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Let $Ax = b$, a system of linear equations, $A$ is matrix of rational numbers, and this system has only one complex solution.

Prove this solution is also rational.

Can someone help me please? I prefer not only hints... thanks.

Seirios
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CnR
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    If you assume that $b$ is also rational then this follows immediately from Cramer's rule – Cocopuffs Jul 11 '13 at 16:05
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    If you don't assume $b$ is rational, here's a counterexample: $x = \pi$ (where $A = 1$). – Neal Jul 11 '13 at 16:07
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    As written, the statement is false--let $A = \pmatrix{1 & 0\0 & 1}$ and $\vec{b} = \pmatrix{\pi \ e}$. There is only one solution, it is an element of $\Bbb C^2$, but the solution is not rational. You must assume something about $\vec b$. – apnorton Jul 11 '13 at 16:10
  • if b is rational, how can you prove it by Cramer's rule? A is not necessarily n x n . – CnR Jul 11 '13 at 16:30
  • @ChenR: Any non-redundant (finite) linear system with a unique solution has to be a Cramer system (in particular with square coefficient matrix). So given that there is a unique solution, you can simply drop some equations that are linear combinations of the others, and be left with a square system for which Cramer's rule applies. – Marc van Leeuwen Jul 11 '13 at 16:41
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    Here. – user10676 Jul 11 '13 at 17:07

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