If two real numbers are immesurable can an integer sum between the two get as close to any real number as we like?

Question

Say for example we have $\pi$ and $1$. Can the sum $m\pi + n$ for $ m, n\in \mathbb{Z}^+$ get as close to a real number as we like? At first I tried using the fact that you could have $n = -floor(m \pi)$ and then use the fact that the sum would always end up between 0 and 1 and that it could never repeat for a specific $m$ to apply a pigeon hole argument. I don't believe this is a good strategy because there are also uncountable holes. Thanks.

Answer 1

Yes.

If $\alpha\in\Bbb R$ is irrational, then $S:=\Bbb Z+\alpha\Bbb Z$ is dense in $\Bbb R$. Assume otherwise, i.e., we find $u\in\Bbb R$ and $n\in\Bbb N$ such that $(u,u+\frac1n)\cap S=\emptyset$. Then $S\cap(0,\frac1n)=\emptyset$, for if $0<|x|<\frac1n$, then $x\Bbb Z$ intersects $(u,u+\frac1n)$. Conclude that $s:=\inf(S\cap (0,\infty))$ is $\ge\frac1n$. If $x\in S$, then there eixts $k\in \Bbb Z$ with $ks\le x<(k+1)s$. From $x-ks\in S\cap[0,s)$, we conclude $x=0$, hence $S=s\Bbb Z$. In particular $\alpha=ms$ amd $1=ns$ for some $m,n\in\Bbb Z$. Then $\alpha=\frac mn$, contradicting its irrationality.

Answer 2

Alternative approach: assuming that $\alpha$ is irrational we have that the convergents of its continued fraction fulfill $\left|\alpha-\frac{p}{q}\right|\leq\frac{1}{q^2}$, hence zero is an accumulation point for $E=\mathbb{Z}+\alpha\mathbb{Z}$, since $\left|q\alpha-p\right|\leq\frac{1}{q}$. On the other hand $E$ is an additive subgroup of $\mathbb{R}$, hence $E$ is dense in $\mathbb{R}$.