Real Analysis
Real Analysis
Let $S=[0,1]$. If $x$ and $y$ are in $S$ with $x\neq y$. How can we show that there are $m,n\in \mathbb{N}$ such that $x< \dfrac{m}{2^n} <y$. Can the Archimedean Property be used to prove this? If yes, could anyone provide me an insight to do this?
Re: Real Analysis
Hi sheriph.
I am giving you a hint for the moment so that you have the joy to complete it.
$ x < \dfrac{m}{2^n} < y \ \iff\ 2^n\: x < m < 2^n\:y \:.\: $ For $\:x < y\:$ such integers $\:m,n\:$ necessarily exist since, by the Archimedean property, $\ 2^n\: (y-x) > 1\ $ for large $\:n\:,\:$ so said interval must contain an integer $\:m\:$ since the interval has length $> 1\:.$
I am giving you a hint for the moment so that you have the joy to complete it.
$ x < \dfrac{m}{2^n} < y \ \iff\ 2^n\: x < m < 2^n\:y \:.\: $ For $\:x < y\:$ such integers $\:m,n\:$ necessarily exist since, by the Archimedean property, $\ 2^n\: (y-x) > 1\ $ for large $\:n\:,\:$ so said interval must contain an integer $\:m\:$ since the interval has length $> 1\:.$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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