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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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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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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