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When the interval is between two rational numbers it is easy. But things get complicated when the interval is between two irrational numbers. I couldn't prove that.

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This is false as stated; $\{0\} = [0,0]$ does not contain an irrational number. –  Michael Albanese Jan 10 '13 at 9:34
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My guess is that "interval" here means any set $(a,b)$ where $a < b$. This is, for instance, how "intervals" are defined in Russian textbooks. Maybe it is also the case in some other languages. Am I right? –  Dan Shved Jan 10 '13 at 9:40
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2 Answers 2

Supposing you mean an interval $(x,y)$ of length $y-x=l>0$ (it doesn't matter whether $l$ is rational or irrational), you can simply choose any integer $n>\frac1l$, and then the interval will contain a rational number of the form $\frac an$ with $a\in\mathbf Z$. Indeed if $a'$ is the largest integer such that $\frac{a'}n\leq x$ (which is well defined) then $a=a'+1$ will do. By choosing $n>\frac2l$ you even get two rationals of this form, and an irrational number between those two by an argument you claim to already have.

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Hint: Use this fact that for any irrational $\xi\in\Bbb R$, there is a sequence $\{a_n\}\in\Bbb R $ established by rational numbers such that $$a_n\to\xi$$ Use the general topology on $\mathbb R$ also.

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Nice hint + 1 :-) –  amWhy Feb 19 '13 at 0:21
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