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what's the rationality of the roots of polynomial related to the rationality of its coefficient? Here is the question just came up to me, is that true for a polynomial with irrational coefficient cannot has a rational root. Is any theorem about this?

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Hint $\rm\ \pi x\ $ has a rational root. –  Math Gems Mar 3 '13 at 18:14

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As another example, the polynomial $$ x^2 - (\pi +1) x + \pi $$ has $1$ as a root, and no multiple of it by a nonzero constant has all coefficients rational.

If you want all coefficients to be non-rational, (see comment by @MartinBrandenburg) $$ \pi x^2 - (\pi^2 +\pi) x + \pi^2 $$ will do.

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This has also a rational coefficient. –  Martin Brandenburg Mar 3 '13 at 17:04
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@MartinBrandenburg, ok, thanks, but multyplying by a constant of course you can make any polynomial monic. –  Andreas Caranti Mar 3 '13 at 17:06
    
@AndreasCaranti but if you put it into an equation, its the same as the previous one –  Dylan Zhu Mar 3 '13 at 17:10
    
@AndreasCaranti anyway you have already give the answer what i want, thanks –  Dylan Zhu Mar 3 '13 at 17:11
    
@DylanZhu, you're welcome! –  Andreas Caranti Mar 3 '13 at 17:12

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