If $b \in \mathbb{R}\setminus \mathbb{Q}$ then are there $h$ very close to $0$ such that $b+h \in \mathbb{Q}$?

Question

This question occured to me when I was solving a problem in analysis. Rephrased in a different way: Is it true that given an $b \in \mathbb{R}\setminus \mathbb{Q}$, for all $\varepsilon >0$ there exists $0<|h|<\varepsilon$ such that $b+h \in \mathbb{Q}$?

I'm pretty sure I don't have the tools to answer this question and Google search also doesn't guide me the right direction.

Answer 1

By the Archimedean property, there is a rational number $q$ with $0<q<\varepsilon$ (there is some natural number $n>\frac1\varepsilon$, setting $q=\frac1n$ works).

There must be some multiple of $q$ (say $kq$ for some $k\in\Bbb Z$) between $b$ and $b+\varepsilon$. Then $h=kq-b$ is the number you're looking for.

Answer 2

Since the rationals are dense in $\Bbb R$, there is some $q\in\Bbb Q\cap(b,b+\varepsilon)$. So, take $h=q-b$.