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?
Answer
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.
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