An origami-inspired adventure in number theory…and on limits Jeanine Meyer Purchase College/SUNY http://faculty.purchase.edu/jeanine.meyer

The adventure begins … Mark Kennedy led people coming together to wait for free tickets to Shakespeare in the Park in NYC and fold. He taught the dollar bill rosette.

Overview [Warm-up] Making estimates Dollar bill rosette Estimating thirds Dollar bill rosette Procedure for making off sections. What does procedure do? Iterates to improve original estimate: E  0 Marks off ALL of intermediate sections For what numbers besides 11 does the procedure work? Python Program generating numbers (under 1000) that work Used this set to identify class of primes Proof

Kissy fish http://faculty. purchase. edu/jeanine

Method to divide into thirds "S" method (actually a Z). Form a S or Z shape looking at it sideways and keep adjusting parts until the same size. Variant: Make an estimate and fold to the estimate (dividing the 2/3 side into 2 part). Suppose the initial fold is off by error E: Say it should be A, but instead is A+E. (Can assume it is on one side). This divides the bill into two parts measuring A+E and Length – (A+E) Assuming that folding a section in half is pretty accurate, the next fold produces a mark at ( Length-(A+E) )/2 So this new mark is off by E/2. The error has been halved! Keep going

Dollar bill rosette Directions (crude drawings) at http://faculty.purchase.edu/jeanine.meyer/origami/oriros.htm

Dollar bill rosette first step Fan fold of 22 parts starts with need to divide bill into 11ths. Start with guess for 1/11. Dollar bill is divided into 1 part and 10 parts, even part and odd part. Divide the even part in half to get 6 parts and 5 parts. Continue. 1 and 10 6 and 5 3 and 8 7 and 4 9 and 2 10 and 1 Keeping going. Get back to 1 and 10. So now have mark at each place 1/11, 2/11, etc.

Repeating argument on error Suppose first estimate is off by error E: Say it should be A, but instead is A+E. (Can assume it is on one side). This divides the bill into two parts measuring A+E and Length – (A+E) Assuming that folding a section in half is pretty accurate, the next fold produces a mark at ( Length-(A+E) )/2 So this new mark is off by E/2. The error has been halved! Continuing through the 10 steps, when back to (around) the first mark, the new mark is off by E/210

Continuing folding With improved estimate, goes through one more time making complete folds, all valleys Divide each of the 11 sections by 2, making mountain fold, to get 22 sections in a fan fold Divide into two Fold out 3 panels on each side Fold at the midpoint Make tabs to fold in to attach the two parts. Pop open to be a rosette

Rosette procedure Besides improving the estimate, the procedure gets back to the start (1, N-1) hitting all the divisions in-between. For what numbers does the procedure work? Must be odd number Must be prime Proof? What kind of prime

Suggestion that it must be prime Suppose P is not prime, and equal to F * G, F and G each >1. Note: neither F or G can be even. In the rosette procedure, at some point, the pair of numbers is G and (F-1)*G. What is the next step? G+((F-1)/2)*G and ((F-1)/2)*G Claim: never get to 1 and (F*G) -1.

All primes? Does procedure hit all numbers for 17 NO 1 and 16 9 and 8

What primes? Adventure continues… After learning fold, a long time ago, a mathematician colleague told me that it probably was certain category of numbers, but I forgot… So I wrote Python program to generate numbers that work

Experiment Wrote program to invoke the testing program for all numbers May have checked for primes first … from 3 to 1000

def tryProcedure(N): count = 1 # start with 1 and N-1 currentpos = 1 remainder = N-currentpos while True: if (isEven(currentpos)): # decide which side is even currentpos = currentpos//2 # integer division remainder = N - currentpos else: currentpos = currentpos + remainder//2 # integer division count = count + 1 if (currentpos==1): break if (count==N): print(" ",N,end="") #good return

Numbers from 3 to 1000 3 5 11 13 19 29 37 53 59 61 67 83 101 107 131 139 149 163 173 179 181 197 211 227 269 293 317 347 349 373 379 389 419 421 443 461 467 491 509 523 541 547 557 563 587 613 619 653 659 661 677 701 709 757 773 787 797 821 827 829 853 859 877 883 907 941 947

