\: \\ $$$${a}={b}={c}=\mathrm{2},\:{and}\:{get}\:{abc}=\mathrm{8}. \\ $$
Answered by manxsol last updated on 26/Nov/22
have f(a,b,c) restriccions a+(1/b)=k ⇒g()=a+(1/b)−k=0 b+(1/c)=k ⇒h()=b+(1/c)−k=0 c+(1/a)=k⇒ j()=c+(1/a)−k=0 ▽g()=((∂g/∂x);(∂g/∂y);(∂g/∂z))=(g_x ;g_y ;g_z ) df=λ_1 ▽g+λ_2 ▽h+λ_3 ▽j=0 df(a,b,c,λ_1 ,λ_2 ,λ_3 )= λ_1 (1,−(1/b^2 ),0)+λ_2 (0,1,−(1/c^2 ))+λ_3 (−(1/a^2 ),0,1)=(0,0,0) ((λ_1 ,λ_2 ,λ_3 ,(const)),(( 1),( 0),(−(1/a^2 )),0),((−(1/b^2 )),( 1),( 0),0),(( 0),(−(1/c^2 )),( 1),0) ) (λ_2 /c^2 )=λ_3 ⇒λ_2 =c^2 λ_3 (λ_1 /b^2 )=λ_2 ⇒λ_1 =b^2 λ_2 λ_1 =(λ_3 /a^2 )⇒λ_3 =a^2 λ_1 multiplication exprexions λ_1 λ_2 λ_3 =a^2 b^2 c^2 λ_1 λ_2 λ_3 1=a^2 b^2 c^2 abc=± 1 LQQD
$${have}\:{f}\left({a},{b},{c}\right) \\ $$$${restriccions} \\ $$$${a}+\frac{\mathrm{1}}{{b}}={k}\:\:\Rightarrow{g}\left(\right)={a}+\frac{\mathrm{1}}{{b}}−{k}=\mathrm{0} \\ $$$${b}+\frac{\mathrm{1}}{{c}}={k}\:\:\Rightarrow{h}\left(\right)={b}+\frac{\mathrm{1}}{{c}}−{k}=\mathrm{0} \\ $$$${c}+\frac{\mathrm{1}}{{a}}={k}\Rightarrow\:{j}\left(\right)={c}+\frac{\mathrm{1}}{\mathrm{a}}−\mathrm{k}=\mathrm{0} \\ $$$$\bigtriangledown{g}\left(\right)=\left(\frac{\partial{g}}{\partial{x}};\frac{\partial{g}}{\partial{y}};\frac{\partial{g}}{\partial{z}}\right)=\left({g}_{{x}} ;{g}_{{y}} ;{g}_{{z}} \right) \\ $$$${df}=\lambda_{\mathrm{1}} \bigtriangledown{g}+\lambda_{\mathrm{2}} \bigtriangledown{h}+\lambda_{\mathrm{3}} \bigtriangledown{j}=\mathrm{0} \\ $$$${df}\left({a},{b},{c},\lambda_{\mathrm{1}} ,\lambda_{\mathrm{2}} ,\lambda_{\mathrm{3}} \right)= \\ $$$$\lambda_{\mathrm{1}} \left(\mathrm{1},−\frac{\mathrm{1}}{{b}^{\mathrm{2}} },\mathrm{0}\right)+\lambda_{\mathrm{2}} \left(\mathrm{0},\mathrm{1},−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)+\lambda_{\mathrm{3}} \left(−\frac{\mathrm{1}}{{a}^{\mathrm{2}} },\mathrm{0},\mathrm{1}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$$$\begin{pmatrix}{\lambda_{\mathrm{1}} }&{\lambda_{\mathrm{2}} }&{\lambda_{\mathrm{3}} }&{{const}}\\{\:\:\:\:\:\:\mathrm{1}}&{\:\:\:\:\mathrm{0}}&{−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}&{\mathrm{0}}\\{−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}&{\:\:\:\:\:\mathrm{1}}&{\:\:\:\:\:\mathrm{0}}&{\mathrm{0}}\\{\:\:\:\:\:\mathrm{0}}&{−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}&{\:\:\:\:\:\mathrm{1}}&{\mathrm{0}}\end{pmatrix} \\ $$$$\:\:\frac{\lambda_{\mathrm{2}} }{{c}^{\mathrm{2}} }=\lambda_{\mathrm{3}} \:\:\Rightarrow\lambda_{\mathrm{2}} ={c}^{\mathrm{2}} \lambda_{\mathrm{3}} \\ $$$$\frac{\lambda_{\mathrm{1}} }{{b}^{\mathrm{2}} }=\lambda_{\mathrm{2}} \:\:\Rightarrow\lambda_{\mathrm{1}} ={b}^{\mathrm{2}} \lambda_{\mathrm{2}} \\ $$$$\lambda_{\mathrm{1}} =\frac{\lambda_{\mathrm{3}} }{{a}^{\mathrm{2}} }\Rightarrow\lambda_{\mathrm{3}} ={a}^{\mathrm{2}} \lambda_{\mathrm{1}} \\ $$$${multiplication}\:{exprexions} \\ $$$$\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} \lambda_{\mathrm{3}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \lambda_{\mathrm{1}} \lambda_{\mathrm{2}} \lambda_{\mathrm{3}} \\ $$$$\mathrm{1}={a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$${abc}=\pm\:\mathrm{1} \\ $$$${LQQD} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 26/Nov/22
what is proved? with 2 conditions for 3 variables abc can get any value.
