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A classic exercise is in real analysis is to prove that for a fixed $\alpha\notin \mathbb{Q}$, $$\{ m+ \alpha n\mid m,n \in \mathbb{Z} \}$$ is dense in $\mathbb{R}$. Will this statement still be true if the parameter $\alpha$ is a rational number? I need to stress here that r is a fixed rational number.

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If you set $m=0$ and $n=1$ you get $\mathbb Q$ itself, which is certainly dense. It can't get less dense by adding points for other values of $m$ and $n$. –  Henning Makholm Jan 30 '12 at 0:45
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By the way, I think the "classic" exercise is to prove for every fixed irrational $\alpha$ that $\{m+n\alpha\mid m,n\in\mathbb Z\}$ is dense, which is not what you have written. This will not be true if $\alpha$ is rational -- if $\alpha=p/q$, then every point in the set will be a multiple of $1/q$. –  Henning Makholm Jan 30 '12 at 0:47
    
Thanks! That's actually what I meant to ask... –  Yang Jan 30 '12 at 0:51
    
@Yang: Edited to say what you meant, then. –  Arturo Magidin Jan 30 '12 at 0:59
    
Thank you so much! –  Yang Jan 30 '12 at 1:05

1 Answer 1

up vote 1 down vote accepted

For a fixed $\alpha\in\mathbb{Q}$, write $\alpha = \frac{a}{b}$ with $\gcd(a,b)=1$, $b\gt 0$. Then $$\{m+\alpha n\mid m,n\in\mathbb{Z}\} = \left\{r\in\mathbb{Q}\;\left|\; r=\frac{t}{v},\ t,v\in\mathbb{Z}, \gcd(t,v)=1, v\gt 0, v|b\right\}\right..$$ To prove the right hand side is contained on the left hand side, given any $\frac{t}{v}$ as given, write $\frac{t}{v} = m + \frac{q}{v}$, with $m\in\mathbb{Z}$, $0\leq q\lt v$; then $\gcd(q,v)=1$ (since $t = vm + q$ is relatively prime to $v$). Since $v|b$, we can write $b=vs$. So $\frac{t}{v} = m + \frac{qs}{b}$. Since $\gcd(a,b)=1$, there exist $k,\ell$ such that $ka+\ell b = 1$. Hence $kqsa + \ell qsb = qs$. Hence $$\frac{t}{v} = m+ \frac{qs}{b} = m + \frac{kqsa + \ell qsb}{b} = m + \ell qs + kqs\alpha.$$ For the converse inclusion, any element on the left hand side can be written as $$ m + n\alpha = \frac{bm+an}{b},$$ which when written in lowest terms will be of the form $\frac{t}{v}$ with $v|b$. This gives equality.

However, the right hand side describes a discrete subset of $\mathbb{R}$ (since any two elements of the set are at least $\frac{1}{b}$ apart), hence it is not dense. It is none other than the group $\mathbb{Z}[\frac{1}{b}]$.

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