[Math] Prove that for every real number $\epsilon > 0$, there exists a natural number $n$ such that $0 < 1/2^n < \epsilon$.

real numbersreal-analysis

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.

Best Answer

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

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