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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?

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    $\begingroup$ 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 holds, 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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    $\begingroup$ @PedroTamaroff Please elaborate on the mapping between the conjugates that preserves the sum and products, by an example. If there is some terminology that links it with minimal polynomial, then please detail that too. $\endgroup$ – jiten Jul 28 at 18:45
  • $\begingroup$ @jiten You can check that it preserves sums and products by yourself. $\endgroup$ – Pedro Tamaroff Jul 29 at 2:35
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    $\begingroup$ @PedroTamaroff Please provide a detailed example, as have difficulty; as am not so expert. Else, if you feel it is too much to ask; then give some hint and/or reference. $\endgroup$ – jiten Aug 1 at 8:53

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