Clay Math 2001 Annual Meeting Talk by Andrew Wiles
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- Опубліковано 22 лис 2011
- 2001 Olympiad Event
Friday, July 13, 2001, 2:00 pm
A Celebration of the Universality of Mathematical Thought
On July 13, 2001 the Clay Mathematics Institute organized the closing ceremonies of the International Mathematics Olympiad in Washington, DC, and incorporated this event into its 2001 Annual Meeting. The events brought approximately five hundred of the world's best high school mathematics students in contact with a cross-section of the world's best research mathematicians, including Edward Witten, Andrew Wiles, and Arthur Jaffe. The meeting of the Clay Mathematics Institute took place at the John F. Kennedy Center for the Performing Arts at 2:00 PM on July 13. This ceremony included the presentation of the Clay Research Awards and two inspirational talks by CMI Scientific Advisory Board members Andrew Wiles and Edward Witten. Following the ceremony at the Kennedy Center was a reception and dinner at the National Building Museum in Washington, DC. The dinner involved almost eleven hundred guests, and included talks by Alfred R. Berkeley, III, Chairman of NASDAQ, and Rita Colwell, Director of the National Science Foundation, as well as other forms of entertainment including a live performance by Christopher Thompson, accompanied by Milton Granger, of an excerpt from Fermat's Last Tango.
www.claymath.org/annual_meetin... - Наука та технологія
One of the cleverest men on the planet. He is the type of people we should have running the place and who should be richly rewarded.....forget bent politicians and overpaid footballers.....intelligence should be hero worshipped and command the praise. Our technological civilisation would be screwed without such amazing people.
Wiles worked diligently for 6 years, with the understanding that in the end, he may discover nothing. It's hard to believe people like this exist!
+Ztech as far as i have read and watched about Wiles and his progress, he started working intensively after he has heard about the proof of the epsilon conjecture by Ribet. i think this deciding step towards proofing Taniyama-Kimura made him look forward to solving this, and i guess he then had the clear expectation to solve this theorem.
He is a modest genius, I'm proud of him.
No!!!
No what?
Just no
what do you mean just no?
I have no idea but one thing I do know is that I love you, I have to go to sleep, I had a long day and ill try to tell you tomarrow goodnight:)♥
You know you are scary brilliant when they play play music after your introduction.
"fermats margin had room for this one" lmao
he's inspirational.
After 3 centuries a vietnamese old man found .How to solve fermat!
I think I've found the soul of piere de Fermat
Define
Sx=1+2^2+3^2+4^2+....+x^2.=x(x+1)(2x+1)/6=(2x^3+3x^2+x)/6
Sy=1+2^2+3^2+4^2+....+y^2=y(y+1)(2y+1)/6=(2y^3+3y^2+y)/6
Sz=1+2^2+3^2+4^2+....+z^2=z(z+1)(2z+1)/6=(2z^3+3z^2+z)/6
So
2x^3=6Sx-3x^2-x
2y^3=6Sy-3y^2-y
2z^3=6Sz-3z^2-z
so
x^3=3Sx-3/2x^2-x/2
y^3=3Sy-3/2y^2 - y/2
z^3=3Sz -3/2z^2-z/2
supose
x^3+y^3=z^3
(3Sx-3/2x^2-x/2)+(3Sy-3/2y^2 - y/2) - (3Sz -3/2z^2-z/2)=0
or
(2Sx-x^2-x/3)+(2Sy-y^2 - y/3) - (2Sz -z^2-z/3)=0
or
2Sx+2Sy-2Sz-(x^2+y^2-z^2) =(x/3+y/3-z/3)
because
2Sx+2Sy-2Sz-(x^2+y^2-z^2) is integer
So
(x/3+y/3-z/3) is also integer
or
x=3k
y=3h and
z=3g
define K,h,g are integers
So
27k^3+27h^3=27g^3.
or
k^3+h^3=g^3
had had condition x ^ 3 + y ^ 3 = z ^ 3
Unable to meet the two conditions in the same time
except
x=k,y=h and z=g
but
x=3k
and
k=x
so
x=3x
this is impossible
conclusive
x^3+y^3=/z^3
general
Z^n=/x^n+y^n
using formular
1^a+2^a+3^a+4^a+....+n^a
Thanks a lot !
If you are here, then you owe it to yourself to watch Simon Singhs' documentary about Wiles and FLT. I've watched it a dozen times and still enjoy it. Wiles is an incredible individual. watch?v=7FnXgprKgSE
Wonderful.
Great man...
Sir, you are one of the greatest mathematician. Your approach and skill in tackling the Fermats Last Theorem is simply mind blowing.With age on your side and your remarkable talent, I just wish you could do another miracle for us. Please attempt on the Riemann Conjecture. I'm sure you can come up with something big as well. God bless you!
He is working on Birch Shwinintondire conjecture
You can look on the bright side as well though. There is a lot of teamwork and growth behind these competitions.
Beautiful man.
This man is a GENIUS!
Great lecture, Andrew!
After 3 centuries a vietnamese old man found .How to solve fermat!
