I'm trying to prove a more complicated theorem, but I'm stuck on a part of the proof which I have condensed to the following lemma, which I want verification of:
Lemma: For all irrational $r \in (0, 1)$ and real $0 < m < 1$, there exists a real $x \in (0, m)$ and an integer $n$ such that $n + x$ is a multiple of $r$.
Proof. We want $n + x = rk$ for some integer $k$. Then $$x = rk - n,$$ and $$\text{frac}(x) = \text{frac}(rk - n) = \text{frac}(rk).$$ So we need $$x = \text{frac}(rk)$$ for some integer $k$. If for integer $j$ we have $$j \leq rk < j + 1,$$ then $$x = \text{frac}(rk) = rk - j,$$ which we require for some $0 < x < m$. Then all we need is $$0 < rk - j < m,$$ $$j < rk < j + 1$$ for some integers $k, j$. Manipulating the second equation gives us $$0 < rk - j < m,$$ $$0 < rk - j < 1.$$ Since $m < 1$ we need only look at the first equation. Then we have $$0 < rk - j < m$$ $$\frac{j}{k} < r < \frac{j}{k} + \frac{m}{k}.$$ Then all we need is to find a rational number $\alpha = \frac{j}{k}$ so that $$\alpha < r < \alpha + \beta,$$ where $\beta = \frac{m}{k} > 0$. Because the rationals are dense in the reals, such an $\alpha$ must exist between the two real numbers $r - \beta$ and $r$ for any $\beta > 0$, that is, $$r - \beta < \alpha < r.$$ Such an $\alpha$ satisfies the above requirement that $$\alpha < r < \alpha + \beta,$$ and the proof is complete.
Does this all seem right? The only thing I can think of that might not work is the last part, where I say that $\alpha$ exists between $r - \beta$ and $r$. Does the fact that $\beta$ depends on $\alpha$ matter for this statement?
Otherwise, is there anything I can do to simplify the proof or make it more clear?
