Monday, January 11, 2010

Pierre de Fermat: an anniversary

Pierre de Fermat died exactly 345 years ago, on January 12th, 1665, at the age of 57. His father was a rich leather merchant and a mayor (consul, to be precise).

Fermat was a lawyer - and an amateur mathematician. That's a remarkable description because together with René Descartes, he was one of the two most important mathematicians in the first half of the 17th century. The only other amateur mathematician and amateur lawyer who got equally successful was a Kenya-born community organizer mentioned in a footnote of The Curvature of Constitutional Space. ;-) This chap has even received a Nobel prize.

But let's return from Obama to Fermat.

Probability theory

Together with Blaise Pascal, he was a 17th century co-father of the probability theory. Most famously, they studied the problem of points, a particular exercise in game theory. Imagine a two-player game with many rounds. During each round, each player adds X dollars to the jackpot. The first player who wins N rounds gets the jackpot.

Fine. That's not the problem of points yet. The problem is how the players should divide the jackpot if they're forced to stop before anyone records N victorious rounds. The "fair" division of the jackpot should clearly depend on the numbers of victories so far. Fermat gave the obviously correct solution, improved by Pascal two years later.



Player #1 needs "r" more rounds to win, player #2 needs "s" more rounds to win. Let's imagine they could keep on playing. After "r+s-1" rounds or less, someone would clearly have to win. We can imagine they would continue to play for fun even after it is decided who wins.

So there are "2^{r+s-1}" equally likely post-interruption histories. Fermat counted how many of them are the victories of #1 and how many of them are victories of #2. The ratio of these two integers is the ratio of the probabilities for them to win (the odds), and the jackpot should be divided according to the same ratio.
A clearly valid solution. Pascal found a better method to "count" the histories than Fermat's one-by-one approach which is getting exponentially inefficient. Those 350 years ago, things were not terribly formalized and ordered but you can see that they effectively understood the concept of probabilities and expectation values.

Pierre de Fermat is also a co-father of the analytic geometry, together with René Descartes. Imagine how hard geometry and physics would be without coordinates.

He found a method to find the maxima and minima of functions: this problem was actually stated as a task to find the greatest and smallest ordinates (values of the last coordinate) of the corresponding curved lines (graphs of the functions). According to his own words, Isaac Newton was "directly" building on Fermat's procedures.

When Newton was inventing calculus, you may see that Fermat was obviously one of the giants upon whose shoulders Newton was standing. Fermat did know lots of things about the tangents and vanishing derivatives: Newton could have turned them into a real industry because he used a far more algebraic language to talk about them using a unified language.

Optics

In optics, Fermat discovered his, Fermat's principle. Light travels along lines where the journey takes the least time. This principle explains the straight lines in empty space, equal angles of reflection, and - most non-trivially - the angles of refraction on the boundary of different media with different speeds.

The principle may be deduced from Huygens' principle describing light as interfering waves coming from each new point. Historically, you should realize that Huygens was a (13 years older) contemporary of Newton's, i.e. his principle came after Fermat's principle.

The derivation of Fermat's principle from Huygens' principle mimics the derivation of the classical limit from Feynman's path integral: near the extrema (in the space of curves), the interference is very constructive. Again, this is an acausal comment because Feynman came centuries after Huygens. But I don't have to explain you that. ;-)

Number theory

Despite his huge contributions to geometry, optics, and probability theory, many of us think of number theory - insights about integers and their relationships - when someone says "Fermat". He has spent lots of time by solving Diophantine equations i.e. polynomial equations with integers allowed as the only solutions.

At the margin of his copy of Diophantus' Arithmetica, he formulated his Fermat's Last Theorem. Unfortunately, there was "not enough room" for him to write his "remarkable proof" on the sheet of paper, so I can't rigorously prove that Fermat was either bullshitting or mistaken. ;-)

All great mathematicians after Fermat broke their teeth with this theorem. The theorem was finally proven by Andrew Wiles in the 1990s. Just to remind you, the goal is to prove that
xn + yn = zn
has no uniformly positive integer solutions "(x,y,z,n)" with "n" greater than two. For "n=2", you could invent infinitely many "Pythagorean" counterexamples. The cases "n=4" and even "n=3" and others could have been proven - exponents up to hundreds were known to be impossible decades ago - but the general "n" case was resisting for quite some time.

