Show that set of Real numbers has the same cardinality as a set of Irrational Numbers.

Question

How can I show that sets $| \mathbb R-\mathbb Q |$ and $ | \mathbb R |$ have same cardinality.
My Solution: As long as for element in one set I can match it with unique element
in another set they have same cardinality:
After removing $\mathbb Q $ from $\mathbb R$ I can shift elements of Irrational numbers to correspond to elements of two sets match like this: $ \begin{array}{c|c|c|c|c|c} \mathbb R&0&1&\cdots&n&\cdots\\\hline \mathbb R\setminus\mathbb Q & \sqrt{2} &\sqrt{5} &\cdots & \sqrt{n+1} &\cdots \end{array} $
Note that I use n as Real number for simplicity.

Even if this solution is correct, I feel like Professor won't accept it on exam.
How would formal solution look like?

Answer 1

Define $$f(x)=\begin{cases}x+\sqrt 2\text{ if } x-n\sqrt 2\in\Bbb Q\text{ for some integer }n\ge 0 \\ x\text{ otherwise}\end{cases}$$

Answer 2

Clearly, $\mathbb{R} \setminus \mathbb{Q}$ has at most the same cardinality as $\mathbb{R}$ and is an infinite set. The set can therefore be either countable or of the same cardinality as $\mathbb{R}$.

Assuming the continuum hypothesis, if $|\mathbb{R} \setminus \mathbb{Q}| \neq |\mathbb{R}|$, we must then have that $\mathbb{R} \setminus \mathbb{Q}$ is countable.

But the union of two countable sets is countable, and $(\mathbb{R} \setminus \mathbb{Q}) \cup \mathbb{Q} = \mathbb{R}$. Therefore $\mathbb{R} \setminus \mathbb{Q}$ must have the same cardinality as $\mathbb{R}$.