Prime Gap by Unheralded Mathematician

http://simonsfoundation.org/features/science-news/unheralded-mathematician-bridges-the-prime-gap/

On April 17, 2013, a paper arrived in the inbox of Annals of Mathematics, one of the discipline’s preeminent journals. Written by a mathematician virtually unknown to the experts in his field — a 50-something lecturer at the University of New Hampshire named Yitang Zhang — the paper claimed to have taken a huge step forward in understanding one of mathematics’ oldest problems, the twin primes conjecture.

Editors of prominent mathematics journals are used to fielding grandiose claims from obscure authors, but this paper was different. Written with crystalline clarity and a total command of the topic’s current state of the art, it was evidently a serious piece of work, and the Annals editors decided to put it on the fast track.


Yitang Zhang (Photo: University of New Hampshire)

Just three weeks later — a blink of an eye compared to the usual pace of mathematics journals — Zhang received the referee report on his paper.

“The main results are of the first rank,” one of the referees wrote. The author had proved “a landmark theorem in the distribution of prime numbers.”

Rumors swept through the mathematics community that a great advance had been made by a researcher no one seemed to know — someone whose talents had been so overlooked after he earned his doctorate in 1991 that he had found it difficult to get an academic job, working for several years as an accountant and even in a Subway sandwich shop.

“Basically, no one knows him,” said Andrew Granville, a number theorist at the Université de Montréal. “Now, suddenly, he has proved one of the great results in the history of number theory.”

Mathematicians at Harvard University hastily arranged for Zhang to present his work to a packed audience there on May 13. As details of his work have emerged, it has become clear that Zhang achieved his result not via a radically new approach to the problem, but by applying existing methods with great perseverance.

“The big experts in the field had already tried to make this approach work,” Granville said. “He’s not a known expert, but he succeeded where all the experts had failed.”



“There are a lot of chances in your career, but the important thing is to keep thinking,” Zhang said.

Prime Secret: ζ(s)

Riemann intuitively found the Zeta Function ζ(s), but couldn't prove it. Computer 'tested' it correct up to billion numbers.

$latex \zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots$

Or equivalently (see note *)

$latex \frac {1}{\zeta(s)} =(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})(1-\frac{1}{5^{s}})(1-\frac{1}{p^{s}})\dots$

ζ(1) = Harmonic series (Pythagorean music notes) -> diverge to infinity
(See note #)

ζ(2) = Π²/6 [Euler]

ζ(3) = not Rational number.

1. The Riemann Hypothesis:
All non-trivial zeros of the zeta function have real part one-half.

ie ζ(s)= 0 where s= ½ + bi

Trivial zeroes are s= {- even Z}:
s(-2) = 0 =s(-4) =s(-6) =s(-8)...

You might ask why Re(s)=1/2 has to do with Prime number ?

There is another Prime Number Theorem (PNT) conjectured by Gauss and proved by Hadamard and Poussin:

π(Ν) ~ N / log N
ε = π(Ν) - N / log N
The error ε hides in the Riemann Zeta Function's non-trivial zeroes, which all lie on the Critical line = 1/2 :

All non-trivial zeroes of ζ(s) are in Complex number between ]0,1[ along real line x=1/2

2. David Hilbert:

'If I were to awaken after 500 yrs, my 1st question would be: Has Riemann been proven?'

It will be proven in future by a young man. 'uncorrupted' by today's math.

Note (*):

$latex \zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots = \sum \frac {1}{n^{s}}$ ...[1]

$latex \frac {1}{2^{s}}\zeta(s) =
\frac{1}{2^{s}}(1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots) $

$latex \frac {1}{2^{s}}\zeta(s) =
\frac {1}{2^{s}}+ \frac{1}{4^{s}} + \frac{1}{6^{s}} + \frac{1}{8^s} +\dots$ ... [2]

[1]-[2]:

$latex (1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{3^{s}} + \frac{1}{5^{s}} + \dots + \frac{1}{p^{s}} +\dots $

$latex \text {Repeat with} (1-\frac{1}{3^s}) \text { both sides:} $

$latex (1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{5^{s}} + \frac{1}{7^{s}} + \dots + \frac{1}{p^{s}} +\dots $

Finally,

$latex (1- \frac{1}{p^{s}}) \dots (1- \frac{1}{5^{s}})(1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1$

Or

$latex \zeta(s) = \prod \frac {1}
{1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}$

Note #:
$latex \zeta(s) = \prod \frac {1}
{1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}$

Let s=1
RHS: Harmonic series diverge to infinity
LHS:
$latex \prod \frac {1}{1- \frac{1}{p}}= \prod \frac{p}{p-1}$
Diverge to infinity => there are infinitely many primes p

[caption id="" align="alignright" width="300"] English: Zero-free region for the Riemann_zeta_function (Photo credit: Wikipedia)[/caption]

Fermat 'Prime' Mistake

Fermat Prime $latex F5= 2^{2^{5}}+1 $ composite ?

Proof: 1937 the 'Calculator boy' Zerah Colburn observed:

$latex 641 = 2^4 + 5^4 \mbox {..[1]} $
$latex 641 = 5.2^7 +1 \mbox {..[2]}$
$latex \mbox {[1:]}\: 2^4 + 5^4 = 0 \: mod \:641 $
Divide $latex 5^4 $ both sides
$latex \frac {2^4}{5^4} + 1 = 0 \:mod \:641$
$latex \frac {2^4}{5^4}= -1 \:mod \:641\mbox {..[a]} $

$latex \mbox {[2:]}\: 5.2^7 +1=0 \: mod \:641 $
$latex 2^7 = -1/5 \: mod \:641 $
x2 both sides
$latex 2^8 = -2/5 \: mod \:641 $
Raise power 4 both sides
$latex (2^8)^4 = (-2/5)^4 \: mod \:641 $
$latex 2^{32} = +2^4 / 5^4 \: mod \:641 $
$latex \mbox {[a:]}\: \frac {2^4}{5^4}= -1 \: mod \:641 $
$latex 2^{32} = -1 \: mod \:641 $
$latex 2^{32} +1= 0 \: mod \:641 $
$latex F5= 2^{2^{5}}+1 \mbox \: {divisible \:by \:641} $
=> $latex \mbox {F5 not prime!} $

Note:
$latex F1= 2^{2^{1}}+1 = 5 (prime)$
$latex F2= 2^{2^{2}}+1 = 17 (prime)$
$latex F3= 2^{2^{3}}+1 = 257 (prime)$
$latex F4= 2^{2^{4}}+1 = 65,537 (prime)$

$latex F5= 2^{2^{5}}+1 = 4,294,967,297 = 641 \mbox{x} 6,700,417 (nonprime)$

Euler (in 1732) had proved mentally F5 was not a prime.

What is Ideal ?

Anything inside x outside still comes back inside

=> Zero x Anything = Zero

=> Even x Anything = Even

Mathematically,

1. nZ is an Ideal, represented by (n)

Eg. Even subring (2Z) x anything big Ring Z = 2Z = Even

2. (football) Field F is 'sooo BIG' that

(inside = outside)

=> Field has NO Ideal (except trivial 0 and F)

Why was Ideal invented ? because of 'failure" of UNIQUE Primes Factorization" for this case (example):

6 = 2 x 3
but also
$latex 6=(1+\sqrt{-5})(1-\sqrt{-5})$
=> two factorizations !
=> violates the Fundamental Law of Arithmetic which says UNIQUE Prime Factorization

Unique Prime factors exist called Ideal Primes: $latex \mbox{gcd = 2} , \mbox{ 3}$, $latex (1+\sqrt{-5})$, $latex (1-\sqrt{-5}) $

Greatest Common Divisor (gcd or H.C.F.):
For n,m in Z
gcd (a,b)= ma+nb
Example: gcd(6,8) = (-1).6+(1).8=2
(m=-1, n=-1)

Dedekind's Ideals (Ij):
6 =2x3= u.v =I1.I2.I3.I4 ;
$latex u= (1+\sqrt{-5})$
$latex v=(1-\sqrt{-5})$

Let gcd(2,u) = 2M+N.u
M,N in form of $latex a+b\sqrt{-5}$

1. Principal Ideals:
2M = (2) = multiple of 2

2. Ideals (nonPrincipal) = 2M+N.u

3. Ideal prime factors: 6=2 x 3=u.v
Let
I1= gcd(2, u)
I2=gcd(2, v)
I3=gcd(3, u)
I4=gcd(3, v)
Easy to verify (by definition):
I1.I2=(2)
I3.I4=(3)
I1.I3=(u)
I1.I4=(v)
=> Ij are prime & unique factors of 6=I1.I2.I3.I4
=> Fundamental Law of Arithmetic satisfied!

=>Ij "Ideal"-ly exist! hidden behind 'compound' (2,3,u,v) !

Verify : gcd(2, 1+√-5).gcd(2, 1-√-5)=(2) ?

Proof by definition:

[2m+n(1+√-5)][2m'+n'(1-√-5)]
=[2m+n+n√-5 ][2m'+n'-n'√-5]
= 4mm'+2mn'+2nm'+6nn'
= 2(2mm'+mn'+m'n+3nn')
= 2M
= 2 multiples
= (2) = Principal Ideal

Riemann Hypothesis

Riemann intuitively found the Zeta Function ζ(s), but couldn't prove it. Computer 'tested' it correct up to billion numbers.

$latex zeta(s)=1+frac{1}{2^{s}}+frac{1}{3^{s}}+frac{1}{4^{s}}+dots$

Or equivalently (see note *)

$latex frac {1}{zeta(s)} =(1-frac{1}{2^{s}})(1-frac{1}{3^{s}})(1-frac{1}{5^{s}})(1-frac{1}{p^{s}})dots$

ζ(1) = Harmonic series (Pythagorean music notes) -> diverge to infinity
(See note #)

ζ(2) = Π²/6 [Euler]

ζ(3) = not Rational number.

1. The Riemann Hypothesis:
All non-trivial zeros of the zeta function have real part one-half.

ie ζ(s)= 0 where s= ½ + bi

Trivial zeroes are s= {- even Z}:
s(-2) = 0 =s(-4) =s(-6) =s(-8)...

You might ask why Re(s)=1/2 has to do with Prime number ?

There is another Prime Number Theorem (PNT) conjectured by Gauss and proved by Hadamard and Poussin:

π(Ν) ~ N / log N
ε = π(Ν) - N / log N
The error ε hides in the Riemann Zeta Function's non-trivial zeroes, which all lie on the Critical line = 1/2 :

All non-trivial zeroes of ζ(s) are in Complex number between ]0,1[ along real line x=1/2

2. David Hilbert:

'If I were to awaken after 500 yrs, my 1st question would be: Has Riemann been proven?'

It will be proven in future by a young man. 'uncorrupted' by today's math.

Note (*):

$latex zeta(s)=1+frac{1}{2^{s}}+frac{1}{3^{s}}+frac{1}{4^{s}}+dots = sum frac {1}{n^{s}}$ ...[1]

$latex frac {1}{2^{s}}zeta(s) =
frac{1}{2^{s}}(1+frac{1}{2^{s}}+frac{1}{3^{s}}+frac{1}{4^{s}}+dots) $

$latex frac {1}{2^{s}}zeta(s) =
frac {1}{2^{s}}+ frac{1}{4^{s}} + frac{1}{6^{s}} + frac{1}{8^s} +dots$ ... [2]


[1]-[2]:

$latex (1- frac{1}{2^{s}})zeta(s) = 1+ frac{1}{3^{s}} + frac{1}{5^{s}} + dots + frac{1}{p^{s}} +dots $

$latex text {Repeat with} (1-frac{1}{3^s}) text { both sides:} $

$latex (1- frac{1}{3^{s}})(1- frac{1}{2^{s}})zeta(s) = 1+ frac{1}{5^{s}} + frac{1}{7^{s}} + dots + frac{1}{p^{s}} +dots $


Finally,

$latex (1- frac{1}{p^{s}}) dots (1- frac{1}{5^{s}})(1- frac{1}{3^{s}})(1- frac{1}{2^{s}})zeta(s) = 1$

Or

$latex zeta(s) = prod frac {1}
{1- frac{1}{p^{s}}}= sum frac {1}{n^{s}}$

Note #:
$latex zeta(s) = prod frac {1}
{1- frac{1}{p^{s}}}= sum frac {1}{n^{s}}$

Let s=1
RHS: Harmonic series diverge to infinity
LHS:
$latex prod frac {1}{1- frac{1}{p}}= prod frac{p}{p-1}$
Diverge to infinity => there are infinitely many primes p




[caption id="" align="alignright" width="300"] English: Zero-free region for the Riemann_zeta_function (Photo credit: Wikipedia)[/caption]

God's "irregular heartbeat"

Primes =  God's "IRREGULAR heartbeat".

You never know the pattern to generate the next prime.

Riemann Conjecture's Zeta function.

Hibert's 8th/23 unsolved problems @1900.

Clay's 7 Millennium Problem (USD 1 million prize)

Field: Galois, Dedekind

Dedekind
(1831-1916)

Dedelind was the 1st person in the world to define Field:
"Any system of infinitely many real or complex numbers, which in itself is so 'closed' and complete, that +, - , *, / of any 2 numbers always produces a number of the same system."

Heinrich Weber (1842-1913) gave the abstract definition of Field.

Field Characteristic

1. Field classification by Ernst Steinitz @ 1910
2. Given a Field, we start with the element that acts as 0, and repeatedly add the element that acts as 1.
3. If after p additions, we obtain 0 again, p must be prime number, and we say that the Field has characteristic p;
4. If we never get back to 0, the Field has characteristic 0. (e.g. Complex Field)

Example: GF(2) = {0,1|+} ; prime p = 2
1st + (start with 0):
0 + 1 = 1
2nd (=p) +:
1 + 1 = 0 => back to 0 again!
=> GF(2) characteristic p= 2

Galois Field GF(p)

1. For each prime p, there are infinitely many finite fields of characteristic p, known as Galois fields GF(p).

2. For each positive power of prime p, there is exactly one field.
(This is the only IMPORTANT Theorem need to know in Field Theory)
E.g. GF(2) = {0,1}

Math Game: Chinese 9-Ring Puzzle  (九连环 Jiu Lian Huan)

http://www.google.com.sg/imgres?imgurl=http://info.makepolo.com/uploadfile/2012/0723/20120723100653765.jpg&imgrefurl=http://info.makepolo.com/htmls/6/69/2669.html&h=400&w=533&sz=44&tbnid=ExodLfHv3cQjHM:&tbnh=91&tbnw=121&zoom=1&usg=__hsZaBecpPNdvTvguQbaQftCsXgo=&docid=qXMWtmo8A-vXEM&hl=en&sa=X&ei=f2NaUciqKYrOrQeT2YHgDw&sqi=2&ved=0CEsQ9QEwAg&dur=591

To solve Chinese ancient 9-Ring Puzzle (九连环) needs a "Vector Space V(9,K) over Field K"

finite Field K = Galois Field GF(2) = {0,1|+,*}
and 9-dimension Vector Space V(9,K):
V(0)=(0,0,0,0,0,0,0,0,0) ->
V(j) =(0,0,... 0,1,..0,0) ->
V(9)= (0,0,0,0,0,0,0,0,1)

From start V(0) to ending V(9) = 511 steps.

