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As the topic, why for all open interval in $\mathbb{R^1}$ it must contains both rationals and irrationals numbers, exist any prove?

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1 Answer 1

This must be proved in two stages. First, given an open interval $(a,b)$, we must show that it contains some rational number $q$, then we must show that it contains some irrational number $r$.

@AsinglePANCAKE's answer already has a link to the first stage. So consider this done. That is, there is $q\in{\Bbb Q\cap(a,b)}$. Now, if $b\in{\Bbb Q}$, then let $r=q+(b-q)/\sqrt 2$, which is both irrational and inside $(a,b)$. Alternatively, if $b$ is irrational, then let $r=(q+b)/2$, which is irrational and inside $(a,b)$.

So we always have rational $q$ and irrational $r$ inside $(a,b)$.

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