# For $\pm\sqrt 1\pm\sqrt 2 \pm\sqrt 3 \pm\cdots\pm\sqrt {2009}$, show there is a choice of signs such that it is irrational [closed]

Considering $$\pm\sqrt 1\pm\sqrt 2 \pm\sqrt 3 \pm\cdots\pm\sqrt {2009}$$ where you can replace each $\pm$ with $+$ or $-$. Prove that there is at least one choice of signs such that the number is irrational.

How can I prove that?

## closed as off-topic by Did, heropup, Matthew Conroy, Chill2Macht, Will JagyJul 23 '16 at 1:42

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• Let $a$ be the sum with all positive. Let $b$ be the sum with $+$ for $\sqrt{2}$ and the rest $-$. If both $a$ and $b$ are rational, so is $a+b$. But $a+b=2\sqrt{2}$, irrational. – André Nicolas Jul 22 '16 at 20:28
• As an aside, the proof used in your previous question still applies here. $\sqrt{2003}+(\sum\limits_{i=1}^{2009} a_i \sqrt{i})$ where $a_i\in\Bbb Q$ for all $i$ and $a_{2003}=0$ is guaranteed to be irrational. Your question is simply a special case where all of $a_i$ (except $a_{2003}$ are $+1$ or $-1$) – JMoravitz Jul 22 '16 at 20:28
• The answer here shows that if the choices of signs "respect products" in the sense that the sign of $\sqrt{mn}$ is the product of the signs of $\sqrt m$ and $\sqrt n$ whenever $m$ and $n$ are coprime integers, you do get an irrational number. The non-constructive but very clever two answers posted here show that my answer is too high-tech for the purposes of this question! I only added this comment because you get an explicit choice from there :-) – Jyrki Lahtonen Jul 22 '16 at 20:30
• I am voting for reopening since even in the absence of efforts from the OP, I think the question is interesting and the answers here can be useful to other MSE users. – Jack D'Aurizio Jul 23 '16 at 14:26

Assume there is a choice of signs that makes the expression rational. If not, there are $2^{2009} \ge 1$ choices that make the expression irrational. Let $r$ be the rational. Now if you change the sign on $\sqrt 2$ you have either $r+2\sqrt 2$ or $r- 2 \sqrt 2$. Each of these will be irrational.
Hint: assume the contrary. Then you have a whole bunch of numbers corresponding to different choices of signs ($2^{2009}$ of them), and all these numbers are rational. Then any algebraic combinations of these numbers, such as sums and products, will also be rational. Can you use this to arrive at a contradiction?
You may exploit the fact that $2003$ is a prime number. Take every sign as positive, except the sign of $\sqrt{2003}$, and assume that: $$\sqrt{1}+\sqrt{2}+\ldots+\sqrt{2002}\color{red}{-\sqrt{2003}}+\sqrt{2004}+\ldots+\sqrt{2009} = \frac{p}{q}. \tag{1}$$
Now take a uber-huge prime $P$ such that $P>q$ and every prime in the range $[2,2009]$, with the exception of $2003$, is a quadratic residue $\!\!\pmod{P}$. Such monster exists by Dirichlet's theorem and the quadratic reciprocity theorem. So $\sqrt{1},\sqrt{2},\ldots,\sqrt{2002},\sqrt{2004},\sqrt{2005},\ldots,\sqrt{2009}$, $p$ and $q^{-1}$ can be intepreted as elements of $\mathbb{F}_P$, but $\sqrt{2003}$ does not belong to $\mathbb{F}_P$ by construction, hence $(1)$ is not possible and $$-2\sqrt{2003}+\sum_{k=1}^{2009}\sqrt{k}\not\in\mathbb{Q}.\tag{2}$$ The same argument also works by replacing $2003$ with $1999,1997,1993$ or any prime in the range $[1004,2009]$.
• Just to confirm, this shows in fact that every choice of signs would lead to an irrational number and moreover it would work for every cut-off other than $2009$ (except $1$ of course)? – quid Jul 23 '16 at 14:57