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#1 2006-12-14 01:50:26

Anthony.R.Brown
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FLT DEMONSTRATION By Anthony.R.Brown

FERMATS LAST THEOREM DEMONSTRATION! BY, ANTHONY.R.BROWN (1998 SOLVED )

THE INFINITE MATHEMATICAL PATTERN OF CUBE NUMBERS!

BELOW IS MY DEMONSTRATION OF FERMATS LAST THEOREM,IT IS BASED ON THE INFINITE MATHEMATICAL PATTERN OF THE FIRST (10) CUBE NUMBERS AS A GROUP! YOU WILL SEE THAT THE LAST OR SINGLE NUMBER,FROM EACH OF THE (10) CUBE NUMBERS REPEATS ITSELF AT THE END OF EACH CUBE NUMBER, IN THE COLUMN,GIVEN IN BRACKETS () THIS PATTERN REPEATS ITSELF NO MATTER HOW LARGE OR SMALL THE CUBE NUMBERS ARE.
THE DEMONSTRATION IS! THAT BECAUSE YOU CANNOT ADD ANY TWO CUBE NUMBERS TOGETHER,TO MAKE A THIRD CUBE NUMBER,WITHIN ONE OF THE GROUPS, OR THE SAME WITHIN ANY TWO GROUPS,NO MATTER WHAT SIZE THE CUBE NUMBERS ARE! THEN YOU WILL NEVER BE ABLE TO MAKE A THIRD CUBE NUMBER,FROM ANY TWO CUBE NUMBERS.


Andrew wiles proof? (not possible! 100 %) of ( FLT ) is far too long! And contains math not really needed,with such a simple problem such as ( FLT ) plus only a few people in the world can really understand it.

Fermat stated the words below!

I HAVE DISCOVERED A TRULY MARVELLOUS DEMONSTRATION [ OF THIS GENERAL THEOREM ] WHICH THIS MARGIN IS TOO NARROW TO CONTAIN.

I believe my Demonstration must have been along the same lines as Fermats if not the same.
To understand ( FLT ) the answer is right there at the beginning, just like time itself, everything you need to know is in the first ten cube numbers!
The fact that all the numbers are used from 0 to 9 at the end of the cube numbers,the fact that some of the cubes are odd,the fact that some of the cubes are even,the fact that the cubes as a group look random!
The first ten cubes can be checked amongst themselves, for any two cubes that may make a third,and checked with any other ten as part of their group.
The infinite unique pattern is what makes everything work! Once you understand how cubes are made,then you have found all the cubes!
p.'s except the last!! The last case is unsolvable! for this reason its impossible to have a 100 % Proof! ( and applies to any infinite math problem ) only a Demonstration,as Fermat himself said!




BELOW THE FIRST TEN CUBE NUMBERS.

CUBE =     (1): PATTERN =   (1): COUNT = 1
CUBE =     (8): PATTERN =   (8): COUNT = 2
CUBE =    2(7): PATTERN =  2(7): COUNT = 3
CUBE =    6(4): PATTERN =  6(4): COUNT = 4
CUBE =   12(5): PATTERN = 12(5): COUNT = 5
CUBE =   21(6): PATTERN = 21(6): COUNT = 6
CUBE =   34(3): PATTERN = 34(3): COUNT = 7
CUBE =   51(2): PATTERN = 51(2): COUNT = 8
CUBE =   72(9): PATTERN = 72(9): COUNT = 9
CUBE =  100(0): PATTERN =   (0): COUNT = 10

BELOW THE NEXT TEN.

CUBE =  133(1): PATTERN =   (1): COUNT = 11
CUBE =  172(8): PATTERN =   (8): COUNT = 12
CUBE =  219(7): PATTERN =  2(7): COUNT = 13
CUBE =  274(4): PATTERN =  6(4): COUNT = 14
CUBE =  337(5): PATTERN = 12(5): COUNT = 15
CUBE =  409(6): PATTERN = 21(6): COUNT = 16
CUBE =  491(3): PATTERN = 34(3): COUNT = 17
CUBE =  583(2): PATTERN = 51(2): COUNT = 18
CUBE =  685(9): PATTERN = 72(9): COUNT = 19
CUBE =  800(0): PATTERN =   (0): COUNT = 20

AND THE NEXT TEN ETC.!

