Approximation Game
14 points
4 days ago
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| lcamtuf.substack.com
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cperciva
1 hour ago
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In the same vein, algebraic irrationals (e.g., √2) all have an irrationality measure of two, but the proof of this is fiendishly difficult and netted its discoverer the Fields Medal back in 1958.

It's worth noting that this is a standard method of proving that a value is transcendental -- just show that it has better rational approximations than any algebraic number can have.

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gjm11
1 hour ago
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It's a standard way of proving that transcendental numbers exist, because it's easy to construct very well-approximable numbers, but so far as I know it isn't a common way to prove that a number you were already interested in is transcendental. For pretty much every number you might be interested in, the best-known lower bound on that exponent is 2, which of course isn't good enough to prove transcendence.

At least, that's my understanding of where things stand, but I'm not an expert. Do you have counterexamples?

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cperciva
1 hour ago
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The Champernowne numbers were already known to be irrational, but this makes the proof much easier!

(But to clarify: When I said "proving that a value is transcendental", I was thinking of numbers specifically constructed for that purpose, not of other numbers more generally. 100% of transcendental numbers have irrationally measure 2.)

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