Suppose an equivalence relation is defined as: for $a, b \in \mathbb{R}$, $a R b \iff a - b = n/m$, where we have $n, 0 \neq m \in \mathbb{Z}$. For any natural number $N$ and any $a\in \mathbb{R}$, show that there is a $b\in \mathbb{R}$ so that $b R a$ and $b< \frac{1}{N}$ and $b \geq 0$.
My attempt:
Remember for any natural number $N$, $\frac{1}{N} > 0$. Thus, given $0$ (a real number) and $\frac{1}{N} \in \mathbb{N} \subset \mathbb{R}$, since $\mathbb{Q}$ is dense in $\mathbb{R}$, there is some rational $\frac{s}{t}$ so that $0 \leq \frac{s}{t} < \frac{1}{N}$. Now, since $\frac{s}{t} \in \mathbb{Q}$, for any given real $a$, there is some real $b$ so that $b - a = \frac{s}{t}$. Therefore, we have an equivalence relation and the inequality satisfied.
QED.
Is this proof correct? Did I make a mistake somewhere? Any assistance is appreciated.