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Take three real numbers $x$, $y$, and $\alpha$, of which $\alpha$ is a given irrational number. Can

$$y = \alpha x$$

be solved if the fractional parts of $x$ and $y$ are known (given) but their integer parts are unknown?

Note: this is a restatement of the third question in The power of irrationality: sin(x)+sin(πx)

Edit: If a more general solution is not available, I'm most interested in logarithms of natural numbers, that is, irrational $\alpha$'s of the form ${\rm log}\,2, {\rm log}\,3, \dots$, and their differences.

Edit 2: A similar question, with yet a different perspective, has been asked here: Measure of recurring trajectories on a billiard.

Arc
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1 Answers1

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Suppose $\lfloor x\rfloor$ and $\lfloor y\rfloor$ are the integer parts while $\{x\}= x- \lfloor x\rfloor$ and $\{y\}= y- \lfloor y\rfloor$ are the fractional parts.

So you know $\beta=\alpha\{x\} - \{y\}$ and this is equal to $\lfloor y\rfloor-\alpha\lfloor x\rfloor$.

If there were two solutions then $\lfloor y_1\rfloor-\alpha\lfloor x_1\rfloor = \lfloor y_2\rfloor-\alpha\lfloor x_2\rfloor$ making $\alpha= -\frac{\lfloor y_1\rfloor-\lfloor y_2\rfloor}{\lfloor x_1\rfloor-\lfloor x_2\rfloor}$ which would be rational

So there is at most one solution.

If there is one then you can find it by going through the countable integers one at a time to solve $\lfloor y\rfloor=\beta+\alpha\lfloor x\rfloor$. This may take a long time, and if there is no solution (consider the cardinalities) then the process will not halt.

Henry
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  • Thanks Henry, I went through a similar development, but in principle $\lfloor x\rfloor$ and $\lfloor y\rfloor$ can be arbitrarily large numbers, and trial and error is out of question, in fact I do expect $x$ and $y$ to be very large numbers... I even tried to take the decimal expansion of their fractional and integers parts, but carries are a mess. – Arc Sep 04 '21 at 11:16
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    @Arc Note that if you know more about $\alpha$ than it being just irrational you may be able to get some guaranteed range where if a solution exists it requires the solution to be in that range. Diophantine approximation and continued fractions may be relevant. – JoshuaZ Sep 04 '21 at 13:43
  • @JoshuaZ, thanks for the advice, I have edited the question to provide some more constraints on the values of $\alpha$'s. – Arc Sep 04 '21 at 15:40
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    @Arc Based on your interest in logs, the relevant thing to look up is Baker's theorem https://en.wikipedia.org/wiki/Baker%27s_theorem . – JoshuaZ Sep 04 '21 at 15:44
  • @JoshuaZ, thanks for the heads up on Baker's theorem, I wasn't aware of it and its a must-know, but as far as I understand it only provides a lower limit on a linear combination of logarithmic numbers, and I don't quite get it how it could help solve the question, could you please elaborate further any thoughs you have? Thanks. – Arc Sep 05 '21 at 23:28
  • @Arc Baker's theorem and related ideas can be used to get bounds on how close irrational numbers can be to each other in a certain sense. One result is it gives bounds on how close rational approximations to a number like log 2 can be. I don't know if this immediately connects to your problem, by my strong suspicion is that there's some connection here. – JoshuaZ Sep 05 '21 at 23:30