I think I've found the soul of piere de Fermat
Define
Sx=1+2^2+3^2+4^2+....+x^2.=x(x+1)(2x+1)/6=(2x^3+3x^2+x)/6
Sy=1+2^2+3^2+4^2+....+y^2=y(y+1)(2y+1)/6=(2y^3+3y^2+y)/6
Sz=1+2^2+3^2+4^2+....+z^2=z(z+1)(2z+1)/6=(2z^3+3z^2+z)/6
So
2x^3=6Sx-3x^2-x
2y^3=6Sy-3y^2-y
2z^3=6Sz-3z^2-z
so
x^3=3Sx-3/2x^2-x/2
y^3=3Sy-3/2y^2 - y/2
z^3=3Sz -3/2z^2-z/2
supose
x^3+y^3=z^3
(3Sx-3/2x^2-x/2)+(3Sy-3/2y^2 - y/2) - (3Sz -3/2z^2-z/2)=0
or
(2Sx-x^2-x/3)+(2Sy-y^2 - y/3) - (2Sz -z^2-z/3)=0
or
2Sx+2Sy-2Sz-(x^2+y^2-z^2) =(x/3+y/3-z/3)
because
2Sx+2Sy-2Sz-(x^2+y^2-z^2) is integer
So
(x/3+y/3-z/3) is also integer
or
x=3k
y=3h and
z=3g
define K,h,g are integers
So
27k^3+27h^3=27g^3.
or
k^3+h^3=g^3
had had condition x ^ 3 + y ^ 3 = z ^ 3
Unable to meet the two conditions in the same time
except
x=k,y=h and z=g
but
x=3k
and
k=x
so
x=3x
this is impossible
conclusive
x^3+y^3=/z^3
general
Z^n=/x^n+y^n
using formular
1^a+2^a+3^a+4^a+....+n^a
prajñā prajñā shhhhhush
Andrew Wiles at 2:30
Long live Sir Andrew Wiles !
"In mathematics, a theorem is a statement that has been proven on the basis of previously established statements"
From wikipedia.
What is the Millennium Problem related to the question he discussed?
Seems fermat's last theorem could not be further than any of the millions problems on the clay math institute, yet i could be grossly mistaken
this must be translated and airdd to every (middle and high) math class in the world at the first day of a semester
math is not a sprint
nor a calculation dual
it is a purest form of asking
and trying to answering it
with the same form of purity
He is my idol
Why hasn't this guy got a field's medal
That age cut-off rule doesn't make mathematical sense
+Master Chief Exactly there should be an medal for "solved-this-antique-math-question-prize" since that is also an achivement!
+Master Chief he got a diffrent medal im pretty sure , there's one for age 40 and up.
He won FM in 1994. He guided Manjula Bhargava for PhD. Dr Bhargava also won FM in 2014. That's rare.
but the answer is haft loong ,i don't think because it problem only 3 x,y,z why it is 200 page??
Andrew vs Reimann hypothesis
(N)
x^3=[x(x+1)/2]^2 - [(x-1)x/2]^2.
Defined the function f(z,x,y):
f(z,x,y)=[z(z+1)/2]^2 - { x(x+1)/2]^2 + [y(y+1)/2]^2 }
If
z^3=x^3+y^3 or z^n=x^n+y^n.
Impossible at same time.
Both
f (z,x,y) and f [(z-1),(x-1),(y-1)] equal Zero.
I created a series of numbers as follows;
1,3,6,10,15,21,28,36,45........
Named:
a b ,c ,d ,e ,f ,g ,h, i ,j ,k l, m ,n, o, p........
This is a wrong mathematical propositions:
d^2+g^2=m^2 => e^2+h^2=n^2.
Attention:
d and e,g and h,m and n are consecutive numbers.
So:
z^3=/x^3+y^3.
Because;
z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2.
Electric universe theory i believe is the great unifier of the large with the small, however the theory in itself fails to properly deal with the nature of the very small. I am currently working on a way to understand how thingsc exist on the verymost fundamental level....but progress is coming in surprising ways
Dear everybody.
Set of integers (N)
x^3=[x(x+1)/2]^2 - [(x-1)x/2]^2.
Define :
F(x)=[x(x+1)/2]^2.
So
x^3=F(x)-F(x-1).
If
z^3=x^3+y^3.
So
F(z)-F(z-1).= F(x)-F(x-1).+F(y)-F(y-1).
So
F(z) - [F(x)+F(y)]= F(z-1).- [F(x-1).+F(y-1)].
This is unreasonable.
So
z^n=/x^n+y^n.
ADIEU.
#Math #Mathematics #Science #News #ScienceNews #EarthProudDay Fermat's Last theorem is FUNDAMENTAL in Math and Science and after 380 years we can say in Culture.Fermat-Murgu Impossible Equations Sent Fermat's Last Theorem in FUNDAMENTAL By two Methods - One need a Postulate and I hope soon a Youngest Mathematician will be coming soon with , is about Fermat-Murgu n Media for 3 Integers associated to Fermat Equations - but second method is Definitive, and Second Grade Impossible Equations are absolute Conditionals for General Cases of n as Power and not only for 3 , 4, 5 and so on. See it at :" www.climaticdisorder.com "