# cardinality of $\mathbb R$ is the same as the cardinality of $\mathbb R^2$

A problem in my homework is to prove the cardinality of $\mathbb R$ and the cardinality of $\mathbb R^2$ is the same. I'm in my first semester and I have no clue how to do this, do I need the axiom of choice to prove this? Can I find a bijection, or must I use Cantor-Schroeder-Bernstein? Apparently if we assume choice we can go further and prove the cardinality of any set $A$ is is equal to the cardinality of $A^2$ thank you very much, I need help.

Regards.

• Here's the answer. math.stackexchange.com/questions/243590/… – Math.StackExchange Oct 30 '14 at 1:07
• Do you know already that $|\mathbb R|=|\mathcal P(\mathbb N)|$? – hmakholm left over Monica Oct 30 '14 at 1:07
• no, that question literally came out of the blue, the rest of the homework is full of really simple questions. Thanks – Yorch Oct 30 '14 at 1:09
• If you're using Bernstein, one direction is trivially easy. For the other direction, a hint: Cantor's proof (or one of them) was by interlacing decimal digits ... – hmakholm left over Monica Oct 30 '14 at 1:12
• Yes, it's very independent of choice. The Cantor-Bernstein theorem is independent of choice as well. – Asaf Karagila Oct 30 '14 at 13:39

We want an injection from $\mathbb R^2$ to $\mathbb R$ we take a surjection from $\mathbb R$ to $\mathbb R^2$ and what we want is the right inverse.
As an example of a surjection I took the following: let the number in $\mathbb R$ be $(-1)^n(k+.d_1d_2d_3\dots)$ if $k$ is not of the form $2^a3^b5^c7^d$ send it to zero. if it is send it it to $((-1)^ad_1d_3d_5\dots,((-1)^bd_2d_4d_6\dots)$. And leave $c$ and $d$ digits before adding the decimal point.