The smallest positive integer vector from a positive rational vector Suppose $\mathbf{q} = \left[\begin{array}{cccc}q_1 & q_2& \dots &q_n\end{array}\right]\in \mathbb{Q}_{>0}^n$ is a vector of positive rational numbers with relatively prime numerator and denominator
\begin{equation}
q_i = \frac{\eta_i}{d_i}
\end{equation}
I am searching for $\alpha\in \mathbb{Q}$ such that
\begin{equation}
\mathbf{x} = \alpha\,\mathbf{q}
\end{equation}
is the smallest integer vector $\mathbf{x} \in \mathbb{N}$ with respect to some norm, e.g. 1-norm $||.||_1$.
I assume that $\alpha = \mathrm{lcm}(d_1, d_2, \dots, d_n)$. If this is the case, I would like to come up with the proof, but I am having trouble starting.
EDIT:
I unintentionally suggested the wrong assumption, i.e. $\alpha \neq \mathrm{lcm}(d_1, d_2, \dots, d_n)$.
This can be shown by choosing an example
\begin{equation}
  \mathbf{q} = \left[\begin{array}{cc}\frac{3}{4} & \frac{3}{4}\end{array}\right]
\end{equation}
In this case $\mathrm{lcm}(4, 4) = 4$ and $\alpha = \frac{4}{3}$ gives a smaller vector $\mathbf{x} = \left[\begin{array}{cc}1 & 1\end{array}\right]$
 A: Yes, $\alpha=\text{lcm}[d_1,\ldots,d_n]$.
First, you need that $\alpha$ is a multiple of every $d_i$ in order to $\alpha\mathbf q\in\Bbb N^n$.
Second, if $\beta$ is another common multiple of the denominators, we have $\beta=k\alpha$ for some integer $k\ge 2$. Then, for any norm, we have
$$\|\beta\mathbf q\|=k\alpha\|\mathbf q\|>\alpha\|\mathbf q\|$$
A: The correct $\alpha$ is  
$$\alpha_{min} = \frac{\rm{lcm}(d_1,\ldots,d_n)}{\gcd(\eta_1,\ldots,\eta_n)}.$$
Proof: 
Let $\alpha=\frac{a}b$ be such that $a,b \in \mathbb N, b> 0, \gcd(a,b)=1$ and $x=\alpha q \in \mathbb N^n.$ That implies $bx=aq \in \mathbb N^n$, which means $a\frac{\eta_i}{d_i} \in N, \forall i=1,\ldots,n$. Since $\eta_i$ and $d_i$ are assumed to be coprime, this implies $\frac{a}{d_i} \in N$, so $d_i|a, \forall i=1,\ldots,n$ and hence $\rm{lcm}(d_1,\ldots,d_m)|a$.
Similiarly, $x_i=\frac{a\eta_i}{b d_i} \in \mathbb N$ implies $b|a\eta_i$ and since $\gcd(a,b)=1$ this implies $b|\eta_i, \forall i=1,\ldots,n$ and hence $b|\gcd(\eta_1,\ldots,\eta_n)$. 
So any rational $\alpha$ that fullfills the condition of the problem and is in reduced fraction form must have an enumerator that is as least as big as in the given $\alpha_{min}$ and a denominator that is at most as big as in the given $\alpha_{min}$. That proves minimality of $\alpha_{min}$ (which can be easily checked to also fullfill the condition of the problem.