Experimental computational mathematics Generated list of the numbers that work up to 1000. Took those numbers and put into Google…. Got to https://en.wikipedia.org/wiki/Full_reptend_prime

What is reptend prime base 2? Numbers P for which 2 is a primitive root modulo P This is a procedure, also. …. Number P for which 2 raised to the power N, N going from 0 to P-2, produces the numbers 1 to P-1, modulo P

What is mod Two numbers are equal mod X if the difference is a multiple of X. M = N mod P if and only if (this is the definition) M-N is a multiple of P. Putting it another way: M = N + a*P where a is an integer This sometimes is called clock arithmetic. For clock arithmetic, the P is 12.

Let's try 11 20 is 1 = 1 mod 11 21 is 2 = 2 mod 11 22 is 4 = 4 mod 11 23 is 8 = 8 mod 11 24 is 16 = 5 mod 11 24 = 5 + 1*11 25 is 32 = 10 mod 11 25 = 10 + 2*11 26 is 64 = 9 mod 11 ? 27 is 128 = 7 mod 11 28 is 256 = 3 mod 11 29 is 512 = 6 mod 11

Interesting…. Not just that 11 works: reptend process hits all the intermediate numbers AND the numbers appear to be the numbers in the dollar bill rosette procedure going backwards…..

Plan To prove that the numbers that can work using the folding procedures are the reptend primes base 2, one needs to prove that the numbers for which the folding procedure hits all the intermediate numbers are the same as the numbers for which the reptend process, raising 2 to powers from 0 to number -1 and then taking mod, hits all the intermediate numbers. Instead, I decided to try for a stronger result: the two procedures are the same procedure, though in opposite order.

Reverse folding process Need to determine: if F and P-F go to G and P-G, starting at G, how to get back to F Consider cases of if F is (was) odd or even The pair G and P-G came from F and P-F either by halving F or halving P-F, so Consider 2*G. If 2*G > P, then need to consider 2*(P-G) If 2*G > P, then 2*(P-G) <P. Why? Both ways "work". I use the second way. Two numbers adding up to P. They are different (note P is odd so they can't be equal) so one must be less than ½ P and the other must be more.

Stronger result the two procedures are the same procedure, though need to take folding in reverse order. Need to prove that for k, the first number of the M, P-M pair by (reverse) folding is 2k mod P Note: both processes can continue past P-2. Proof by induction Initial case Induction case Assume true for k, prove for k+1

Initial case Reptend and reverse folding start out with 1 Reptend: 20 =1, so 20 =1 mod P Reverse folding starts with 1.

Induction step Can assume G = 2k mod P which means G = 2k + a*P where a is an integer What is the next case for reverse folding? Two cases: 2*G< P and 2*G > P. Case 1: 2*G< P. So F = 2*G. Expand F F=2*(2k + a*P) = 2k+1 + 2*a*P so F = 2k+1 mod P

Continuing with 2nd case Case 2: 2*G>P. So F and P-F produced next step by halving the P-F term. So the P-F term is equal to 2*(P-G). F = P-2*(P-G) F = P – 2*P + 2*G rearranging terms F = 2*G – P now substitute G F = 2*(2k + a*P) – P F = 2k+1 + 2*a*P – P F = 2k+1 + (2*a-1)*P F = 2k+1 mod P

Conclusion of proof The reptend process of raising 2 to a power and then calculating mod P produces the same numbers as the reverse folding process. Because both processes yield the same results at each step, both either satisfy criteria of hitting all the intermediate points between 1 and P-1 or neither does. Note: the fact that folding resembled reptend for 11 and that the numbers under 1000 matches are not proofs. What I meant by a stronger result

Recap Origami inspired adventure folding improving approximations programming online investigation proof

References http://faculty.purchase.edu/jeanine.meyer/origamiAdventure.docx My books frequently contain origami-related material Programming 101 (Processing), Apress, June, 2018 Chapter 6: Origami flower; Chapter 8 (extra) Directions. HTML5 and JavaScript Projects, Apress, 2nd Edition scheduled for publication very soon Chapter 7: Using HTML & JavaScript for Origami Directions