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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}$.

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Your title does not seem to ave any connection with the actual question. –  Mariano Suárez-Alvarez Feb 21 '13 at 9:08
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2 Answers

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.

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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 $2$.

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Interesting argument, but what you've proved is not actually what you claim to have proved. Rather, from your first displayed line, if $\alpha_2\notin\mathbb{Q}$ then it must be that $n\frac{p}{q}+m=0$ and $k=0$, which more or less means there is a unique way of solving this problem. –  Sean Eberhard Feb 21 '13 at 12:51
    
@SeanEberhard thanks. You're right. I ignored the possibility of having $n\frac{p}{q} + m =0$. I fail to see how this implies uniqueness. –  Teddy Feb 21 '13 at 13:27
    
when $\alpha_2\notin\mathbb{Q}$, you've showed that the only solutions to $n\alpha_1 + m\alpha_2 \in \mathbb{Z}$ are of the form $n=tq$ and $m=-tp$ with $t\in\mathbb{Z}$, assuming that $(p,q)=1$. You can of course check that these are solutions (say by cross-multiplying in the equation $\frac{\alpha_1}{\alpha_2} = \frac{p}{q}$). –  Sean Eberhard Feb 21 '13 at 18:20
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