Last time I studied these things in detail was at the high school - so I only know the "somewhat different and simpler" proof for "n=4" only. The "n=4" case is the only non-prime value of "n" where you have to produce a separate proof. All other composite exponents "n=pq" can be eliminated trivially if you have proven the theorem for "n=p" or "n=q". That's because "x^{pq}=(x^p)^q".

The equation above looks purely "discrete" or "number-theoretic" except that it is very useful to look at the animal geometrically. Wiles proof uses a lot of algebraic geometry and homomorphisms between groups including "GL(2,Z_{N})". While modern mathematicians such as Wiles could have used some truly powerful tools, it's fair to say that the general "spirit" - the deep and fruitful relationship between number theory, algebra, and geometry - was known to Fermat himself.

2 comments:

  1. My solution for Pierre De Fermat 's last Theorem.
    The conditions:
    x,y,z,n are the integers and >0. n>2.
    z^n=/x^n+y^n.

    Suppose:
    z^n=x^n+y^n
    With m<n
    So
    z^m+a=x^m+y^m=b
    So
    a and b are two integers.
    So
    (z^m+z^n)+a=(x^m+x^n)+(y^m+y^n)=c
    So
    c is an integer.
    Because
    z^m+z^n=z^m[1+z^(n-m)]=d^m

    I had invented a string which is very important to solve Flt in general case.
    Define
    z^k=w^t-1
    and
    z^(k-1)=e^(t+1)-1
    z^(k-2)=f^(t+2)-1
    z^(k-3)=g^(t+3)-1
    z^(k-4)=h^(t+4)-1.
    .....
    So
    z=(w^t-1)/[e^(t+1)-1]=[e^(t+1)-1]/[f^(t+2)-1]=[f^(t+2)-1]/g^(t+3)-1]=[g^(t+3)-1]/h^(t+4)-1.]=.....
    ....
    ....
    This is a string.
    Impossible all the numbers w,e,f,g,h,....are the integers.
    Example
    z=2
    So
    1+2^3=3^2.
    So
    2^3=3^2-1.
    2^2=u^3-1
    2^1=r^4-1
    2^0=v^5-1
    2^(-1)=s^6-1
    ....
    Impossible
    all the numbers
    3,u,r,v,s are the integers
    ....
    So
    Exist an irational number p.
    1+z^(n-p)=p^q
    So
    z^q[1+z^(n-q)]=(zp)^q
    Define
    The numbers zp and q are d and m.
    So
    z^m[1+z^(n-m)]=d^m
    Because d is an irratjonal number so i define
    d^m=(z+u)^m=(j^1/m)^m
    So
    z+u=j^1/m is an irrational number.

    I should use this important equation in my proof.
    Now starting by simple case.

    m=2
    Suppose:
    z^n=x^n+y^n
    So
    z^2+a=x^2+y^2=b
    So
    a and b are two integers.
    So
    (z^2+z^n)+a=(x^2+x^n)+(y^2+y^n)=c.
    c is an integer.
    Because
    z^2+z^n=z^2[1+z^(n-2)]=(e^1/2)^2 is an integer.However e^1/2 is an irrational number. Had proved.
    Because had proved.
    z+u=r^1/m is an irrational number.
    So define
    (z+f)=e^1/2
    Because
    a=b-z^2.
    And
    a=c-(z+f)^2
    So
    b-z^2=c-(z+f)^2.
    So
    b-z^2=c-z^2-f^2-2zf
    So
    2zf=c-b - f^2.
    So
    z=(c-b)/2f - f/2
    Because
    z+f=e^1/2
    So
    z=(c-b)/[2(e^1/2-z)] - (e^1/2-z)/2.
    So
    z=(c-b)/(2e^1/2-2z ) - e^1/2/2+z/2
    So
    z/2=(c-b)/(2e^1/2-2z ) - (e^1/2)/2
    Because
    (c-b) is an integer.
    (2e^1/2-2z ) is an irrational number.
    And
    (e^1/2)/2 is an irrational number.
    Example:
    z/2=8/(52^1/2-7) - (3.25^1/2)
    Impossible
    So
    z is an irrational number.

    And
    another simple case.
    m=3
    Suppose
    z^n=x^n+y^n
    So
    z^3+a=x^3+y^3
    So
    z^n+z^3+a=x^n+x^3+y^n+y^3=c
    Had proved
    z^n+z^3=(z+f)^3=(e^1/3)^3
    So
    z+f=e^1/3.
    Because
    a=b-z^3=c-(z+f)^3
    So
    b-z^3=c-z^3-3z^2f-3zf^2-f^3
    b-z^3=c-z^3-3(zf)(z+f)-f^3
    3zf=(c-b)/(z+f)-f^3/(z+f)
    So
    Because
    f=e^1/3-z
    So
    z=(c-b)/3(e^1/3-z)(e^1/3) - (e^1/3-z)^2/3e^1/3
    So
    z=(c-b)/(3e^2/3-3ze^1/3) - (e^1/3)/3+2/3z - z^2/3e^1/3
    So
    z/3=(c-b)/(3e^2/3-3ze^1/3) - (e^1/3)/3 - z^2/(e^1/3)
    So
    z is an irrational number.

    General case.
    m is any integr.
    to continue about
    z^m[1+z^(n-m)]=d^m
    Because
    d is an irrational number
    Define
    (z+f)^m=d^m=(e^1/m)^m
    So
    (z+f)=e^1/m is an irrational number.
    Similar the cases
    m=2 and m=3....
    So
    z^n=/x^n+y^n

    ISHTAR.

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  2. My solution for Pierre De Fermat 's last Theorem.

    The conditions:

    x,y,z,n are the integers and >0. n>2.

    z^n=/x^n+y^n.

    Define:

    z^(n-m+a)=q^(m-a)-1

    z^(n-m+a-1))=w^(m-a+1)-1

    z^(n-m+a-2)=s^(m-a+2)-1

    z^(n-m+a-3)=v^(m-a+3)-1

    ....

    z^(n-m+a-a)=r^(m-a+a)-1

    .....

    So

    z=[q^(m-a)-1]/[w^(m-a+1)-1]=[w^(m-n+1)-1]/[s^(m-a+2)-1]=[s^(m-a+2)-]/[ v^(m-a+3)-1] =......=...

    This is a special string .Impossibe all the numbers q,w,s,v and r are the integers

    So

    there exists at least one case :

    z^(n-m+a-a)=r^(m-a+a)-1

    r is an irrational number

    So

    1+z^(n-m)=r^m

    Because

    z^m+z^n=z^m[1+z^(n-m)]

    r is an irrational number and z is an integer.

    So there exists at least one case:

    z^m+z^n=d^m

    d is a an irrational number

    This is an important equation that was used in general proof.

    Special case

    m=2

    Suppose:

    z^n=x^n+y^n

    So

    z^2+a=x^2+y^2=b

    So

    a and b are two integers.

    So

    (z^2+z^n)+a=(x^2+x^n)+(y^2+y^n)=c.

    c is an integer.

    So

    z^2+z^n=z^2[1+z^(n-2)]=d^2

    Appear two cases

    d is an integher or an irrational number.

    Case d is an integer. Watching at general case later.

    Case d is an irrational number . Continuing:

    Define

    (z+f)=d=e^1/2

    e^1/2 is an irrational number.

    Because

    a=b-z^2.

    And

    a=c-(z+f)^2

    So

    b-z^2=c-(z+f)^2.

    So

    b-z^2=c-z^2-f^2-2zf

    So

    2zf=c-b - f^2.

    So

    z=(c-b)/2f - f/2

    Because

    z+f=e^1/2

    So

    z=(c-b)/[2(e^1/2-z)] - (e^1/2-z)/2.

    So

    z=(c-b)/(2e^1/2-2z ) - e^1/2/2+z/2

    So

    z/2=(c-b)/(2e^1/2-2z ) - (e^1/2)/2

    Because

    (c-b) is an integer.

    (2e^1/2-2z ) is an irrational number.

