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New Proof Solves 80-Year-Old Irrational Number Problem
- xelxebar 7y agoFor those that just want the meat, this is about the Duffin-Schaeffer conjecture [0]. Here's the relevant paper: https://arxiv.org/abs/1907.04593 https://arxiv.org/abs/1907.04593 For those that want a little more, the Duffin-Schaeffer conjecture is about rational approximations of real numbers. The setup is this: Given some real number, we're concerned with its rational approximations. To this we choose some function that assigns an "acceptable error" for every possible denominator. Thus, we want to look at the set of "acceptable approximations." A priori, we can intuit that if our errors are too strict, we might not have any---or only a handful---that meet our criteria. The Duffin-Schaeffer makes this intuition precise, giving a condition that tells us whether our set of acceptable approximations will be finite or infinite. [0]:https://en.wikipedia.org/wiki/Duffin-Schaeffer_conjecture https://en.wikipedia.org/wiki/Duffin-Schaeffer_conjecture
- signalsmith 7y agoSlightly off-topic, but for anyone who hasn't encountered them: using continued fractions (https://en.wikipedia.org/wiki/Continued_fraction#Best_rational_approximations https://en.wikipedia.org/wiki/Continued_fraction#Best_ration...), you can easily get the sequence of optimal approximations - where "optimal" means "better than anything else with a smaller denominator". (e.g. pi -> 3, 22/7, 333/106, 355/113, ...) This paper is about a different situation (Duffin-Schaeffer), where you don't just want the minimum denominator, but instead have more structured/custom constraints for whether a approximation passes/fails, based on the denominator.
- contravariant 7y agoNote that that sequence of approximations of pi is not the entire sequence of optimal approximations. That one requires a bit more work. It is a sequence of optimal approximations though.