The Conjugate Roots Theorem for Irrational Roots states that for a polynomial $f(x)$ with integer coefficients, if a root of the equation $f(x) = 0$ is expressed as $a+\alpha$, where $a\in\mathbb{Q}$ and $\alpha\in\mathbb{R}-\mathbb{Q}$ are rational and is irrational, then $a-\alpha$ is also a root of the equation. However, some additional explanation should be added to this theorem so that the following contradiction, which I found, will not occur.

Let $\alpha+1=\beta\in\mathbb{R}-\mathbb{Q}$.

Then, $a+\alpha=(a-1)+\beta$

Since the root of $f(x)=0$ is expressed as $(a-1)+\beta$, by the above theorem,

$(a-1)-\beta=(a-2)+\alpha$ is also a root of the equation $f(x)=0$

Could you explain why this contradiction occurred and how to interpret this theorem to prevent the contradiction?

  • 1
    $\begingroup$ You want a root of the form $a+b\sqrt D$. But this might not always be possible. Consider $\sqrt 2+\sqrt 3$, which is a root of $x^4-10x+1$. It's "conjugate" is not. The map $a+b\sqrt D\mapsto a-b\sqrt D$ has the property that it preserves sums and products, thus the theorem you state hold, since it fixes rationals (or integers). $\endgroup$ – Pedro Tamaroff Nov 7 '13 at 3:43
  • $\begingroup$ Thanks for making me convinced why this contradiction occurred. $\endgroup$ – Math.StackExchange Nov 7 '13 at 4:02

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