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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$\begingroup$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 $x=r+\dfrac{1}{n\sqrt2}\notin\mathbb Q$
$$\left|x-r\right| = \left| r+\frac{1}{n\sqrt2}-r\right|= \dfrac{1}{n\sqrt2} \lt \varepsilon $$
$\endgroup$ 8 $\begingroup$Let $q$ be rational. Let $a_n = q-1/(n\cdot\pi)$. This sequence converges to $q$ and consists of irrationals only.
$\endgroup$ $\begingroup$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 $\,\Bbb R = \Bbb Q.$
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$$\pi$ is irrational, so also $k\pi$ is irrational for any rational $k$. Find any number $a<\epsilon/\pi$. If $a$ is irrational, $r+a$ is also irrational. If $a$ is rational, $r+\pi a$ is irrational.
$\endgroup$ $\begingroup$Say they give you a rational $q$, and you are to show that there is an irrational $r$ nearer than $\epsilon$ from $q$. Select $n$ natural such that $\frac{1}{n} < \epsilon$, then $q + \frac{\sqrt{2}}{2 n}$ is irrational, and between $q$ and $q + \epsilon$.
$\endgroup$ $\begingroup$Let $x$ be an irrational and $n$ a positive integer, then $$ 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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