So this question is actually $3$ parts.

  • a) Prove $n < 2^n$ which I have already done.

  • b) Prove that for every real number $\epsilon > 0$, there exists a natural number $n$ such that $0 < \frac{1}{2^n} < \epsilon$.

  • c) A number of the form $\frac{m}{2^n}$ where $m$ and $n$ are integers with $n \geq 0$ is called a dyadic rational number. Prove that for any $x, y \in\mathbb{R}$ with $x < y$, there exists a dyadic rational number $r \in \mathbb{R}$ such that $x < r < y$.

I've already solved part a. I'm not sure where to go on part b and c. I'm thinking I need to use the Archimedean property, but I"m not sure how.

  • $\begingroup$ Note that $\frac{1}{2^n}<e$ is equivalent to $\frac1e<2^n$. $\endgroup$
    – Arthur
    Jan 30 '17 at 22:24

If you showed that $n < 2^n$ for every positive integer $n$, then you've also shown that

$\frac{1}{2^n} < \frac{1}{n}$

for all positive integers. Given $\varepsilon$, it should be easy now to find $n$ large enough to satisfy the relation in part $b)$

For part $c)$, pick $n$ large enough that $q: = \frac{1}{2^n} < y -x$. Then consider multiples $mq$ of this $q$. In particular, consider the smallest multiple $mq$ in which $x < mq$. By choice, $(m-1)q \leqslant x$, so that

$mq = (m-1)q + q < x + (y-x) = y$

Thus $x < mq < y$. Of course, $mq = \frac{m}{2^n}$ is dyadic

  • $\begingroup$ Okay, we have a corollary to the Archimedean principle stating "For any positive real number e, there exists a natural number n such that 0 < 1/n < e." So, for part b, I'm just making recognizing that 1/n < 1/2^n, so we know that 0 < 1/n < 1/2^n < e, which ends the proof for that one. I'm not sure how to format this properly by the way. $\endgroup$
    – user21
    Jan 30 '17 at 22:35
  • $\begingroup$ It's not $\frac{1}{n} < \frac{1}{2^n}$, but rather the other way around: $\frac{1}{2^n} < \frac{1}{n}$. The second of these inequalities (which is the true one) is what you have already proven. Next, to have $\frac{1}{n} < \varepsilon$ is the same as having $1 < n \varepsilon$. This is where the archimedian principle comes in: there must be an integer multiple of $\varepsilon$ that surpasses $1$. $\endgroup$
    – joeb
    Jan 30 '17 at 22:39
  • $\begingroup$ Okay. Using the corollary, we know that 1/2^n < e though, right? We just don't know that 0 < 1/2^n. That's what I'm trying to prove. Correct? $\endgroup$
    – user21
    Jan 30 '17 at 22:45
  • $\begingroup$ If you want to use the corollary, then yes you have $1/2^n < 1/n < \varepsilon$. Proving that $0 < 1/2^n$ should be easy. You know that $\frac{1}{2}$ is positive. So multiplying it to itself any number of times must result in a positive number, and thus $\frac{1}{2^n} = \frac{1}{2} \cdots \frac{1}{2} > 0$ (with $n$ factors in the product of halves). Stay with my edits, I keep making typos. $\endgroup$
    – joeb
    Jan 30 '17 at 22:51
  • $\begingroup$ Yeah, that makes sense. Thanks. Also, we have a proposition stating that if 0<x and x<y, then 0<y^-1 and y^-1 < x^-1. Couldn't I just use that proposition to say that since 0<n and n<2^n, then 0<1/2^n and 1/2^n < 1/n and then combine it with the corollary to say that 0 < 1/2^n < e? $\endgroup$
    – user21
    Jan 30 '17 at 23:06

Hint :Part B :for fixed $e$$$\dfrac{1}{2^n}<\epsilon \to 2^n>\dfrac1\epsilon \to\\log_{2}{2^n}>\log_{2}{\dfrac1\epsilon} \to n>\log_{2}{\dfrac1\epsilon} \to n \geq \lfloor \log_{2}{\dfrac1\epsilon}\rfloor +1$$


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