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x + y * 3√2 + z * 3√4 = 0 (Posted on 2009-10-15) Difficulty: 3 of 5
Each of x, y and z is a rational number such that: x + y * 3√2 + z * 3√4 = 0.

Prove that x=y=z=0 is the only possible solution to the above equation.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Non-rigorous argument | Comment 1 of 14

My argument uses three assumptions that are likely true:

1. The cube root of an integer that is not a perfect cube is irrational.

2. The product of a nonzero rational number and an irrational number is irrational.

3. The sum of two irrational numbers is also irrational.

Suppose x + cubrt(2)*y + cubrt(4)*z = 0 has a nontrivial solution, in that at least one of x, y, z is nonzero.  Then -x = cubrt(2)*y + cubrt(4)*z, which is irrational by the above three assumptions.  But x is rational, so -x must also be rational -- a contradiction.  Therefore the original equation has no nontrivial solution, hence x=y=z=0 is the only solution.

Edited on October 15, 2009, 4:00 pm
  Posted by Jim Keneipp on 2009-10-15 14:06:26

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