Why $\mathbb Z(\sqrt2)$ is not a lattice?

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The set $\mathbb Z(\sqrt2) = \{a + b\sqrt2 : a, b \in \mathbb Z\}$ is not a lattice, according to the book of Robeldo =

because when you replace $a, b \in \mathbb Z$ by $a, b \in \mathbb R$ we do not obtain all of $C$

but only a $1$-dimensional real space (in this case just $\mathbb R$).

In other words, there are no two points $w_1, w_2$ in $\mathbb Z$ whose coordinates are linearly independent in $R^2$.

This is hard to get for me.

How can I know that if I replace $a, b \in \mathbb Z$ by $a, b \in \mathbb R$ we do not obtain all of $C$?

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Because $\sqrt{2} \in {\mathbb R}$, ${\mathbb R}[\sqrt 2] = \{ a + b \sqrt2 \mid a, b \in {\mathbb R}\}$ is equal to $\mathbb R$ and not to $\mathbb C$.

(I'm not sure what is exactly meant by "In other words, there are no two points $w_1, w_2$ in $\mathbb Z$ whose coordinates are linearly independent in $R^2$.")