I want to prove that set of all dyadic rational numbers in $[0,1]$ is dense in $[0,1]$ but I do not want to prove it using binary expansion of a number. Is there any other proof for the same?
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Hi, any reference/document on dyadic rational numbers and its properties? thanks – Rémy Hosseinkhan Boucher Feb 23 '21 at 11:10
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HINT $\rm\quad r < \dfrac{m}{2^n} < s\ \iff\ 2^n\: r < m < 2^n\:s\:.\: $ For $\rm\:r < s\:$ such integers $\rm\:m,n\:$ necessarily exist since, by the Archimedean property, $\rm\ 2^n\: (s-r) > 1\ $ for large $\rm\:n\:,\:$ so said interval must contain an integer $\rm\:m\:$ since the interval has length $> 1\:.$
Bill Dubuque
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Given $\epsilon>0$, take $n\in\mathbb{N}$ such that $2^{-n}<\epsilon/2$. Then any interval of length $\epsilon$ contains at least one point of the form $k\,2^{-n}$, $0\le k\le 2^n$.
Julián Aguirre
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+1 The key fact here is that the sequence $2^{-n}$ converges to zero in $[0,1]$, which is "obvious" but a bit tricky to prove from first principles. – hardmath Nov 22 '11 at 15:23
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If $x \in [0,1]$ then $2^n x$ is some real number. Let $y_n$ be the integer part of this number. Then ${\displaystyle|{y_n \over 2^n} - x| \leq {1 \over 2^n}}$.
Zarrax
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