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Sum = Cube Root 2, Triplet Not (Posted on 2009-07-23) Difficulty: 3 of 5
Prove that there does not exist any triplet (F, G, H) of rational numbers, that satisfies this equation.

F + G*√H = 3√2

No Solution Yet Submitted by K Sengupta    
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proof with assumption | Comment 1 of 2

I'm assuming that I can assume that the cube root of 2 is irrational and thus not have to prove it.

let

f+g*sqrt(h)=2^(1/3)    cube both sides

f^3+f^2*g*sqrt(h)+f*g^2*h+g^3*h*sqrt(h)=2

(f^2*g+g^3*h)*sqrt(h)=2-f^3-f*g^2*h

sqrt(h)=(2-f^3-f*g^2*h)/(f^2*g+g^3*h)

now since all parts of the right hand side are rational then the right hand side is rational thus sqrt(h) must be rational.

If sqrt(h) is rational then so is f+g*sqrt(h) but that would mean that 2^(1/3) is rational which we know to not be true, thus there can not be any rational solutions to the equation

f+g*sqrt(h)=2^(1/3)


  Posted by Daniel on 2009-07-23 12:34:12
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