Irrationals are Everywhere Dense in Reals/Normed Vector Space

Theorem
Let $\struct {\R, \size {\, \cdot \,} }$ be the normed vector space of real numbers.

Let $\R \setminus \Q$ be the set of irrational numbers.

Then $\R \setminus \Q$ is everywhere dense in $\struct {\R, \size {\, \cdot \,} }$

Proof
Let $x \in \R$.

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

Either $x \in \Q$ or $x \in \R \setminus \Q$.

Suppose $x \in \R \setminus \Q$.

Let $y := x$.

Then:


 * $\size {x - y} < \epsilon$

Suppose $x \in \Q$.

Let $n \in \N : n > \dfrac {\sqrt 2} \epsilon$

Let $y := x + \dfrac {\sqrt 2} n$

Then $y \in \R \setminus \Q$.

Furthermore:

In both cases $x$ was arbitrary.

Hence:


 * $\forall x \in \R : \exists \epsilon \in \R_{\mathop > 0} : \exists y \in \R \setminus \Q : \size {x - y} < \epsilon$

By definition, $\R \setminus \Q$ is dense in $\R$.