Real Numbers represented as a coordinate plane?

Question

Not sure if I’m using the right terminology here, but has anybody recontextualized Real Numbers into a coordinate plane? Is there a name for it?

I understand that what I’m looking for is similar to the Cartesian coordinate system, but instead of two real number axis, the grid itself IS a representation of real numbers. Something kind of like this:

$$\begin{array}{|c|c|} \hline \frac{-5}{5} & \frac{-4}{5} & \frac{-3}{5} & \frac{-2}{5} & \frac{-1}{5} & \frac{0}{5} & \frac{1}{5} & \frac{2}{5} & \frac{3}{5} & \frac{4}{5} & \frac{5}{5}\\\hline \frac{-5}{4} & \frac{-4}{4} & \frac{-3}{4} & \frac{-2}{4} & \frac{-1}{4} & \frac{0}{4} & \frac{1}{4} & \frac{2}{4} & \frac{3}{4} & \frac{4}{4} & \frac{5}{4}\\\hline \frac{-5}{3} & \frac{-4}{3} & \frac{-3}{3} & \frac{-2}{3} & \frac{-1}{3} & \frac{0}{3} & \frac{1}{3} & \frac{2}{3} & \frac{3}{3} & \frac{4}{3} & \frac{5}{3}\\\hline \frac{-5}{2} & \frac{-4}{2} & \frac{-3}{2} & \frac{-2}{2} & \frac{-1}{2} & \frac{0}{2} & \frac{1}{2} & \frac{2}{2} & \frac{3}{2} & \frac{4}{2} & \frac{5}{2}\\\hline \frac{-5}{1} & \frac{-4}{1} & \frac{-3}{1} & \frac{-2}{1} & \frac{-1}{1} & \frac{0}{1} & \frac{1}{1} & \frac{2}{1} & \frac{3}{1} & \frac{4}{1} & \frac{5}{1}\\\hline \frac{-5}{0} & \frac{-4}{0} & \frac{-3}{0} & \frac{-2}{0} & \frac{-1}{0} & \frac{0}{0} & \frac{1}{0} & \frac{2}{0} & \frac{3}{0} & \frac{4}{0} & \frac{5}{0} \\\hline \frac{-5}{-1} & \frac{-4}{-1} & \frac{-3}{-1} & \frac{-2}{-1} & \frac{-1}{-1} & \frac{0}{-1} & \frac{1}{-1} & \frac{2}{-1} & \frac{3}{-1} & \frac{4}{-1} & \frac{5}{-1}\\\hline \frac{-5}{-2} & \frac{-4}{-2} & \frac{-3}{-2} & \frac{-2}{-2} & \frac{-1}{-2} & \frac{0}{-2} & \frac{1}{-2} & \frac{2}{-2} & \frac{3}{-2} & \frac{4}{-2} & \frac{5}{-2}\\\hline \frac{-5}{-3} & \frac{-4}{-3} & \frac{-3}{-3} & \frac{-2}{-3} & \frac{-1}{-3} & \frac{0}{-3} & \frac{1}{-3} & \frac{2}{-3} & \frac{3}{-3} & \frac{4}{-3} & \frac{5}{-3}\\\hline \frac{-5}{-4} & \frac{-4}{-4} & \frac{-3}{-4} & \frac{-2}{-4} & \frac{-1}{-4} & \frac{0}{-4} & \frac{1}{-4} & \frac{2}{-4} & \frac{3}{-4} & \frac{4}{-4} & \frac{5}{-4}\\\hline \frac{-5}{-5} & \frac{-4}{-5} & \frac{-3}{-5} & \frac{-2}{-5} & \frac{-1}{-5} & \frac{0}{-5} & \frac{1}{-5} & \frac{2}{-5} & \frac{3}{-5} & \frac{4}{-5} & \frac{5}{-5}\\\hline \end{array}$$

This would then "simplify" to:

$$\begin{array}{|c|c|} \hline -1 & -0.8 & -0.6 & -0.4 & -0.2 & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1\\\hline -1.25 & -1 & -0.75 & -0.5 & -0.25 & 0 & 0.25 & 0.5 & 0.75 & 1 & 1.25\\\hline -1.\overline6 & -1.\overline3 & -1 & -0.\overline6 & -0.\overline3 & 0 & 0.\overline3 & 0.\overline6 & 1 & 1.\overline3 & 1.\overline6\\\hline -2.5 & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 & 2 & 2.5\\\hline -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5\\\hline \frac{-5}{0} & \frac{-4}{0} & \frac{-3}{0} & \frac{-2}{0} & \frac{-1}{0} & \frac{0}{0} & \frac{1}{0} & \frac{2}{0} & \frac{3}{0} & \frac{4}{0} & \frac{5}{0} \\\hline 5 & 4 & 3 & 2 & 1 & 0 & -1 & -2 & -3 & -4 & -5\\\hline 2.5 & 2 & 1.5 & 1 & 0.5 & 0 & -0.5 & -1 & -1.5 & -2 & -2.5\\\hline 1.\overline6 & 1.\overline3 & 1 & 0.\overline6 & 0.\overline3 & 0 & -0.\overline3 & -0.\overline6 & -1 & -1.\overline3 & -1.\overline6\\\hline 1.25 & 1 & 0.75 & 0.5 & 0.25 & 0 & -0.25 & -0.5 & -0.75 & -1 & -1.25\\\hline 1 & 0.8 & 0.6 & 0.4 & 0.2 & 0 & -0.2 & -0.4 & -0.6 & -0.8 & -1\\\hline \end{array}$$

Answer 1

What you are displaying is, essentially, the real projective line, denoted $\mathbb RP^1$.

Usually, though, one uses the reciprocals of the fractions that you depicted, because each point $(x,y)$ of the Cartesian coordinate plane is then represented by the ratio $\frac{y}{x}$ which is equal to the slope of the line passing through $(0,0)$ and $(x,y)$. What you have depicted instead uses the ratio $\frac{x}{y}$.

One other interesting feature to notice: there's not just real numbers in your diagram, there is also $\infty = \frac{y}{0}$ for any $y \ne 0$. And, by the way, the central entry $\frac{0}{0}$ should not be there because it does not represent anything sensible, neither a finite number nor $\infty$ (and there does not exist a unique line passing through $(0,0)$).

Answer 2

If you want to do it thoroughly, you have to reverse the numbers to the right of the decimal point. This way you can be certain to enumerate every possible real number (though it will go off into infinity).

So, to enumerate the reals (just like the rationals, folks), you can simply start counting at 0.0, then 0.1, 0.2 to 0.9 and what's after 9 is 10, but you write it in reverse:

0.01, 0.11, 0.21 (twelve in reverse), 0.31 (thirteen). I call this the "reverse odometer", but eventually(hmm?) you will enumerate every real number between 0.0 and 1.0.