Is there any sequence that covers all the irrational numbers? Like $\forall x \in I$, $\exists n\in \mathbb{N}$ s.t. $a_n = x$.
I think the answer is not, but I need help to prove this, if you have some argument please comment on it.
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4The set of irrational numbers is uncountable, therefore it is not possible. – Vitor Sep 24 '21 at 18:27
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3Cantor's diagonal argument – TonyK Sep 24 '21 at 18:33