Irrationals are Everywhere Dense in Reals/Normed Vector Space

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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 $\R \setminus \Q$ are 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 $\displaystyle n \in \N : n > \frac {\sqrt 2} \epsilon$

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

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

Furthermore:

\(\displaystyle \size {x - y}\) \(=\) \(\displaystyle \size {\frac {\sqrt 2} n}\)
\(\displaystyle \) \(<\) \(\displaystyle \epsilon\)

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 $\R$.

$\blacksquare$


Sources