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I don't know how to prove that Kronecker's theorem is false if $\alpha_{1}$, $\alpha_{2}$, $\dots$, $\alpha_{M}$ are not independent.

Kronecker's theorem Suppose that $\alpha_{1}$, $\alpha_{2}$, $\dots$, $\alpha_{M}$ are independent real numbers. Then given real numbers $\beta_{1}$, $\beta_{2}$, $\dots$, $\beta_{M}$ and $\epsilon>0$ we can find integers $N$, $m_{1}$, $m_{2}$, $\dots$, $m_{M}$ such that $$|N\alpha_{j}-\beta_{j}-m_{j}|<\epsilon$$ for each $1\leq j\leq M$.

Where if $\alpha_{1}$, $\alpha_{2}$, $\dots$, $\alpha_{M}$ satisfy

$\sum_{j=1}^{M} n_{j}\alpha_{j}\notin{\mathbb Z}$ for integers $n_{j}$ not all zero

then we say that they are independent real numbers.

It's easy if $M=1$, since then $\alpha_1$ is rational and $N\alpha_1$ only takes finitely many values modulo 1. But I can't get anything done even for $M=2$. I've tried drawing a picture but I'm struggling to get anywhere.

porkramen
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1 Answers1

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If $\alpha_1,\cdots, \alpha_M$ are not independent, then they satisfy $$ n_0+\sum_{j=1}^{M} n_{j}\alpha_{j}=0,$$ for integers $n_i$ not all zero.

Then the set $\{( N\alpha_1,\cdots, N\alpha_M)|N\in \mathbb{N}\}$ will be a subset of $$ \{(x_1,\cdots, x_n)| \sum_{j=1}^{M} n_{j}x_j =m \textrm{ for some } m\in\mathbb{Z}\}. $$

The above set is not a dense subset of $\mathbb{R}^M$.

Sungjin Kim
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