Bijection between $\mathbb N^+ \times \mathbb R^+$ and $\mathbb R^+$ Let $\mathbb N^+$ denote the set of natural numbers bigger than $0$ and let $\mathbb R^+$ denote the set of real numbers bigger than $0$.
Is there a way to write down an explicit bijection between $\mathbb N^+ \times \mathbb R^+$ and $\mathbb R^+$?
 A: Yes:
Consider $f(x)=\dfrac{x}{1+x}$, it is a bijection $\mathbb R^+\to\mathbb [0,1[$. Now, $g(n,x)=n+x$ is a bijection like you want.
A: Well, there exist the following bijections:


*

*An explicit bijection between $\mathbb R$ and $(0,1)$ (namely, scaled the co-tan function)

*An explicit bijection between $\mathbb R^+$ and $\mathbb R$ (the $\exp$ function)

*An explicit bijection between $(0,1)$ and $(0,1]$


This means there exists an explicit bijection between $\mathbb R^+$ and any semiclosed interval of $(a,a+1]$ length $1$, call that $f_{a}$. Now just create a bijection from $\mathbb N^+\times \mathbb R^+$ to $\mathbb R^+$ from the individual $f_a$s
A: You can replace $\mathbb{N}^+$ by $\mathbb{N} = \{ 0 , 1 , 2 ... \}$. Also $\psi : \mathbb{R}^+ \cong [0,1)$.
Then the required bijection can be
$$ \mathbb{N} \times \mathbb{R}^+ \ni  (n,r) \longmapsto n+\psi(r) \in \mathbb{R}^+ \ .$$
A: There are many bijections between $(0,1)$ and $\mathbb R$. Using them you can get easily bijection between $\mathbb N\times\mathbb R$ and $\mathbb R\setminus\mathbb Z = \bigcup_{n\in\mathbb Z} (z,z+1)$.
So it only remains to modify the bijection to include integers. This can be done using "Hilbert hotel type" argument. (Something similar to this, with the difference that now you want to add infinitely many elements.)
EDIT: I have worked with $\mathbb R$ and not $\mathbb R^+$. As other answers explain, there is a bijection between $\mathbb R$ and $\mathbb R^+$. So this problem can be solved, too.
