# Mapping $\Bbb N\to\Bbb Q$

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?

• What difference is there between $M_1$ and $M_2$? – Saucy O'Path Feb 22 at 5:31
• $M_1$ and $M_2$ yield different curves because they have different parameters – Ultradark Feb 22 at 5:37
• 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. – Saucy O'Path Feb 22 at 5:40