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I need to prove that

$$\exists a,b \in \mathbb{R} \setminus \mathbb{Q} : a + b, ab \in \mathbb{Q}$$

Any ideas? I, unfortunately, don't have one yet. The most obvious way with equations in integers (using the definition of a rational number) with irrational coefficients doesn't seem to bear any fruit.

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  • $\begingroup$ Hint: $0$ is a rational number. $\sqrt 2$ is not. $\endgroup$ – Ittay Weiss Oct 5 '15 at 15:41
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If we carefully choose $s,p\in\mathbb{\mathbb{Q}}$ ($s^2>4p$, for starters), the roots $a,b$ of the quadratic polynomial $$ x^2-sx+p $$ are real numbers but they cannot be rational numbers by the rational root theorem.

On the other hand, they fulfill $a+b=s$ and $ab=p$ by Viète's theorem.

So there are plenty of couples of irrational numbers with both their sum and their product being rational numbers: for instance, $(-\sqrt{2},\sqrt{2}),\left(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right),(2-\sqrt{3},2+\sqrt{3}),\ldots$

The interesting part is that if $a$ and $b$ are irrational numbers, but both $a+b$ and $ab$ are rational numbers, then $a$ and $b$ are conjugated algebraic numbers of degree $2$ over $\mathbb{Q}$: the proof is hidden in the previous lines.

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Hint. Think about the quadratic formula for a quadratic equation with integer coefficients but two irrational roots.

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Take $a=\sqrt 2$ and $b=-\sqrt 2$ for the first one and $a=b=\sqrt 2$ for the second one.

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  • $\begingroup$ come on.... don't just give the answer. $\endgroup$ – Ittay Weiss Oct 5 '15 at 15:41
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Hint

How about $a=\sqrt{2}$. What can you add to $\sqrt{2}$ to get an integer? What can you multiply by it to get an integer?

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  • $\begingroup$ there are no two cases. It's just one condition OP is interested in: both $a+b$ and $ab$ need to be rational. Your hint is still good though. $\endgroup$ – Ittay Weiss Oct 5 '15 at 15:44
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$ab = s$ so $a = s/b$

$b + s/b = t$ so $b^2 - tb + s = 0$. Any irrational root to a quadratic equation will do it and only roots to a quadratic equations will do it. Right?

Or am I missing something?

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