# Find integer $k$ such that $k \alpha_i \bmod{1}$ are simultaneously small for all $i$

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.

• Yes. This is because the infinite cyclic subgroup generated by $(\alpha_1,\dots,\alpha_m)$ in $(\mathbf{R}/\mathbf{Z})^m$ cannot be closed (since the latter is compact), and hence it accumulates at zero. – YCor Oct 1 '19 at 8:32
• No. Check out $\alpha_1=1-\alpha_2=\sqrt3$. If you want the $k\alpha_i$ to be just close to integers (from either side) --- then yes, e.g., by the reasons @YCor mentions. – Ilya Bogdanov Oct 1 '19 at 8:36
• The answer is yes under the assumption that $1,\alpha_1,\dots,\alpha_m$ are linearly independent, by Kronecker's theorem – Wojowu Oct 1 '19 at 8:42
• @IlyaBogdanov Ah, I see, I tend to work in the quotient (and hence interpreted as "$d(\alpha_i,\mathbf{Z})\le\varepsilon$"). Indeed taking the representative in $[0,1[$ yields a different conclusion. Given such a trivial counterexample I guess my interpretation was the OP's intent, so hopefully OP will clarify. – YCor Oct 1 '19 at 9:20
• It's necessary. I've done so, adding the context which makes the comments understandable. – YCor Oct 5 '19 at 8:41

The answer is yes, by Kronecker's theorem. Namely, the assumption that $$(1,\alpha_1,\dots,\alpha_m)$$ is linearly independent over $$\mathbf{Q}$$ means that $$(\alpha_1,\dots,\alpha_m)$$ generates a dense sub(semi)group of the torus $$(\mathbf{R}/\mathbf{Z})^m$$, and hence this subsemigroup meets $$[0,\varepsilon]^m$$ (infinitely many times).
In the original setting (assuming only $$\alpha_i$$ irrational for some $$i$$), it is still true that it meets $$[-\varepsilon,\varepsilon]^m$$ infinitely many times: indeed this assumption ensures that the closure of the sub(semi)group is non-discrete and accumulates at zero. (And using that every nonempty closed subsemigroup of a compact group is a subgroup.) But for $$m\ge 2$$, taking $$\alpha_2=-\alpha_1$$ irrational shows that it doesn't work with $$[0,\varepsilon]^m$$ instead.