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...?
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 $x=r+\dfrac{1}{n\sqrt2}\notin\mathbb Q$
$$\left|x-r\right| = \left| r+\frac{1}{n\sqrt2}-r\right|= \dfrac{1}{n\sqrt2} \lt \varepsilon $$
"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
– terrible at math Feb 06 '14 at 20:45this makes PERFECT sense to me, i just see a question like this and never know where to begin, I guess i just need more practice and exposure.
– terrible at math Feb 06 '14 at 21:04Let $q$ be rational. Let $a_n = q-1/(n\cdot\pi)$. This sequence converges to $q$ and consists of irrationals only.
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 $\,\Bbb R = \Bbb Q.$
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.
$\pi$ 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.
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 $q + \epsilon$.
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 $$ x-\varepsilon<x-\frac{1}{n}<\frac{\lfloor nx\rfloor}{n}=\frac{m}{n}\le x. $$
