How to prove this function is bijective?

Question

Funtion $$f:\mathbb{N}^2\rightarrow\mathbb{N},\quad(x,y)\mapsto\frac{(x+y)(x+y+1)}{2}+x$$

Note that given $z=f(x,y)\in\mathbb{N}$,$\,\,$you can find $x$ and $y$ by bounded minimization.

Answer 1

It is not bijective because

$f(0,0)=f(0,-1)=0$

The map is surjetive because

$x=f(x,-x)$

If the domain is $\mathbb{N}^2$ to $\mathbb{N}$ you can fix $y$ and for every $x_1,x_2$ such that $f(x_1,y)=f(x_2,y)$ you have that $x_1=x_2$ because if, for example, $x_1< x_2$ you have that

$\frac{(x_2+y)^2}{2}-\frac{(x_1+y)^2}{2}+\frac{1}{2}[-(x_1+y)+(x_2+y)]=$

$=x_1-x_2<0$

But

$\frac{1}{2}(x_2-x_1)(x_2+x_1)+\frac{1}{2}(x_2-x_1)>0$

And it is not possibile.

Answer 2

We can factor $f$ as $$\mathbb N\times\mathbb N\overset{f_1}\longrightarrow Z\overset{f_2}\longrightarrow \mathbb N$$ where $Z = \{(x,z)\in\mathbb N\times\mathbb N \,|\, z\ge x\}$ and the maps are given by \begin{align*} f_1(x,y) &= (x,x+y), \\ f_2(x,z) &= \frac{z(z+1)}2 + x. \end{align*} We see that $f_1$ is a bijection with inverse $(x,z)\mapsto(x,z-x)$, so we only have to show that $f_2$ is a bijection. For this note that $z(z+1)/2$ is a triangle number, so that $$ f(x,z) = \sum_{k=1}^z k + x. $$ Now given any $n\in\mathbb N$, there is a unique $z\in\mathbb N$ such that $n$ sits between the $z$-th and $(z+1)$-th triangle number: $$ \sum_{k=1}^z k \le n < \sum_{k=1}^{z+1} k = \sum_{k=1}^z k + (z+1) $$ Letting $x = n - \sum_{k=1}^z k$ we obtain $$ n = \sum_{k=1}^z k + x, $$ where $x<z+1$ by the choice of $z$. So we have $n=f_2(x,z)$.