Republic of Mathematics

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Geometric proofs of the irrationality of square-roots for select integers arxiv.org/pdf/2410.14434

@dmarain @jamestanton Whoops!!

@jamestanton @republicofmath no wonder I was confused. You and I were responding to different questions! When you gave an example for one over six and one over seven, I was still responding to representing 1/3 using values from one to six or from one to seven!

It is possible to express each of 1/2, 1/3, and 1/4 as a ratio of two integers where each of the digits 1, 2, 3, .... (up to some value) are used exactly once. Is there an easier way to express 1/3?

@jamestanton @iconjack Is base 9 robust? 1/7 is the issue: 1/2 = 1/2 (base 9) 1/3 = 375/1246 (base 9) 1/4 = 1542/6378 (base 9) 1/5 = 756/4213 (base 9) 1/6 = 2/13 (base 9) 1/7 = ?? /???(base 9) 1/8 = 417/3652 (base 9)

Base b is "robust" if each fraction 1/2,1/3,..,1/(b-1) is a ratio of two integers expressed in base b that use each digit 1, 2, 3,.... (up to some value) exactly once. (1/b cannot be so expressed. Why?) @iconjack shows base 10 is robust. Base 3 and base 4 are robust. Base 5? 6?

Hi! Next month I'm running a learning experience for K-12 teachers called "21st Century Mathematics". Its goal is to help teachers to connect the math they teach with some math that has been discovered in the 21st century. Info+register: justinlanier.org/21st-century-m… Pass it along!

@jamestanton @iconjack Base 8 is robust: 1/2 = 1/2 (base 8) 1/3 = 347/1265 (base 8) 1/4 = 275/1364 (base 8) 1/5 = 653/4127 (base 8) 1/6 = 245/1736 (base 8) 1/7 = 421/3567 (base 8)

@jamestanton @iconjack Is base 7 robust: 1/2 = 1/2 (base 7) 1/3 = 3/12 (base 7) 1/4 = 54/312 (base 7) 1/5 = 2/13 (base 7) 1/6 = ?? ?

@jamestanton @iconjack 1/3 seems to be problematic bases 5 and 6, but 1/3 = 3/12 (base 7)

@jamestanton @iconjack 1/5 = 41/325 (base 6)

@jamestanton @iconjack 1/4 = 2/13 (base 5) but 1/3 = ??

@jamestanton @iconjack Statement for 1/6 should be 1/6 = 571/3426

@dmarain @jamestanton 1/6 = 571/3426 and 1/7 = 3/21

@republicofmath @jamestanton Yep, I missed that one. Great job! you agree that there’s no solution for six or seven?

Dominic Welsh's important contributions to probability and combinatorics: arxiv.org/pdf/2404.13942

@jamestanton Express each 1/n, n= 1, ..., 9 in the simplest form as a ratio of two integers where each of the digits 1, 2, 3, .... (up to some value) are used exactly once. What about 1/n for n>= 10?

@jamestanton And 1/9 = 6381/57429

@jamestanton Yes: 1/3 = 1746/5238

Data from 350,757 coin flips provide strong evidence that when some (but not all) people flip a fair coin, it tends to land on the same side it started. arxiv.org/pdf/2310.04153…

N>1 cups all upright. A "move" consists of turning all but 2 cups over. (So, each move N-2 cups change state upright/upside-down.) For which N is it possible to make all N cups upside-down? [General theory about N cups and turning N-k over at a time?]