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If $x \in \mathbb{R}$ and $n \in \mathbb{N}$ then there exists $p, q \in \mathbb{Z}$ such that $$\left|x − \frac{p}{q}\right| < \frac{1}{n}.$$

Do we use the Archimedian Principle to prove this?

alkabary
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Susan
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  • Since the rationals are dense in $\mathbb{R}$, then if $x \in \mathbb{R}$ for any $\varepsilon > 0 $ there exist $r=r(\varepsilon)\in \mathbb{Q}$ such that $$ |x-r|<\varepsilon $$ Therefore, take $\varepsilon =1/n$ and you are done. Note that the density of $\mathbb{Q}$ in $\mathbb{R}$ is proved using the Archimedean Principle – Alonso Delfín Jul 05 '15 at 22:44
  • possible duplicate of Proving the rationals are dense in R –  Jul 05 '15 at 22:57

4 Answers4

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Consider nx. Let A be the largest integer ≤ nx. (we have implicitly used the Archimedean Principle here). Then, of course, nx ≤ A + 1 It follows that either nx - A < 1 or nx = A + 1 . In the first case, A/n is the rational number you seek, in the second case it is (A+1)/n.

lulu
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Arquimedian principle says that for every $\delta>0$ there is a $n$ such that $\frac{1}{n}< \delta$. Consider $C = \{\frac{k}{n}, k \in \Bbb{Z}\}$. There is a $k^*$ such that

$$ \frac{k^*}{n}\leq x < \frac{k^* +1}{n}$$ this implies that $$\bigg|x - \frac{k^*}{n} \bigg| \leq \frac{1}{n}$$

remark: take $k^*: k^*\leq n x < k^* + 1$.

Edit:Note that we do use the Arquimedian principle to find an integer $k^*+1$that is greater than $nx$

Conrado Costa
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  • Also the pigeon-hole principle. – marty cohen Jul 05 '15 at 21:42
  • Thanks for your help! I'm wondering if you could clarify how you get to the absolute value line? It seems to me you can only go to the abs. value when you have equivalent statements on both sides of x (-m < x < m)? – Susan Jul 05 '15 at 21:51
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You can also use the fact that Rationals are dense in the Reals. If $x \in \mathbb Q$, you're done. If not, take the decimal expansion of $x$, i.e.,

$$x=a.a_0a_1,.....,a_n... $$

Then cut of the expansion at some fixed $a_k$ to get a Rational approximation by $10^{-n}$ by the rational $\frac {aa_0a_1....a_k}{10^n}$. Then you can use that $10^{-n} <\frac{1}{n}$.

Gary.
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Here's a cleaner version of Gary's argument. Pick $x\in\Bbb R$ and let $A=\{a\in\Bbb R: |x-a|<1/n\}$. Clearly $A=(x-1/n,x+1/n)$ and since $\Bbb Q$ is dense in $\Bbb R$, there is a rational number $r$ such that $R\in A$.

Tim Raczkowski
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