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I want to set up a map from $\Bbb N\to\Bbb Q.$

Take $\Phi_S(x)=e^{(S/\ln(1-x))}$

and $M_T(1-x)=\Phi_S(x); S,T\in\Bbb N.$

Set $\Phi_S(x)=M_T(x)$ to obtain algebraic $x$ coordinates.

If $x$ happens to be irrational round to finitely many digits to obtain a rational.

Then plug $x$ back into either equation to get the height, which is transcendental.

Round the transcendental to finitely many digits to obtain a rational.

For example, $S=T=2$ would get mapped to the point $ (1/2, 0.055833), $

Indeed, all $S=T$ would get mapped to $(1/2, H),$ where $H$ is the height.

So, $\Bbb N$ would be mapped to the vertical strip $x=1/2.$

This strip corresponds to the line outside the unit square: $y=x; x\in \Bbb N.$

In fact all vertical lines in the unit square are mapped to diagonal lines with positive slope, by a rotation.

For $S\ne T$ you'd get points such as $(2,3)$ and $(3,2)$ which would lie symmetrically on either side of $x=1/2.$

Essentially this scheme sets up a grid in the unit square and associates each point to two natural numbers.

Is the structure of the lattice of natural numbers preserved under this transformation?

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  • $\begingroup$ What difference is there between $M_1$ and $M_2$? $\endgroup$ – Saucy O'Path Feb 22 at 5:31
  • $\begingroup$ $M_1$ and $M_2$ yield different curves because they have different parameters $\endgroup$ – Ultradark Feb 22 at 5:37
  • $\begingroup$ It doesn't look like that. It looks like $M_1(x)$ and $M_2(x)$ are both equal to $\Phi_S(1-x)$ for the same $S$ (completely unrelated to $T$), and that the lower case numbers are there just for show. $\endgroup$ – Saucy O'Path Feb 22 at 5:40

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