Suppose that $\lambda_{1}, \lambda_{2}, \lambda_{3}\in\mathbb{C}\setminus\{0\}$ and that $\frac{\lambda_2}{\lambda_1}, \frac{\lambda_3}{\lambda_1}\in\mathbb{R}^{+}\setminus\mathbb{Q}$ such that the set $\{1, \frac{\lambda_2}{\lambda_1}, \frac{\lambda_3}{\lambda_1}\}$ is rationally independent.
Does anyone have any suggestions to prove the following$$\overline{\mathbb{R}\left(1,\frac{\lambda_2}{\lambda_1}, \frac{\lambda_3}{\lambda_1}\right)+\mathbb{Z}^{3}}=\mathbb{R}^{3} \hspace{1cm}?$$
I tried using sequences, I have not been able to get the same sequence of integers to satisfy two of the three convergences.Also, from hypothesis, $\frac{\frac{\lambda_3}{\lambda_1}}{\frac{\lambda_2}{\lambda_1}}\not\in\mathbb{Q}$.
Here
$\mathbb{R}^{3}=\{ (x_{1}, x_{2}, x_{3}) : x_{i}\in\mathbb{R},\hspace{0.1cm}\forall j=1,2,3 \}$.
$\mathbb{R}\left(1,\frac{\lambda_2}{\lambda_1}, \frac{\lambda_3}{\lambda_1}\right)=\{ r\left(1,\frac{\lambda_2}{\lambda_1}, \frac{\lambda_3}{\lambda_1}\right)=\left(r,r\frac{\lambda_2}{\lambda_1}, r\frac{\lambda_3}{\lambda_1}\right) : r\in\mathbb{R}\}$.
A set $A \subset \mathbb{R}$ is said to be rationally independent if given $a_1, \ldots, a_n \in A$ and $(k_1, \ldots, k_n) \in \mathbb{Z}^{n} \setminus \{(0,\ldots,0)\}$ imply $\sum\limits_{i=1}^{n} k_i a_i \neq 0$.
Thanks