Proof Irrational number $x$, given $N\in \mathbb{N}$ exists $\u03f5>0$ so all rationals $\in {V}_{\u03f5}(x)$ have denominator larger than $N$.

antennense
2022-07-11
Answered

Proof Irrational number $x$, given $N\in \mathbb{N}$ exists $\u03f5>0$ so all rationals $\in {V}_{\u03f5}(x)$ have denominator larger than $N$.

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