proof that $\mathbb{R}=\mathbb{Q}\cup\mathbb{Q}^{'}$

Question

I'm not sure if this is correct or not, I'm trying to find a proof or disproof that $\mathbb{R}=\mathbb{Q}\cup\mathbb{Q}^{'}$, that means, the set of the real numbers is the union of the rational and irrational numbers, if possible I'd like a more explicative answer since I'm not that good in math, thanks in advance!

Answer 1

$\mathbb Q =\{x\in \mathbb R|$ there are integers $n,m$ so that $x = \frac mn\}$.

$\mathbb Q' = \{x\in \mathbb R|$ there are not any integers $n,m$ so that $=\frac mn\}$

So $\mathbb Q\cup \mathbb Q' = \{x\in \mathbb R|$ there are integers $n,m$ so that $x = \frac mn\}\cup \{x\in \mathbb R|$ there are not any integers $n,m$ so that $=\frac mn\}=$

$\{x\in \mathbb R|$ there are integers $n,m$ so that $x=\frac mn$ or there are not any integers $n,m$ sothat $x=\frac mn\}=$.

$\{x\in \mathbb R|$ there are or are not integers $n,m$ where $x=\frac mn\}=$

$\{x\in \mathbb R| x$ can be any real number whather it can be written as $x=\frac mn$ for some integers $m,n$ or not$\}=$

$\{x\in \mathbb R|x$ is a real number$\}=$

$\{x\in \mathbb R\}=$

$\mathbb R$.

Answer 2

There is nothing to prove: an irrational number is by definition a real number that is not a rational number, so by definition every real number is either rational or irrational (and not both).