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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?

quid
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ITChristian
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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
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    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
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    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
  • By Galois, all $2^{2009}$ choices of signs produce irrationals, don't they? While things like $\sqrt 2+\sqrt 8-\sqrt{18}$ can cancel, nothing can do so with $\sqrt p$ when $p>2009$ (or even $9p>2009$) – Hagen von Eitzen Nov 22 '20 at 10:02

3 Answers3

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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.

Ross Millikan
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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?

Dan Shved
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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 $[1004,2009]$.

Jack D'Aurizio
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  • 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
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    @quid: with almost no efforts, this shows that if the only negative/positive sign occurs at the square root of a prime number, the whole sum is an irrational number. But ultimately you're right, the argument can be refined to show that actually every choice of signs leads to an irrational sum. – Jack D'Aurizio Jul 23 '16 at 14:59