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Suppose that $r,s$ and $d$ are rational numbers and that $\sqrt{d} $ is irrational. Assume that $r + s\sqrt{d}$ is a root of $x^3-3x-1$. Prove that $3r^2s+s^3d-3s = 0$ and that $r-s\sqrt{d}$ must also be a root of $x^3-3x-1$.

I have tried substituting the assumed root into the given polynomial by utilising the remainder theorem equating it to zero. I always get extra terms that I am not sure how to cancel out. I would appreciate any help!

DeepSea
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allquiet1984
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2 Answers2

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Substituting in $x=r+s\sqrt{d}$ and rearranging, you get $$ r^3+2rs^2d-3r-1+\sqrt{d}(3r^2+s^3d-3s)=0 $$ If you assume $3r^2+s^3d-3s\neq0$, then you get $$ \sqrt{d}=\frac{-(r^3+2rs^2d-3r-1)}{3r^2+s^3d-3s} $$ contradicting the irrationality of $\sqrt{d}$.

Mike Earnest
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  • Thank you very much for this! This has taught me an important lesson to remember the use of proving by contradiction. Really appreciate it Mike! – allquiet1984 Apr 25 '18 at 01:05
  • ah so its simply that because we assume √d to be irrational, the expression must be undefined since it cannot be expressed as a fraction. So set the denominator to zero? – user71207 Jul 18 '21 at 01:51
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Suppose that $r,s$ and $d$ are rational numbers and that $\sqrt{d} $ is irrational. Assume that $r + s\sqrt{d}$ is a root of $x^3-3x-1$. Prove that $\;\ldots$

You could in fact prove anything from that premise. That's because no root of $\,x^3-3x-1\,$ is of the form $\,r+s\sqrt{d}\,$ with $\,r,s,d \in \mathbb{Q}\,$, so the premise is false, and therefore any statement could be logically derived from it, including for example that $\,1=-1\,$.

To see why no such root exists, it's enough to note that the minimal polynomial of $\,r+s\sqrt{d}\,$ over the rationals is $\,(x-r-s\sqrt{d})(x-r+s\sqrt{d})\,$, and that would need to divide $\,x^3-3x-1\,$. In that case, however, the third root of the cubic would have to be a rational (by Vieta's relations), but there are no rational roots (by the rational root theorem).

dxiv
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