$${what}\:{is}\:{proved}? \\ $$$${with}\:\mathrm{2}\:{conditions}\:{for}\:\mathrm{3}\:{variables} \\ $$$${abc}\:{can}\:{get}\:{any}\:{value}. \\ $$
Commented by manxsol last updated on 26/Nov/22
excuse me,Sr. W.how should I do it
$${excuse}\:{me},{Sr}.\:{W}.{how}\:{should}\:{I}\:{do}\:{it} \\ $$
Commented by manxsol last updated on 26/Nov/22
or the condition is missing a≠b≠c
$${or}\:{the}\:{condition}\:{is}\:{missing}\:{a}\neq{b}\neq{c} \\ $$
Commented by Agnibhoo98 last updated on 26/Nov/22
yes
$$\mathrm{yes} \\ $$
Commented by manxsol last updated on 26/Nov/22
Sr. W, my procedure is correct???
$${Sr}.\:{W},\:{my}\:{procedure}\:{is}\:{correct}??? \\ $$
Commented by mr W last updated on 26/Nov/22
something is wrong in the question. an additional condition is needed. a≠b≠c could be a such condition.
$${something}\:{is}\:{wrong}\:{in}\:{the}\:{question}. \\ $$$${an}\:{additional}\:{condition}\:{is}\:{needed}. \\ $$$${a}\neq{b}\neq{c}\:{could}\:{be}\:{a}\:{such}\:{condition}. \\ $$
Commented by mr W last updated on 26/Nov/22
i can′t judge, because i don′t know what you want to prove.
$${i}\:{can}'{t}\:{judge},\:{because}\:{i}\:{don}'{t}\:{know} \\ $$$${what}\:{you}\:{want}\:{to}\:{prove}. \\ $$
Commented by manxsol last updated on 26/Nov/22
i understand, I will reflect on[the question. thank you very much for you attention
$${i}\:{understand},\:{I}\:{will}\:{reflect}\:{on}\left[{the}\:{question}.\:{thank}\:{you}\:{very}\:{much}\:{for}\:{you}\:{attention}\right. \\ $$
Answered by Agnibhoo98 last updated on 27/Nov/22
a + (1/b) = b + (1/c) or a − b = (1/c) − (1/b) = ((b − c)/(bc)) .... (1) b + (1/c) = c + (1/a) or b − c = (1/a) − (1/c) = ((c − a)/(ca)) .... (2) a + (1/b) = c + (1/a) or a − c = (1/a) − (1/b) = ((b − a)/(ab)) .... (3) Multiplying (1), (2), (3) (a − b)(b − c)(a − c) = (((b − c)(c − a)(b − a))/((abc)^2 )) or (abc)^2 = (((a − b)(b − c)(a − c))/((a − b)(b − c)(a − c))) or (abc)^2 = 1 or abc = ±1
$${a}\:+\:\frac{\mathrm{1}}{{b}}\:=\:{b}\:+\:\frac{\mathrm{1}}{{c}} \\ $$$${or}\:{a}\:−\:{b}\:=\:\frac{\mathrm{1}}{{c}}\:−\:\frac{\mathrm{1}}{{b}}\:=\:\frac{{b}\:−\:{c}}{{bc}}\:\:….\:\left(\mathrm{1}\right) \\ $$$${b}\:+\:\frac{\mathrm{1}}{{c}}\:=\:{c}\:+\:\frac{\mathrm{1}}{{a}} \\ $$$${or}\:{b}\:−\:{c}\:=\:\frac{\mathrm{1}}{{a}}\:−\:\frac{\mathrm{1}}{{c}}\:=\:\frac{{c}\:−\:{a}}{{ca}}\:….\:\left(\mathrm{2}\right) \\ $$$${a}\:+\:\frac{\mathrm{1}}{{b}}\:=\:{c}\:+\:\frac{\mathrm{1}}{{a}} \\ $$$${or}\:{a}\:−\:{c}\:=\:\frac{\mathrm{1}}{{a}}\:−\:\frac{\mathrm{1}}{{b}}\:=\:\frac{{b}\:−\:{a}}{{ab}}\:….\:\left(\mathrm{3}\right) \\ $$$$\mathrm{Multiplying}\:\left(\mathrm{1}\right),\:\left(\mathrm{2}\right),\:\left(\mathrm{3}\right) \\ $$$$ \\ $$$$\left({a}\:−\:{b}\right)\left({b}\:−\:{c}\right)\left({a}\:−\:{c}\right)\:=\:\frac{\left({b}\:−\:{c}\right)\left({c}\:−\:{a}\right)\left({b}\:−\:{a}\right)}{\left({abc}\right)^{\mathrm{2}} } \\ $$$${or}\:\left({abc}\right)^{\mathrm{2}} \:=\:\frac{\left({a}\:−\:{b}\right)\left({b}\:−\:{c}\right)\left({a}\:−\:{c}\right)}{\left({a}\:−\:{b}\right)\left({b}\:−\:{c}\right)\left({a}\:−\:{c}\right)} \\ $$$${or}\:\left({abc}\right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${or}\:{abc}\:=\:\pm\mathrm{1} \\ $$$$\: \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *