p-adic numbers

"Stopped understanding anything past 1:30 but still watched the entire thing." (Commentary to the Video)

"Real numbers are like the sun, p-adics are like the stars. Sun blocks out stars during day, humans sleep at night and dont see stars, even though they are just as important." (Quote by mathematician Kazuya Kato)

5²=	25	Always ends with itself digits.
25²=	625
625²=	390,625
390,625²=	152,587,890,625	Doesnt end with itself, just take the digits that do.
90,625²=	8,212,890,625
8,212,890,625²=	67,451,572,418,212,890,625	Already 10 same digits.

Can be repeated, its like converging on a number with infinite digits to the left (of the decimal point), with $\mathrm{n}²=\mathrm{n}$.

- Written in base 10
- Addition: digit by digit from right to left: ...111 + ...222 = ...333

- Multiplication: repeatedly multiply multiplicand with digits of multiplier from right to left, products have increasing numbers of zeroes from right, thus can be incrementally added for the result.

...456 x ...321 ...456 ...120 ...800 + ⋮ ...376

- Multiplying ...857142857143 with 7 (digits from right to left) yields ...00001 (only zeroes left of 1) which is 1, thus the 10-adic number is $\frac{1}{7}$. Thus there are rational numbers, fractions in the 10-adic numbers without use of the divided by symbol.
- Whats 10-adic number is $\frac{1}{3}$ ? Take ...??????? x 3 = ...00001 and calculate digits of multiplicand from right to left: ...666667.

- 10-adic numbers like ...6667. look similar to infinite number going to right of the decimal point, e.g. 0.999... Is this 1? Take .999...=k, multiply by 10 -> 9.99..=10k, subtract both equations -> 9=9k -> k=1
- from left to right to right to left, was does ...999. equal? set ...999.=m, multiply by 10 -> ...990=10m, substract second from first equation -> 9=-9m -> m=-1 -> ...999=-1
- test: ...999+1=...000=0
- similarly e.g. ...993=-7
-> 10-adic numbers contain negative numbers without needing a negative sign
- with 10-adic numbers to do substraction, add negative of the number
- to find negative of any 10-adic number, one can multiply it with ...999 (-1), or perform 2 steps:
- Take 9´s complement (difference between each digit and 9), then add 1. Example: 1/7:...857142857143, -1/7:...142857142856+1, adding them gives 0

- first number has n²=n, why is it problematic? n(n-1)=0, solutions: n=1, n=0, but also above ...0625 x ...0624 = ...000
- breaks tool of mathematicians to solve equations, e.g. x³+5x²=4x+20, move terms to one side, set to zero and factor them: (x+2)(x-2)(x+5)=0, one of the terms must be 0, x=-2,x=2,x=-5.
- with 10-adic numbers factored terms must not be zero, because 10 is a composite number, not a prime, 10=5x2
- find 2 10-adic numbers ...???x...???=0, rightmost digits can be 0, but also 5x4=0
- prime-adic/p-adic numbers fix this, e.g. 3-adic; digits can be 0,1,2; for ...???x...???=0 a rightmost digit must be 0
- 3-adic number ...121 means ...+1x3^2+2x3^1+1x3^0, 3-adic is infinite expansion in powers of 3; -1=...222, test ...222+1=0

- p-adic number are tool for fields medalists solving problems, e.g. 1637 Pierre de Fermat reads book 'Arithemtica' by greek Diophantus.
- Diophantos looked for solutions for polynomial equations phrased in geometrc terms, like the Pythagorean (1000bc) theorem x²+y²=z².
- set of solutions in real numbers is infinite cone, Diaphantos wanted whole numbers or fractions.
- babylonian clay tablet from 2000bc already had list of pythagorean triples.
- Fermat writes next to the pythagorean discussion: x^n+y^n=z^n has no integer solutions for n>2, i have a marvelous proof but it doesnt fit in the margin.
- This "Fermats last theorem" was unproven for 358 years.
- To solve it, new numbers had to be invented, the p-adics. They provide a systematic method to also solve other problems of 'Arithmetica'.

- Find 3 squares, whose areas added create a bigger square, and where area of a square is sidelength of the square right to it: x^2+x^4+x^8=y^2.
- Real number solution is easy, e.g. x=1 -> 3=y² -> y=sqrt(3), x-y-plot of real number solutions can be made.
- Diophantus wanted rational solutions, whole numbers or fractions. Where to find them?
- In late 1800s mathematician Kurt Hensel tried to find solutions in the form of an expansion of increasing powers of primes.
- E.g. with prime 3: x=x0 + x1 x 3^1 + x2 x 3^2 + ..., same with y. Each coefficient would be 0,1 or 2.
- Insert expressions into equation, looks difficult but can be simplified.

