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Question Number 7847 by madscientist last updated on 20/Sep/16
is there a proof of a relationship    between ϕ,π,e where π=3.14,ϕ=1.618   and e=2.718 such that ε is some  oporator,ε=−+×÷  ϕεπεe=0 or πεeεϕ=0 or eεϕεπ=0
$${is}\:{there}\:{a}\:{proof}\:{of}\:{a}\:{relationship}\:\: \\ $$$${between}\:\varphi,\pi,{e}\:{where}\:\pi=\mathrm{3}.\mathrm{14},\varphi=\mathrm{1}.\mathrm{618}\: \\ $$$${and}\:{e}=\mathrm{2}.\mathrm{718}\:{such}\:{that}\:\varepsilon\:{is}\:{some} \\ $$$${oporator},\varepsilon=−+×\boldsymbol{\div} \\ $$$$\varphi\varepsilon\pi\varepsilon{e}=\mathrm{0}\:{or}\:\pi\varepsilon{e}\varepsilon\varphi=\mathrm{0}\:{or}\:{e}\varepsilon\varphi\varepsilon\pi=\mathrm{0} \\ $$
Commented by FilupSmith last updated on 20/Sep/16
ϕ∗π∗e≠0         (i am using ∗ instead of ε)  π∗e∗ϕ≠0  e∗ϕ∗π≠0     ϕ, π and e > 0  π>e>ϕ     π−e >0            π−ϕ >0  e−π<0              e−ϕ>0  ϕ−π<0             ϕ−e<0      multiplying, dividing, etc  wont make it equal zero
$$\varphi\ast\pi\ast{e}\neq\mathrm{0}\:\:\:\:\:\:\:\:\:\left(\mathrm{i}\:\mathrm{am}\:\mathrm{using}\:\ast\:\mathrm{instead}\:\mathrm{of}\:\varepsilon\right) \\ $$$$\pi\ast{e}\ast\varphi\neq\mathrm{0} \\ $$$${e}\ast\varphi\ast\pi\neq\mathrm{0} \\ $$$$\: \\ $$$$\varphi,\:\pi\:\mathrm{and}\:{e}\:>\:\mathrm{0} \\ $$$$\pi>{e}>\varphi \\ $$$$\: \\ $$$$\pi−{e}\:>\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\pi−\varphi\:>\mathrm{0} \\ $$$${e}−\pi<\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}−\varphi>\mathrm{0} \\ $$$$\varphi−\pi<\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\varphi−{e}<\mathrm{0} \\ $$$$\: \\ $$$$\:\mathrm{multiplying},\:\mathrm{dividing},\:{etc} \\ $$$$\mathrm{wont}\:\mathrm{make}\:\mathrm{it}\:\mathrm{equal}\:\mathrm{zero} \\ $$$$ \\ $$
Commented by madscientist last updated on 20/Sep/16
thank you
$${thank}\:{you}\: \\ $$
Answered by prakash jain last updated on 02/Oct/16
see coments
$$\mathrm{see}\:\mathrm{coments} \\ $$

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