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Question Number 48129 by kaivan.ahmadi last updated on 19/Nov/18

how we can show _3_(√2) on axis?

$$\mathrm{how}\:\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:\:\:_{\mathrm{3}_{\sqrt{\mathrm{2}}} } \:\:\mathrm{on}\:\mathrm{axis}? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Nov/18

trying to solve... 1)first plot graph y=x^3 2)plot point p(0,2) on yaxis 3)draw straight line y=2 which is parallel to x axis 4)let straight line y=2 cuts curve at point q the co−oridinate of point q is (2^(1/3) ,2) 5)draw straight line parralel to yaxis from point q 6)let this straight line cuts x axis at point say M so co ordinate of M is(2^(1/3) ,0)

$${trying}\:{to}\:{solve}... \\ $$$$\left.\mathrm{1}\right){first}\:{plot}\:{graph}\:{y}={x}^{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right){plot}\:{point}\:{p}\left(\mathrm{0},\mathrm{2}\right)\:{on}\:{yaxis} \\ $$$$\left.\mathrm{3}\right){draw}\:{straight}\:{line}\:{y}=\mathrm{2}\:{which}\:{is}\:{parallel} \\ $$$${to}\:{x}\:{axis} \\ $$$$\left.\mathrm{4}\right){let}\:{straight}\:{line}\:{y}=\mathrm{2}\:\:{cuts}\:{curve}\:{at}\:{point}\:{q} \\ $$$$\:{the}\:{co}−{oridinate}\:{of}\:{point}\:{q}\:{is}\:\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} ,\mathrm{2}\right) \\ $$$$\left.\mathrm{5}\right){draw}\:{straight}\:{line}\:{parralel}\:{to}\:{yaxis}\:{from} \\ $$$${point}\:{q} \\ $$$$\left.\mathrm{6}\right){let}\:{this}\:{straight}\:{line}\:{cuts}\:{x}\:{axis}\:{at}\:{point}\:{say} \\ $$$${M}\:\:{so}\:{co}\:{ordinate}\:{of}\:{M}\:{is}\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} ,\mathrm{0}\right) \\ $$

Commented by MJS last updated on 20/Nov/18

this works but there′s no way to exactly construct it

$$\mathrm{this}\:\mathrm{works}\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{way}\:\mathrm{to}\:\mathrm{exactly} \\ $$$$\mathrm{construct}\:\mathrm{it} \\ $$

Commented by kaivan.ahmadi last updated on 20/Nov/18

thank you sir. would you show it geometricaly?

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{would}\:\mathrm{you}\:\mathrm{show}\:\mathrm{it}\:\mathrm{geometricaly}? \\ $$

Commented by MJS last updated on 20/Nov/18

it′s impossible. it′s the old problem of “doubling the cube” which has been proven impossible in the 19^(th) century

$$\mathrm{it}'\mathrm{s}\:\mathrm{impossible}. \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{old}\:\mathrm{problem}\:\mathrm{of}\:``\mathrm{doubling}\:\mathrm{the}\:\mathrm{cube}'' \\ $$$$\mathrm{which}\:\mathrm{has}\:\mathrm{been}\:\mathrm{proven}\:\mathrm{impossible}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{19}^{\mathrm{th}} \:\mathrm{century} \\ $$

Commented by kaivan.ahmadi last updated on 20/Nov/18

thanks for ur answer

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{ur}\:\mathrm{answer} \\ $$

Commented by MJS last updated on 21/Nov/18

you can maybe use the approximation (2)^(1/3) ≈1.26 which is .00008 units or .006% greater. 1.26^3 =2.000376

$$\mathrm{you}\:\mathrm{can}\:\mathrm{maybe}\:\mathrm{use}\:\mathrm{the}\:\mathrm{approximation}\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\approx\mathrm{1}.\mathrm{26}\:\mathrm{which} \\ $$$$\mathrm{is}\:.\mathrm{00008}\:\mathrm{units}\:\mathrm{or}\:.\mathrm{006\%}\:\mathrm{greater}.\:\mathrm{1}.\mathrm{26}^{\mathrm{3}} =\mathrm{2}.\mathrm{000376} \\ $$

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