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I can't find a rigorous proof but I have a feeling it's true.

Informal argument: Suppose $a+b$ and $ab$ are rational, $a$ and $b$ are irrational (since just one can't be irrational). Then $a$ and $b$ must have irrational "parts" that "cancel out" eg. $a = \frac4\pi$, $b = \pi$, $ab = 4$, which is rational. But then $a + b$ ends up being irrational, since the denominator has the irrational part squared and the numerator does not.

Obviously, this isn't a proper proof at all, but I can't think of any examples of numbers where this argument wouldn't hold true. Can someone provide a proof or counterexample?

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    $\begingroup$ $\sqrt2$ and $-\sqrt2$. $\endgroup$
    – Sawarnik
    Commented Jul 9, 2014 at 9:29
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    $\begingroup$ Just a remark: what you can say though, is that $ a,b $ are algebraic over $ \mathbb{Q} $, since they are zeros of the polynomial $ x^{2} - 2(a+b)x +ab $, which has by assumption rational coefficients. $\endgroup$
    – m.g.
    Commented Jul 9, 2014 at 9:35
  • $\begingroup$ @Sawarnik I feel stupid now... $\endgroup$ Commented Jul 9, 2014 at 21:51

2 Answers 2

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It is not true:

$$ a = 1 - \sqrt 2, \\ b = 1 + \sqrt 2 \\ a + b = 2, \\ a b = 1 - 2 = -1 $$

$a + b$ and $a b$ are rational, but $a$ and $b$ are not rational.

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If $s=a+b$ and $p=ab$ are rational, then $a$ and $b$ are roots of the polynomial $X^2-sX+p$ which has rational coefficients.

Then $a$ and $b$ are irrational if and only if the discriminant $D=s^2-4p$ is not a perfect square in $\mathbb{Q}$. Thus $$ a=\frac{s}{2}+\frac{\sqrt{D}}{2},\qquad b=\frac{s}{2}-\frac{\sqrt{D}}{2} $$ (or conversely), so your conjecture is true, provided we interpret irrational part with a loose meaning. To wit, what's the irrational part of $\sqrt{2}$? Is it $\sqrt{2}$ or $\sqrt{2}-1$, as $\sqrt{2}=1+(\sqrt{2}-1)$?

More rigorously, when $a$ and $b$ are irrational, there is a field automorphism in $\mathbb{Q}[a,b]$ sending $a$ to $b$.

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