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May 13 at 20:01 comment added Oscar Lanzi Make that $\mathbb{Q}[\sqrt{-30}]$.
May 13 at 19:01 comment added Oscar Lanzi In some cases tighter bounds on $\pi$ enter the reckoning. For example, $\pi^2>480/49$ enters into rendering the class group of $\mathbb{Q}[\sqrt{30}]$.
May 13 at 7:29 comment added Junyan Xu This answer mathoverflow.net/a/26504/3332 apparently requires tighter bounds to prove certain number fields have no unramified extensions. (found through math.stackexchange.com/q/699128/12932)
May 13 at 3:42 comment added Timothy Chow This was the example that came to my mind when I first read Terry's question. When I took a first course in algebraic number theory many years ago, applying this inequality was the only time during the whole course that I had to pull out my calculator. For example, sometimes we used the Minkowski formula to prove that the class number of a particular number field was 1. But I just took a quick look at the assigned homework questions, and none of them needed much numerical accuracy for $\pi$.
May 12 at 18:01 history edited JoshuaZ CC BY-SA 4.0
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May 12 at 17:46 history edited JoshuaZ CC BY-SA 4.0
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May 12 at 17:27 comment added Terry Tao Nice! In retrospect I guess I can see that the bounds $2 < \pi < 4$ (which geometrically correspond to the optimal ways to inscribe or circumscribe a disk by a square) would play some non-trivial role in mathematics. Perhaps the bound $\pi>3$ (which geometrically corresponds to the optimal way to inscribe a disk by a regular hexagon) would also show up for similar reasons. But I wonder how much of modern mathematics would actually be lost if for some reason the best numerical accuracy we had for $\pi$ was $2 < \pi < 4$...
May 12 at 16:55 history answered JoshuaZ CC BY-SA 4.0