prove that set of reals numbers and complex numbers are equipotent. I have to prove that set of  reals R and set of complex C are equipotent. 
  " i know that set A and B are equipotent iff there is one to one mapping of A onto B. "
    please anyone give me answer of this...
 A: Assuming you know that $\mathbb R$ is equipotent with $P(\mathbb N)$, the power set of $\mathbb N$, show the following lemmas:

  
*
  
*For any set $X$,  $P(X\times\{1,2\})$ is equipotent with $P(X)\times P(X)$. 
  
*If $X$ is equipotent with $Y$ then $P(X)$ is equipotent with $P(Y)$.
  
*$\mathbb C$ is equipotent with $\mathbb R\times\mathbb R$.
  
*$\mathbb N\times\{1,2\}$ is equipotent with $\mathbb N$.
  

The we get: $$\mathbb C\cong \mathbb R\times\mathbb R\cong P(\mathbb N)\times P(\mathbb N) \cong P(\mathbb N\times \{1,2\})\cong P(\mathbb N)\cong\mathbb R$$
Where we read $A\cong B$ as "$A$ is equipotent to $B$."
A: Let $x$ and $y$ be two real numbers and let $z$ be the complex number given by $x+iy$. Suppose the fractional part of $x$, $\lfloor x\rfloor$ has decimal expansion (always choose the greedy expansion which does not have a trailing string of $9$s) given by $0.x_1x_2x_3\ldots$ and let the natural number $|n_x|$ be such that $n_x+\lfloor x\rfloor =x$. Similarly let $n_y+\lfloor y\rfloor =y$.
We define $g\colon\mathbb{C}\to\mathbb{R}$ by $$g(z)=g(x+iy) = \epsilon_x \epsilon_y \underbrace{11\ldots 11}_{|n_x|} \underbrace{00\ldots00}_{|n_y|}.x_1y_1x_2y_2x_3y_3\ldots x_iy_i\ldots$$ where $\epsilon_x=0$ if $x$ is non-negative and $2$ if $x$ is negative, and $\epsilon_y=0$ if $y$ is non-negative and $3$ if $y$ is negative.
As an example, consider the complex number $w=x+yi=\pi-8.689i$. From the definition we need to know
$$\begin{array}{rcl}\lfloor x\rfloor & = & 14159265359\ldots\\
\lfloor y\rfloor& = &6891\\
n_x & = & 3\\
n_y & = & 8\\
x & \geq & 0\\
y & < & 0\end{array}$$
and so
$$\begin{array}{rcl}g(w) & = & 311100000000.1648195090206050305090\ldots\end{array}$$
I leave you to verify that this is an injective function. The rest of the necessary steps were covered in the comments, namely finding an injection $\mathbb{R}\to\mathbb{C}$ and then making use of the Cantor-Schroender-Berstein theorem.
A: Hint:
This is equivalent to proving that $\mathbb{R}^2$ and $\mathbb{R}$ are equipotent.
Consider the following mapping  $\pi:\mathbb{R}\to\mathbb{R^2}$
We take $a\in\mathbb{R}$ and we write it in binary notation (this works in any base, but binary makes it simpler):
$a = a_n 2^n + a_{n-1}2^{n-1} + \ldots + a_0 2^0 + b_12^{-1}+b_22^{-1}\ldots$
(we donsider $a_t = 0$ for $t > n$)
Then we construct $(b,c) \in \mathbb{R}^2$ as follows:
$$b = \sum_{k=0}^\infty a_{2k}2^k + b_{2k+2}2^{-k}$$
$$c = \sum_{k=0}^\infty a_{2k+1}2^k + b_{2k+1}2^{-k}$$
It's quite easy to prove that it's both injective and exhaustive. I'll leave that to you.