    And

    (e^1/2)/2 is an irrational number.

    Example:

    z/2=8/(52^1/2-7) - (3.25^1/2)

    Impossible

    So

    z is an irrational number.

    So

    z^n=/x^n+y^n.

    And another simple case.

    m=3

    Suppose

    z^n=x^n+y^n

    So

    z^3+a=x^3+y^3

    So

    z^n+z^3+a=x^n+x^3+y^n+y^3=c

    So

    z^3+z^n=z^3[1+z^(n-3)]=d^3

    Appear two cases

    d is an integher or an irrational number.

    Case d is an integer. Watching at general case later.

    Case d is an irrational number . Continuing:

    Define

    (z+f)=d=e^1/3

    e^1/3 is an irrational number.

    Because

    a=b-z^3=c-(z+f)^3

    So

    b-z^3=c-z^3-3z^2f-3zf^2-f^3

    b-z^3=c-z^3-3(zf)(z+f)-f^3

    3zf=(c-b)/(z+f)-f^3/(z+f)

    So

    Because

    f=e^1/3-z

    So

    z=(c-b)/3(e^1/3-z)(e^1/3) - (e^1/3-z)^2/3e^1/3

    So

    z=(c-b)/(3e^2/3-3ze^1/3) - (e^1/3)/3+2/3z - z^2/3e^1/3

    So

    z/3=(c-b)/(3e^2/3-3ze^1/3) - (e^1/3)/3 - z^2/(e^1/3)

    So

    z is an irrational number.

    So

    z^n=/x^n+y^n

    General case

    m is any integer..

    Suppose

    z^n=x^n+y^n

    So

    z^m+a=x^m+y^m=b

    So

    z^m+z^n+a=x^m+x^n+y^m+y^n=c

    So

    z^m+z^n=z^m[1+z^(n-m)]=d^m

    Because there exists at least one case:

    z^m+z^n=d^m

    d is an irrational number.

    Define

    (z+f)=d=e^1/m

    e^1/m is an irrational number.

    Because

    a=b-z^m=c-(z+f)^m

    So

    b-z^m=c-z^m-Q

    So

    Q=(c-b) is an integer.

    Q=mz^(m-1)f+m(m-1)/2z^(m-2)f^2+....+...+f^m.

    So

    Q=P+f^m

    So

    P=Q-f^m

    So

    P/zf=(c-b)/zf - f^(m-1)/z

    So

    mz^(m-2)+m(m-1)/2z^(m-3)f+....+...=(c-b)/zf - f^(m-1)/z

    Because

    f=e^1/m-z

    So

    mz^(m-2)+m(m-1)/2*z^(m-3)e^1/m - m(m-1)/2z^(m-2)+....+....=(c-b)/(ze^1/m-z^2) - (e^(m-1)-(m-1)(m-2)/2e^(m-2)z+....-..(-1)^(m-1)z^(m-1)

    So

    mz^(m-2)+m(m-1)/2*z^(m-3)e^1/m - m(m-1)/2z^(m-2)+....+..+(-1)^(m-1)z^(m-1)=(c-b)/(ze^1/m-z^2) - (e^(m-1)-(m-1)(m-2)/2e^(m-2)z+....-.).

    So

    m(m-1)/2*z^(m-3)e^1/m+G(e^1/m)+m(m-1)/2z^(m-2)+(-1)^(m-1)z^(m-1)+z^r+z^t+..mz^(m-2)=(c-b)/(ze^1/m-z^2) - (e^(m-1)-(m-1)(m-2)/2e^(m-2)z+....-.)..

    Because

    [m(m-1)/2*z^(m-3)e^1/m+G(e^1/m)] is an irrational number

    And

    [m(m-1)/2*z^(m-2)+(-1)^(m-1)z^(m-1)+z^r+z^t+..mz^(m-2)] is an integer

    And

    (c-b)/(ze^1/m-z^2) and (e^(m-1)-(m-1)(m-2)/2*e^(m-2)z+....-.). are two irrational numbers.

    So

    Impossible

    So

    z is an irrational number.

    So

    z^n=/x^n+y^n.

    ISHTAR.

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