Gaussian Rationals are Everywhere Dense in Complex Numbers

Theorem
Let $\struct {\C, \cmod {\, \cdot \,}}$ be the normed vector space of complex numbers.

Let $\Q \sqbrk i = \set {a + i b: a, b \in \Q}$ be the set of Gaussian rational numbers.

Then $\Q \sqbrk i$ is everywhere dense in $\struct {\C, \cmod {\, \cdot \,}}$.

Proof
Let $z = x + i y \in \C$ be a complex number.

Let $q = a + i b \in \Q \sqbrk i$ be a Gaussian rational number.

Then:

We have that rationals are everywhere dense in reals.

Then:


 * $\forall x \in \R: \forall \epsilon_x \in \R_{> 0}: \exists a \in \Q: \size {x - a} < \epsilon_x$


 * $\forall y \in \R: \forall \epsilon_y \in \R_{> 0}: \exists b \in \Q: \size {y - b} < \epsilon_y$

Let $\epsilon \in \R_{> 0}$.

Let $\epsilon_x, \epsilon_y \in \R_{> 0}$ be such that $\epsilon_x + \epsilon_y < \epsilon$.

Then:


 * $\exists q \in \Q \sqbrk i: \cmod {z - q} < \epsilon$

$z$ and $\epsilon$ were arbitary.

Therefore:


 * $\forall z \in \C: \forall \epsilon \in \R_{> 0}: \exists q \in \Q \sqbrk i: \cmod {z - q} < \epsilon$

By definition, $\Q \sqbrk i$ is everywhere dense in $\C$.