Tell me more ×  As the topic, why for all open interval in
$\mathbb{R^1}$ it must contains both rationals and irrationals
numbers, exist any prove?  share|improve this question add
comment  1 Answer  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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