Show that set of Real numbers has the same cardinality as a set of Irrational Numbers. 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&amp;0&amp;1&amp;\cdots&amp;n&amp;\cdots\\\hline
\mathbb R\setminus\mathbb Q &amp; \sqrt{2} &amp;\sqrt{5} &amp;\cdots &amp; \sqrt{n+1} &amp;\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?
Define
$$f(x)=\begin{cases}x+\sqrt 2\text{ if } x-n\sqrt 2\in\Bbb Q\text{ for some integer }n\ge 0 \\ x\text{ otherwise}\end{cases}$$
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 $\mathbb{R}$.