The distance between the integral multiples of two numbers Given a positive real number $r \in \left(0,\infty\right)$, define $\left[r\right]$ to be the set of positive integral multiples of $r$:
$$
\left[r\right] := \left\{nr\ :\mid\ n \in \mathbb{N}_1\right\}
$$
Given two positive real numbers $a, b \in \left(0,\infty\right)$, define the distance between them, $d\left(a,b\right)$, to be
$$
d\left(a,b\right) := \inf_{x\in\left[a\right],\ y\in\left[b\right]}\left|x - y\right|
$$
For example, if $a, b \in \mathbb{N}_1$, then $d\left(a,b\right) = 0$, since $ab \in \left[a\right]\cap\left[b\right]$.
Are there any two numbers $a,b\in \left(0,\infty\right)$ with $d\left(a,b\right) > 0$?

No. If $a/b$ is rational, say, $a/b=p/q$, then $qa=pb$, so the distance is $0$.
On the other hand if $a/b$ is irrational, then the fractional parts of the integer multiples of $a/b$ are dense in $[0,1]$, so in particular there are fractional parts arbitrarily close to $0$ -- which correspond to multiples of $a$ arbitrarily close to some multiple of $b$. Again the infimum of the distances is $0$.

Here's an alternative phrasing that doesn't depend on knowing that the fractional parts of the multiples of an irrational number are dense:
Choose any $\varepsilon>0$, and I will show that $d(a,b)<\varepsilon$. Assume wlog that $\varepsilon < a$.
Consider the sequence $(na\bmod b)_n$. Its values lie in $[0,b]$ which is compact, and therefore it has a convergent subsequence $(n_ia \bmod b)_i$ (where the $n_i$s are some strictly increasing sequence of naturals). Because this subsequence is in particular Cauchy, there is an $N$ such that
$$|(n_{N+1}a \bmod b)-(n_Na \bmod b)|<\varepsilon$$
But that means that for some $k_1, k_2 \in \mathbb Z$ we have
$$ |(n_{N+1} a - k_1 b) - (n_N a - k_2 b)| = |(n_{N+1}-n_N)a - (k_1-k_2)b | < \varepsilon $$
Because $n_{N+1}-n_N \ge 1$ we have $(n_{N+1}-n_N)a\in [a]$. Also, we assumed that $\varepsilon < a$, so $k_1-k_2$ needs to be positive, and therefore $(k_2-k_2)b \in [b]$.
This shows that $d(a,b) < \varepsilon$.
