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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

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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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.


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




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


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