- 17 base 3 -> ???, one way: 17/3 remainder is 2 -> ??2, then 17/9 (next higher power of 3) remainder 8, subtract found before 2: 8-2=6 which is 2x3^1 -> ?22, then 17/27 remainder is 17, subtract accounted for 8: 17-8=9 = 1x3^2 -> 122.
- Form of modular arithmetic, where numbers reset back to 0 once the reach value "modulus".
- Hours of a clock work with modulus of 12. E.g. hour 10 + 4 hours = 14 modulo 12 = hour 2.
- Modular arithmetic is about finding th remainder. 36 modulo 10 or 36 mod 10 is 6, 25 mod 5 is 0.

- Modular arithmetic allows to work out coefficients in expansions one at a time. First solving equation mod 3, then mod 9, mod 27,...
- With mod 3, all higher terms are divisible by 3, thus 0. -> x=x0, y=y0, x0^2+x0^4+x0^8=y0^2.
- x0, y0 can be 0,1,2; make table, possible solutions for x0,y0 are (0,0), (1,0), (2,0).
- Solution (0,0) is straightforward, but doesnt count for Diophantus.
- Expand (1,0), solve equation mod 9. Higher terms get 0, x=1+x1 x 3^1, y=0+y1 x 3^1.
- Enter into polynomial equation, multiply out, terms multiplied by 9 get 0. -> 3+15x1=0. x1=1 is a solution, 3+15=18=0mod9.
- Solve equation mod 27, higher terms get 0, x0=1, x1=1, y0=0, y1 unknown for now.
- Equation simplifies to only have x2,y2. Multiply out, mod 27 gets 0. -> 9+18x2=0, x2=1 because 9+18=0.
- First 3 coefficients are 1, next would be also 1. Thus the 3-adic number ...111 is a solution.

- ...111 is 1x3^0 + 1x3^1 + 1x3^2 + ..., each term is 3x the term before. Its a geometric series. 1+L^1+L^2+...=1/(1-L)
- L is meant to be less than 1 else sum doesnt converge. For now, see what happens, L=3, x=-1/2.
- Enter into polynomial equation, put terms over common denominator: 64/256+16/256+1/256=81/256
- 81/256 is square so rational solution is found, sidelengths: 1/2, 1/4, 1/16, 9/16
- Solution found with p-adic numbers and geometric series formula. How does it work?

- Geometry of p-adics is different from that of real numbers. They dont exist on a number line.
- Can be visualized with a growing tree, 3-adic is tree with 3 base cylinders, each infinitely tripely branching.
- Looks like Sierpinski gasket. Each number is stack of infinite cylinders getting narrower.
- This reflects relative contributions each successive cylinder makes to the value of the 3-adic number.
- Coefficients of higher powers of 3 actually make finer and finer adjustments.
- Calculating coefficients of Diaphanus square problem is zooming in more and more accurately on the solution.
- Normally the size of a number is determined by digits left to the decimal point, here all numbers have infinitely left digits.
- Distance between 2 numbers, look at lowest level where they differ. First digit, separation is 1. If they differ at 27th place, they differ by 1/27.
- In world of p-adics big is small and vice versa. If 2 3-adics start to differ at 3^3, their difference is about 3^-3.
- With swapping big and small all math laws work in the usual way. This is why the geometric sum worked out and didnt blow up.
- Notion of size fits criteria for absolute value ||x||: non-negative, positive definite (||0||=0), multiplicative ||x*y||=||x||*||y||, triangle inequality ||x+y||<=||x||+||y||.
- Geometry makes p-adics disconnected, compared to real numbers, useful for finding rational solutions to an equation, there are fewer p-adic solutions in the neighborhood of a rational solution. Solving Diophantos equation with real numbers would fail, because there are too many real solutions, that get in the way.
- Groundbraking papers from 1995 by Andrew Wiles and by Wiles and Richard Taylor proved Fermats last theorem with p-adic numbers.
- Proof was first tried with 3-adic then 5-adic. Three,five trick: somehting didnt worked with 3 but with 5. Each prime gives an unrelated numbersystem.
- Quote by mathematician Kazuya Kato: "Real numbers are like the sun, p-adics are like the stars. Sun blocks out stars during day, humans sleep at night and dont see stars, even though they are just as important."