Two reals with rational quotient, Z-span the integers

Question

Given two real numbers $\alpha_1,\alpha_2\in\mathbb{R}\cap(0,1)$, with rational ratio $\frac{\alpha_1}{\alpha_2} \in \mathbb{Q}$,

show there exist $m,n\in \mathbb{Z}$, such that $n\alpha_1 + m\alpha_2 \in \mathbb{Z}$, but

$n\alpha_1 \notin \mathbb{Z}$, or

$ m\alpha_2 \notin \mathbb{Z}$.

Answer 1

Hint: Write out what it means for $\displaystyle \frac{\alpha_1}{\alpha_2}\in\mathbb{Q}$, cross multiply, and then divide by appropriately large integers to get non-integers.

Answer 2

This is actually wrong, I now realize. For assume $n\alpha_1 + m\alpha_2 = k\in \mathbb{Z}$. Let $\frac{\alpha_1}{\alpha_2}=\frac{p}{q}\in \mathbb{Q}$. Then $\alpha_1 = \frac{p}{q} \alpha_2$, and $$ n\alpha_1 + m\alpha_2 = n\frac{p}{q} \alpha_2 + m \alpha_2= (n\frac{p}{q} + m) \alpha_2=k.$$ This implies $\alpha_2\in\mathbb{Q}$. A similar argument works for $\alpha_1$, so actually $\alpha_1, \alpha_2\in \mathbb{Q}$ is a necessary condition.

It's wrong even for rationals. Take $\alpha_1=\frac{1}{2}$, $\alpha_2=\frac{1}{3}$. Assume $n\frac{1}{2} + m\frac{1}{3} = 1$. This is equivalent to $$ n = -\frac{2}{3}m +2.$$ But for $n$ to be an integer, $m$ must be a multiple of $3$, which leaves $n$ a multiple of $2$.