Find a bijection from $\mathbb{R}^3$ to $\mathbb{R}$.
I was asked to find the bijection and I've gone in this following way :
Let, $$(x,y,z) (\ne(0,0,0)) \in \mathbb{R}^3$$ $$(x,y,z) \to (\frac{1}{1+e^x},\frac{1}{1+e^y},\frac{1}{1+e^z}) \to (0.x_1x_2... ,0.y_1y_2... , 0.z_2z_2... ) \to (0.x_1y_1z_1x_2y_2z_2... ) = a \to \log(\frac{1}{a} -1)$$ And, $$f(0,0,0) = 0$$
Here, $$\frac{1}{1+e^x} \in (0,1)$$ and I am taking only one decimal representation of any real number in $(0,1)$.
So $f: \mathbb{R}^3 \to \mathbb{R}$ defined by $f(x,y,z) = \log(\frac{1}{a} -1)$ one-one map and onto also because I've considered unique decimal representation of a real number in $(0,1)$.
Am I correct? If not then can someone give any constructive bijection (Not geometrical)?