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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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Solution fixing the proof Comment 2 of 2 |
(In reply to proof with assumption by Daniel)

There are one or two things missing from the previous proof.

First, here is a similar (incorrect) proof that there does not exist any triplet (F, G, H) of rational numbers, that satisfies the equation.

      F + G*√H =√2

Incorrect Proof: 

Square both sides, giving
2fg*sqrt(h) = 2 - f^2 - g^2h
sqrt(h) = (2 - f^2 - g^2h)/2fg

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^(1/2) is rational which we know to not be true, thus there can not be any rational solutions to the equation.

Except this is a wrong conclusion, because (0,1,2) satisfies the equation.  We went wrong (did you notice?) when dividing by 2fg. This operation is not valid if f = 0 or g = 0 


Daniel's third step is still OK.


Since the right hand side is rational, so is sqrt(h), unless (f^2*g+g^3*h) = g(f^2+g^2*h) = 0.  

If g = 0, then F = 3√2, which is a contradiction.

If (f^2+g^2*h) = 0, then h is negative, which is also a contradiction.

Therefore sqrt(h) is rational, which makes F + G*√H rational, which is a contradiction.

Therefore, there does not exist any triplet (F, G, H) of rational numbers, that satisfies F + G*√H = 3√2 

  Posted by Steve Herman on 2009-07-23 23:14:12
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