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I have this problem:

$$\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$$

Prove that you need to change ONLY a sign (to convert a $+$ to $-$) of a single square root, for the sum to be rational.

EDIT: My math book was wrong. This exercise is not correct.

Can you help me please? I really don't know how to do it.

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closed as off-topic by Did, Dietrich Burde, Batominovski, hardmath, Zain Patel Jul 22 '16 at 22:25

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    $\begingroup$ That doesn't sound very likely, are you sure? $\endgroup$ – Gregory Grant Jul 22 '16 at 19:57
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    $\begingroup$ I, too, would like to call shenanigans. How are you sure? $\endgroup$ – tilper Jul 22 '16 at 20:00
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    $\begingroup$ Interesting. When I get home later today I can play around with this on Maple. $\endgroup$ – tilper Jul 22 '16 at 20:04
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    $\begingroup$ The sum contains a term of $\sqrt{2003}$. But $2003$ is prime, so none of the other terms in the sum add up to $\sqrt{2003}$ or to any multiple of it. There is no way to make the sum rational while that $\sqrt{2003}$ is in there. $\endgroup$ – MJD Jul 22 '16 at 20:04
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    $\begingroup$ This math book wouldn't happen to be called "the mathematics of the Simpsons" would it? $\endgroup$ – abiessu Jul 22 '16 at 20:06
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Impossible! Let $K$ be the field extension of $\mathbb{Q}$ by $\sqrt{k}$, for $k\in\{1,2,\ldots,2009\}\setminus\{2003\}$. Then, $\sqrt{2003}$ is not contained in $K$ due to a result by I. Boreico. Hence, $\sqrt{2003}$ can not be a $\mathbb{Q}$-linear combination of $\sqrt{k}$ for $k\in\{1,2,\ldots,2009\}\setminus\{2003\}$. In fact, no matter how many signs you flip, it is never possible to make the sum a rational number.

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  • $\begingroup$ The result you refer to is much older than that - sometimes credited to Besicovitch. You can find a proof in my answer here. $\endgroup$ – Bill Dubuque Jul 22 '16 at 21:23

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