- By hypothesis, $x\in\mathbb R$ is a real number and $\epsilon > 0$ any (arbitrarily small) positive real number.
- We are going to find a rational number $q\in\mathbb Q$ with $q\in(x-\epsilon, x+\epsilon)$ using a constructed convergent rational sequence.
- First, we choose two rational numbers $\frac {a_0}1,\frac{b_0}1\in\mathbb Q$ with some integers $a_0 < x < b_0.$ This is possible because of the Archimedean axiom.
- Note that $x\in [a_0,b_0],$ where $[a_0,b_0]$ is a closed interval.
- Now, we construct for $n=1,2,\ldots$ new closed intervals by the following rule:
- if $\frac{a_n+b_n}2 > x$, we set $a_{n+1}:=a_n$ and $b_{n+1}=\frac{a_n+b_n}2,$
- otherwise, we set $a_{n+1}:=\frac{a_n+b_n}2,$ and $b_{n+1}:=b_n.$

- By construction, the sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are rational sequences with $x\in [a_{n},b_{n}]$ for all $n:=1,2,\ldots.$
- Since the length of the $n$th interval is $(b_0-a_0)/2^n,$ the sequences are also real convergent sequences with the limit $$x=\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n.$$
- Although the limit $x$ does not have to be a rational number
^{1}itself, all sequence members $a_n,b_n$ are, by construction, rational numbers for all $n\in\mathbb N.$ - From the definition of convergence, it follows that, taking $(a_n)_{n\in\mathbb N}$ as an example, there is an index $N\in\mathbb N$ such that $|a_n-x| < \epsilon$ for all $n > N.$

- By setting $q:=a_N\in (x-\epsilon, x+\epsilon),$ we have found a rational number that lies arbitrarily dense with the real number $x.$

^{1} Take $x=\sqrt{2}$ as an example.

q.e.d

| | | | created: 2020-07-05 15:20:23 | modified: 2020-07-05 15:20:40 | by: *bookofproofs* | references: [8311]

[8311] **Modler, F.; Kreh, M.**: “Tutorium Analysis 1 und Lineare Algebra 1”, Springer Spektrum, 2018, 4. Auflage