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Show that arbitrarly close to any rational number there is a real (non-rational) number. In other words, show that to each real $\varepsilon>0$ and each rational $r\in\mathbb Q$ there exists $x\in\mathbb R\setminus\mathbb Q$ with $\left|x-r\right|\lt\varepsilon $

No idea how to prove this one. Perhaps I can define some sort of sequence and show it converges...?

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marked as duplicate by Magdiragdag, nbubis, user61527, Davide Giraudo, Lost1 Feb 6 '14 at 21:22

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For every $n\in\mathbb N$ you have

$$\sqrt2\notin \mathbb Q \Rightarrow \dfrac{1}{n\sqrt2}\notin\mathbb Q$$

Now let $\varepsilon >0$, then $n$ can be found such that

$$\dfrac{1}{n\sqrt2} \lt \varepsilon$$

Now for arbitrary $r\in\mathbb Q$ and given $\varepsilon>0$ chose $x=r+\dfrac{1}{n\sqrt2}\notin\mathbb Q$

$$\left|x-r\right| = \left| r+\frac{1}{n\sqrt2}-r\right|= \dfrac{1}{n\sqrt2} \lt \varepsilon $$

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  • $\begingroup$ these types of proofs always make sense to me but confuse me too: "Now let $\epsilon > 0$, then $n$ can be found such that" is this an obvious fact? how do we know that? I'm never sure of what is needed to be proven or not in these sorts of things $\endgroup$ – terrible at math Feb 6 '14 at 20:45
  • $\begingroup$ @terribleatmath For any real number $r$ there is a natural number $n$ such that $n>r$, so the statement in my proof can be reformulated as: "there is an $n$ such that $n> \dfrac{1}{\varepsilon\sqrt2}$" $\endgroup$ – user127.0.0.1 Feb 6 '14 at 20:48
  • $\begingroup$ To find such an $n$ you just have to "round up" to the next integer $\endgroup$ – user127.0.0.1 Feb 6 '14 at 20:50
  • $\begingroup$ it may seem like a silly question, but how did you get fromm the second to last step to the last step? mainly, the absolute value sign? $\endgroup$ – terrible at math Feb 6 '14 at 20:54
  • $\begingroup$ I'am not sure what exactly you mean with your question, but I made a little edit. Is it more clear now? $\endgroup$ – user127.0.0.1 Feb 6 '14 at 20:57
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Let $q$ be rational. Let $a_n = q-1/(n\cdot\pi)$. This sequence converges to $q$ and consists of irrationals only.

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Hint $\ $ If not there exists a nonempty interval $I$ containing only rationals. Repeatedly shifting $I$ left or right by a fixed rational less than the length of $I$ covers the real line with rationals, so $\,\Bbb R = \Bbb Q.$

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If it weren't true, then you could find an open interval around a rational number which contains only rational numbers. But this contradicts the countability of the rationals, since every interval of positive length contains uncountably many points.

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  • $\begingroup$ this makes sense, but 2 things: we haven't talked about open intervlas in the class yet so i'm not sure I could use that, and we're to prove it the way set out in this problem, the convergence to this $\epsilon$ $\endgroup$ – terrible at math Feb 6 '14 at 20:30
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$\pi$ is irrational, so also $k\pi$ is irrational for any rational $k$. Find any number $a<\epsilon/\pi$. If $a$ is irrational, $r+a$ is also irrational. If $a$ is rational, $r+\pi a$ is irrational.

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Say they give you a rational $q$, and you are to show that there is an irrational $r$ nearer than $\epsilon$ from $q$. Select $n$ natural such that $\frac{1}{n} < \epsilon$, then $q + \frac{\sqrt{2}}{2 n}$ is irrational, and between $q$ and $q + \epsilon$.

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Let $x$ be an irrational and $n$ a positive integer, then $$ nx-1<\lfloor nx\rfloor\le nx $$ and thus $$ x-\frac{1}{n}<\frac{\lfloor nx\rfloor}{n}\le x $$ This means that within $\dfrac{1}{n}$ away from $x$ there is arational $\dfrac{\lfloor nx\rfloor}{n}$

In particular, if we want the difference less that $\varepsilon>0$, we first note that if $$ n=\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1, $$ then $$ n>\frac{1}{\varepsilon} \quad \Longrightarrow\quad \varepsilon >\frac{1}{n}, $$ and hence $$ x-\varepsilon<x-\frac{1}{n}<\frac{\lfloor nx\rfloor}{n}=\frac{m}{n}\le x. $$

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