Bijection between $\{x^2: x \in R \}$ and $\{x: x \in R\}$

Question

Let's consider two sets: $A := \{x^2: x \in R\}$ and $B:= \{x: x \in R\}$.

In my opinion its very intuitive that those two sets have exactly the same cardinality. In other words there has to be a bijection $g$ between $A$ and $B$ but I couldn't find proper form of $g$. I tried to pick $g(x) = x$ or $g(x) = \sqrt x$ but none of them works (second example doesn't work because domain differs).

Could you please help me finding this bijection?

Answer 1

Let $A_0=[0,1]$ and $A_n=(n,n+1]$ for $n\in \Bbb Z^+.$ Let $B_0=[0,1]$. For $n\in\Bbb Z^+$ let $B_{2n}=(n,n+1]$ and let $B_{2n-1}=[-n,1-n)$. For each non-negative integer $n$, map $A_n$ bijectively onto $B_n.$

Answer 2

$A := \{x: x \in \mathbb{R}\ \text{ and }\ x\geq0\}$ and $B:= \mathbb{R}.$

Now we need to find a bijective map from $A$ to $B$. After fiddling around with some sketches, I found the bijective function $\ g: A\to B\ $ defined by:

$ f(x)= \begin{cases} \frac{n}{2} + (x\mod1), &\ x\in [n,n+1),\quad \text{for each even integer}\ n\geq 0\\ -\left(\frac{n+1}{2}\right) + (x\mod1), &\ x\in [n,n+1),\quad \text{for each odd integer}\ n\geq 0\\ \end{cases} $

I'm not sure this is the clearest way to write that function, so someone feel free to amend this to make it simpler/read better.