Context:
A friend chooses 2 rational numbers to use, R1 and R2.
The friend chooses an irrational number, I1
The friend uses this equation to get I2
R1 + (R2 * I1) = I2
The friend gives I1 and I2 to you, and tells you to decode I1 and I2 into R1 and R2, without guessing every rational number, and it has to be done in a reasonable amount of time.
Given:
These statements have been proved by me.
Every combination of R1 and R2, given the same I1, always gives a unique I2 (1 to 1 correlation between any 2 rational numbers and I2)
if I1 is irrational, and R1 and R2 are rational, then I2 has to be irrational
Question:
How do you find R1 and R2, in a polynomial amount of time?
Proof that if I1 is irrational, and R1 and R2 are rational, then I2 is always Irrational
Suppose I2 is rational
$\frac{p}{q}+\frac{a}{b}I_{1}=I_{2}$
$\frac{p}{q}+\frac{a}{b}I_{1}=\frac{c}{d}$
$\frac{a}{b}I_{1}=\frac{c}{d}-\frac{p}{q}$
$\frac{a}{b}I_{1}=\frac{cq-pd}{dq}$
$I_{1}=\frac{bcq-bpd}{dq}$
This says that if we suppose I2 is rational, then I1 can be shifted around to equal a rational number, which can not happen, therefore this is proven by contradiction.
Proof that for any I2, there is only one set of R1 and R2 that can be used to produce it.
suppose that R1 and R2 can be modified to create the same I2 value,
also suppose that, $\frac{p}{q}+\frac{a}{b}I_{1}=I_{2}$
$\left(\frac{p}{q}+\frac{m}{n}\right)+\left(\frac{a}{b}+\frac{c}{d}\right)I_{1}=I_{2}$
$\left(\frac{p}{q}+\frac{m}{n}\right)+\left(\frac{a}{b}+\frac{c}{d}\right)I_{1}=\frac{p}{q}+\frac{a}{b}I_{1}$
$\frac{p}{q}+\frac{m}{n}+\frac{a}{b}I_{1}+\frac{c}{d}I_{1}-\frac{p}{q}-\frac{a}{b}I_{1}=0$
$\frac{m}{n}+\frac{c}{d}I_{1}=0$
$\frac{m}{n}=-\frac{c}{d}I_{1}$
$-\frac{md}{nc}=I_{1}$
This says that if we modify R1 or R2 or both in any way, then you can shuffle the terms around so that a rational number equals I1, therefore this is proven by contradiction.
Extra Information
This problem can technically be solved in a $ω_0$ amount of time by guessing every single $R_2$ with cantor's diagonal argument, and checking if $R_1$ is rational ${I_2} - {I_1}{R_2}$∈ℚ, but that is not what I am looking for, I don't want to solve this with any method that takes a $ω_0$ amount of time
The number of rational numbers is countable infinity, while the number of irrational numbers is uncountable infinity, therefore you can encode as many rational numbers as you want in an irrational number.
${R_1}+{R_2}{I_1}+{R_3}{I_2}...{R_n}{I_{n-1}}={I_n}$
$I_1$ is chosen in advance, it can be any irrational number, but for the sake of simplicity, try starting with $\sqrt{2}$.
We are assuming that the computer running the decoding algorithm takes the numbers in as ${I_1}$ and $R_1+{I_1}{R_2}$, and does operations on them to get to $R_1$ and $R_2$ respectively. The computer cannot just 'look' at the R1 and R2 as defined in I2, it has to procure the numbers with math.