My textbook says the following:
Let $a$ be any irrational number. Now consider $a(2n+1)\pi$ for various $n\in\Bbb{N}$. For any $\epsilon>0$, it is possible to choose an $n\in\Bbb{N}$ and a number $b\in\Bbb{N}$ such that $|b-a(2n+1)\pi|<\epsilon$. In other words, it is possible to make $a(2n+1)\pi$ "almost" an integer with an arbitrary degree of precision.
This appears likely. However, I can't seem to think up a proof. Any hints for such a proof would be great.
In general: no. If $a<0$, consider $\epsilon=3\pi|a|$. Then for all $n\in\mathbb N$, $b\in\mathbb N$, we have $|b-a(2n+1)\pi|\ge 3\pi |a|=\epsilon$.
Otherwise (e.g. if you switch to $\mathbb Z$ instead of $\mathbb N$), the irrationality of $a$ does not play a role. The answer depends on whether or note $\pi a$ is irrational. If $a\pi$ is rational, we may have bad luck (namely when the denominator in shortest terms is even). Assume therefore that $a\pi$ is irrational. By the density of $\{\,ka\pi\bmod 1\mid k\in\mathbb Z\,\}$ in $[0,1]$ we can surely find $b$ and $k$ such that $|b-ak\pi|<\epsilon$. However, can we ensure that $k$ is odd? Yes: Find $k$ with $0<ak\pi\bmod 1<\epsilon$. If $k$ is odd, we are done. Otherwise, the numbers $rak\pi\bmod 1$, $r\in \mathbb N$, walk trhogh $[0,1]$ in steps of size $<\epsilon$, hence for suitable $r$ differ from $a\pi\bmod 1$ by less than $\epsilon$. We conclude that $(rk-1)a\pi$ differes from an integer by less than $\epsilon$.
