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}{cccccc}
\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?
Asked
Active
Viewed 1,526 times
0
Baimyrza Shamyr
 211
 2
 12

2What is supposed to be the image of $2$ with your shift? – md5 Jan 06 '17 at 13:21

2Your problem is a special case of the one dealt with in [this question](http://math.stackexchange.com/questions/355049/letabeanyuncountablesetandletbbeacountablesubsetofaprove). – Brian M. Scott Jan 06 '17 at 13:24

I'm afraid you are right  your attempt is very far from being a proof. For instance, what is the image of $\sqrt 2$ ? (Perhaps this is what @md5 meant to ask.) It can't be $\sqrt 2$, because that is already the image of $0$. – TonyK Jan 06 '17 at 13:25

@md5 shift is just an example. I agree it is bad example, but the best came to my mind. – Baimyrza Shamyr Jan 06 '17 at 13:27

2@AlexMacedo it's not a duplicate of that one. – Jan 06 '17 at 13:28

The number of algebraic irrationals is countable, the big group is the number of the irrationals that are transcendental. – Masacroso Jan 06 '17 at 13:47
2 Answers
3
Define $$f(x)=\begin{cases}x+\sqrt 2\text{ if } xn\sqrt 2\in\Bbb Q\text{ for some integer }n\ge 0 \\ x\text{ otherwise}\end{cases}$$
ajotatxe
 62,632
 2
 52
 101
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 $\mathbb{R}$.
Hans Hüttel
 4,098
 2
 12
 22