$\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$ change a sign to be rational

Question

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.

Interesting. When I get home later today I can play around with this on Maple. — , Jul 22 '16 at 20:04
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. — MJD, Jul 22 '16 at 20:04
This math book wouldn't happen to be called "the mathematics of the Simpsons" would it? — abiessu, Jul 22 '16 at 20:06
Ok. I asked my math teacher by phone and she said the exercise is wrong and to change them to another but based on the same. I am editing the question. Or shoud I post a new question? Thank you. — ITChristian, Jul 22 '16 at 20:07
@MMPP I would ask a new question since there is already a good answer to the currently posed question. — Eff, Jul 22 '16 at 20:09
My answer to your previous question (http://math.stackexchange.com/questions/1867912/for-pm-sqrt-1-pm-sqrt-2-pm-sqrt-3-pm-cdots-pm-sqrt-2009-show-there-is-a) proves that, too. — Jack D'Aurizio, Jul 23 '16 at 14:11

score 11 · Accepted Answer · answered Jul 22 '16 at 20:06

11

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.

answered Jul 22 '16 at 20:06

Batominovski

50,341

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

$\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$ change a sign to be rational

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