Finding a point in an infinite space

[Radar]

So the problem instead is to determine x and y to n decimal places, given a finite amount of questions (however many you want, but not infinite). Can that be done for all numbers?

In that case, the answer is most certainly no, you can't do this for all numbers. You could do it for approximately 0% of numbers in existence. That's basically the definition of computable numbers, so of course it wouldn't work for uncomputable numbers, and most numbers are uncomputable.


The set of irrationals is uncountably infinite, but the amount of digits in an irrational number is countably infinite. So you can still brute force the problem by asking questions until each digit is known with only a countable amount of questions.

You are actually disproving the point you make in the second paragraph with what you plainly state in the last one. Since one only needs countable infinity of questions to obtain any irrational number exactly, then getting just any finite number of digits will require finite amount of questions.

In other words, if you prescribe a given precision p, you silce the whole plane into p by p squares. Since you have countable infinity of such squares and the problem is simplified to just finding the right square, then with finite amount of questions you will always reach the target.

edit:

Warty Goblin's method also "bisects" the space of possible answers, in the sense that after each question the possible number of solutions is halved. (His method was asking if successive digits in the binary representation of each number are 0 or 1.) I don't believe it works for continuous spaces though. (Maybe one could add a step checking whether the number was less than a particular power of two, and then go to successively more precise digits?)

—Caerulea

Warty Goblin's method does bisect the the plane every time and you can easily extend it, so it would work for continuum simply by asking not just about higher digits but the fracions as well. The only thing it does not do is to give a good approximation until it obtains the highest digit.