For my proof, I'm hypothesizing that $\displaystyle s$ is the solution that is known to be rational.

So the rationals are closed under multiplication, addition and subtraction (and division, with nonzero divisors.) If $\displaystyle x$ is rational, then that closure ensures a rational $\displaystyle r$, but $\displaystyle x$ isn't necessarily rational. I know I'm missing something, but I've been doing math all day and I'm a little brain-burned. Can anyone nudge me past the whole non-rational $\displaystyle x$ hurdle?

Thanks!