Is the Cantor Pairing function guaranteed to generate a unique real number for all real numbers? I recently learned that for natural numbers, the Cantor Pairing function allows one to output a unique natural number from any combination of two natural numbers. According to wikipedia, it is a computable bijection
$$f : \mathbb N \times \mathbb N \rightarrow \mathbb N$$
$$f(x,y) := \frac 12 (x+y)(x+y+1)+y$$
Will it generate a unique value for all real (non-integer) number values of $x$ and $y$? I believe there is no inverse function if using non-integer inputs, but I just want to know if the output $f(x,y)$ will still be unique.
Please forgive me if this isn't a worthwhile question, I do not have a mathematics background.
Edit: I'm interested in the case where we constrain $x$ and $y$ to real numbers $>0$.
 A: Even for positive reals the answer is no, the result is not unique.  You can choose any $x,y,$ compute $f(x,y)$, then choose any $x'\lt x$  and solve $\frac 12(x'+y')(x'+y'+1)+y'=f(x,y)$ for $y'$  The only reason for the $x'$ restriction is to make sure you get a positive square root.  For example, let $x=3,y=5,x'=2$.  We have $f(3,5)=41$ so want $\frac 12(2+y')(3+y')+y'=41$, which has solutions $y'=\frac 12(-7\pm\sqrt{353})\approx -12.8941,5.8941$ so $f(3,5)=f(2,\frac 12(-7+\sqrt{353}))$ in the positive reals.  You can allow any of $x,y,x'$ to be other than integers.  $y'$ will usually not be integral.
A: A function on two variables $x$ and $y$ is called a polynomial function if it is defined by a formula built up from $x$, $y$ and numeric constants (like $0, 1, 2, \ldots$) using addition,multiplication. So Cantor's pairing function is a polynomial function.
If $f(x, y)$ is a polynomial function, then $f$ cannot be an injection of $\Bbb{R}\times\Bbb{R}$ into $\Bbb{R}$ (because of o-minimality).
A: Observe that $f(-1,1)=f(-2,1)$
