Is there some theorem that shows that rational and irrational numbers together can be used to completely occupy a continuous infinite line? i.e. that the real numbers are the union of the two sets.
I am misunderstanding? It is as it is because that is how we define it. Let $\displaystyle \mathbb{R}$ be a universal set and $\displaystyle \mathbb{Z}\subset\mathbb{R}$. Then define $\displaystyle \mathbb{Q}\equiv\left\{x:\exists{a,b}\in\mathbb{Z} \backepsilon~x=\frac{a}{b}\right\}$, and then define $\displaystyle \mathbb{I}\equiv\left\{x:\not\exists{a,b}\in\mathb b{Z}\backepsilon~x=\frac{a}{b}\right\}\implies\mat hbb{I}=\mathbb{Q}^c$
So by defintion of complement and universal set
$\displaystyle \mathbb{Q}\cup\mathbb{I}=\mathbb{Q}\cup\mathbb{Q}^ c=\mathbb{R}$
Slight correction:
I am misunderstanding? It is as it is because that is how we define it. Let $\displaystyle \mathbb{R}$ be a universal set and $\displaystyle \mathbb{Z}\subset\mathbb{R}$. Then define $\displaystyle \color{red}\mathbb{Q}\equiv\left\{x:\exists{a,b}\b ackepsilon~b\neq0\in\mathbb{Z}\backepsilon~x=\frac {a}{b}\right\}$, and then define $\displaystyle \color{red}\mathbb{I}\equiv\left\{x:\not\exists{a, b}\backepsilon~b\neq0\in\mathbb{Z}\backepsilon~x=\ frac{a}{b}\right\}\implies\mathbb{I}=\mathbb{Q}^c$
So by defintion of complement and universal set
$\displaystyle \mathbb{Q}\cup\mathbb{I}=\mathbb{Q}\cup\mathbb{Q}^ c=\mathbb{R}$