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A classical result shows that if $\alpha$ is irrational, then $\{k \alpha \bmod{1}\}_{k \in \mathbb{Z}}$ is dense over $[0,1]$. Can we extend this result as follows?

Suppose $\alpha_1,\dots,\alpha_m$ are such that $(1,\alpha,\dots,\alpha_m)$ is linearly independent over $\mathbb{Q}$. For any $\varepsilon > 0$, there exists an integer $k$ such that $k \alpha_i \bmod{1} \leq \varepsilon$ for all $1 \leq i \leq m$.

Edit: it was originally asked assuming only that all $\alpha_i$ are rational. As observed in the comments, for $m\ge 2$ and $\varepsilon<\frac12$, $\alpha_2=-\alpha_1$ yields an obvious counterexample.

A classical result shows that if $\alpha$ is irrational, then $\{k \alpha \bmod{1}\}_{k \in \mathbb{Z}}$ is dense over $[0,1]$. Can we extend this result as follows?

Suppose $\alpha_1,\dots,\alpha_m$ are such that $(1,\alpha,\dots,\alpha_m)$ is linearly independent over $\mathbb{Q}$. For any $\varepsilon > 0$, there exists an integer $k$ such that $k \alpha_i \bmod{1} \leq \varepsilon$ for all $1 \leq i \leq m$.

Edit: it was originally asked assuming only that all $\alpha_i$ are irrational. As observed in the comments, for $m\ge 2$ and $\varepsilon<\frac12$, $\alpha_2=-\alpha_1$ yields an obvious counterexample.

A classical result shows that if $\alpha$ is irrational, then $\{k \alpha \bmod{1}\}_{k \in \mathbb{Z}}$ is dense over $[0,1]$. Can we extend this result as follows?

Suppose $\alpha_1,\dots,\alpha_m$ are all irrational. For any $\epsilon > 0$, there exists an integer $k$ such that $k \alpha_i \bmod{1} \leq \epsilon$ for all $1 \leq i \leq m$.

A classical result shows that if $\alpha$ is irrational, then $\{k \alpha \bmod{1}\}_{k \in \mathbb{Z}}$ is dense over $[0,1]$. Can we extend this result as follows?

Suppose $\alpha_1,\dots,\alpha_m$ are such that $(1,\alpha,\dots,\alpha_m)$ is linearly independent over $\mathbb{Q}$. For any $\varepsilon > 0$, there exists an integer $k$ such that $k \alpha_i \bmod{1} \leq \varepsilon$ for all $1 \leq i \leq m$.

Edit: it was originally asked assuming only that all $\alpha_i$ are rational. As observed in the comments, for $m\ge 2$ and $\varepsilon<\frac12$, $\alpha_2=-\alpha_1$ yields an obvious counterexample.