Prime Numbers * +

Prime Numbers

God multiplies Primes: Fundamental Law of Arithmetic (Euclid):

Any number (n) can be UNIQUELY Prime Factorized: n = P1 x P2 x ...Pn

e.g. 12= 2 x 2 x 3,
28=2 x 2 x7

Men add Primes: Goldbach Conjecture.
2n = P1+P2

e.g. 12= 5 + 7, 28 = 17 + 11, 30 = 11+19

Find Prime numbers

Primes (Landau's trick)
List down systematically the primes below n < 4,000
Start with 2
2x2= 4 => take 3 (<4)
< 2x3 =6 => 5
< 2x5 =10 =>7
< 2x7 =14 => 13
...
23, 43, 83, 163, 317, 631, 1259, 2503, (4001)

Prime and Perfect Square

For all primes p ≠2, (a,b ∈Z)

p= a² + b² <=> p ≡ 1 mod 4


(2=1² + 1²)
5= 1² + 2² = 1 + 4 ≡ 1 mod 4
13= 2² + 3² = 4 + 9 ≡ 1 mod 4
17=1² + 4² = 1 + 16 ≡ 1 mod 4
29= 2² + 5² = 4 + 25 ≡ 1 mod 4
37= 1² + 6² = 1 +36 ≡ 1 mod 4

Notes:

1) Perfect squares (4, 9, 16, 25... ) ≡ 0 or ≡ 1 mod 4
2) Prime (4n+1) = a² + b²   (Euler took 7 yrs to prove)

3) Gauss expanded the proof to quadratic reciprocity (2 prime numbers p & q are linked by mod 4)

Prime Number

Prime Number

Prove we can always find an interval [p, q] wherein no prime exists, for any p, q ∈ N
Proof:
For any n ∈ N
Let H= (n+1)! = 1 x 2 x ... x(n+1)
Let G = {H+2, H+3,. . . , H+(n+1) }
=> G composites (trivial)

Choose [p, q] such that  :
1. p the closest prime < H+2,
2. q the closest prime > H+(n+1)
=> no prime ∈ [p, q]

---n----------------n!-(H+1)-p[(H+2)---------H+(n+1)]q------------------

Fermat Little Theorem

Fermat ℓittle Theorem (FℓT)

∀ m ∈ N,
1) p prime => $Latex m^p$= m mod (p)
Note: Converse False
[Memorize Trick: military police = military in the mode of police]
Note:
p prime, ∀m,
if p | m,
=> m ≡ 0 mod (p) ...(1)
=> multiply p times:
$Latex m^p$≡ 0 mod (p) ...(2)

Substitute (1) to (2): m ≡ 0
=> p prime, ∀ m
$Latex m^p$≡ m mod (p)
=> No need (m, p) = 1 [co-prime]

E.g. (m, p) = (3, 2) =1
$Latex 3^2$= 9 ≡ 3 mod (2) ≡ 1 mod (2)
9 ≡ 1 mod (2)

E.g. (m, p) = (6, 2), p =2 |6
$Latex 6^2$= 36 ≡ 6 mod (2) ≡ 0 mod (2)
6 ≡ 0 mod (2)

WRONG EXAMPLE:  p = 4 = non prime
$Latex 2^4$= 16 ≡ 2 mod (4)
but 16 ≡ 0 mod (4)
Equivalent:
p prime => $Latex m^{p-1}$= 1 mod (p)

by Contra-positive:
3) $Latex m^{p-1}$ ≠ 1 mod (p)
=> p non-prime

Apply: Prove 39 non-prime?
Take m =2, p= 39
Prove $Latex m^{p-1}$ ≠ 1 mod (p)?
a) $Latex 2^{38} =2^{36}.2^2$ .
b) $Latex 2^6$= 64 ≡ 25 mod (39)
c) $Latex 25^2$ = 625 ≡ 1 mod (39)
From (b) & (c):

$Latex (2^6)^6$ = $latex 2^{36}$ ≡ $latex 25^6$ mod(39)≡$latex (25^2)^3$ ≡$Latex 1^3$ ≡ 1
From (a): $Latex 2^{38}$ = $Latex 2^{36}.2^2$ ≡ 1x4 ≠ 1 mod(39)
=> 39 non-prime