CUBE =  926(1): PATTERN =   (1): COUNT = 21
CUBE = 1064(8): PATTERN =   (8): COUNT = 22
CUBE = 1216(7): PATTERN =  2(7): COUNT = 23
CUBE = 1382(4): PATTERN =  6(4): COUNT = 24
CUBE = 1562(5): PATTERN = 12(5): COUNT = 25
CUBE = 1757(6): PATTERN = 21(6): COUNT = 26
CUBE = 1968(3): PATTERN = 34(3): COUNT = 27
CUBE = 2195(2): PATTERN = 51(2): COUNT = 28
CUBE = 2438(9): PATTERN = 72(9): COUNT = 29
CUBE = 2700(0): PATTERN =   (0): COUNT = 30



EXAMPLE BELOW : TRY TO FIND IF THE TWO NUMBERS BELOW ADD TOGETHER MAKE A CUBE NUMBER!

2685619 + 4173281 = 6858900

Following what I have said in my Demonstration,all cube numbers are made in the same way,the infinite pattern is unique in the fact that the numbers 0 to 9 are only used once as in PATTERN A, at the end of the cubes numbers,within each group of any ten! plus more proof further on.
What it is like is a Mathematical engine that produces the cube numbers in unique groups of ten,starting from the first group of ten onwards Infinitely.
The sum of the two numbers above ( 6858900 ) is not a cube number! the reason its not a cube number is because it does not fit in any of the groups of ten,that are produced!
The nearest cube number too ( 6858900 ) is ( 6859000 ) this cube number is part of the 19 th Group given below,and is just before the start of the 20 th Group.
I will explain more how the Demonstration works below!


19 th Group

CUBE = 5929741 : PATTERN A =   (1)
CUBE = 6028568 : PATTERN A =   (8)
CUBE = 6128487 : PATTERN A =  2(7)
CUBE = 6229504 : PATTERN A =  6(4)
CUBE = 6331625 : PATTERN A = 12(5)
CUBE = 6434856 : PATTERN A = 21(6)
CUBE = 6539203 : PATTERN A = 34(3)
CUBE = 6644672 : PATTERN A = 51(2)
CUBE = 6751269 : PATTERN A = 72(9)
CUBE = 6859000 : PATTERN A =   (0)

20 th Group

CUBE = 6967871 : PATTERN A =   (1)
CUBE = 7077888 : PATTERN A =   (8)
CUBE = 7189057 : PATTERN A =  2(7)
CUBE = 7301384 : PATTERN A =  6(4)
CUBE = 7414875 : PATTERN A = 12(5)
CUBE = 7529536 : PATTERN A = 21(6)
CUBE = 7645373 : PATTERN A = 34(3)
CUBE = 7762392 : PATTERN A = 51(2)
CUBE = 7880599 : PATTERN A = 72(9)
CUBE = 8000000 : PATTERN A =   (0)


MORE INFORMATION BELOW ON HOW MY DEMONSTRATION WORKS!

The next pattern is the group of numbers that are produced, by taking away the cube numbers starting with the largest and working backwards,as in the first group below,so 1000 - 729 = (271),729 - 512 = (217),512 - 343 = (169),343 - 216 = (127),216 -125 = (91),125 - 64 = (61),64 - 27 = (37),27 - 8 = (19),8 - 1 = (7).
What we now have is the second unique pattern (B),that shows us we have a group of related cube numbers,the end numbers are made up of 7,9,7,1,1,7,9,7,1 in brackets ()

1 st Group PATTERN B

CUBE = 1
DIFFERENCE =   7 : PATTERN B = (7)
CUBE = 8
DIFFERENCE =  19 : PATTERN B = (9)
CUBE = 27
DIFFERENCE =  37 : PATTERN B = (7)
CUBE = 64
DIFFERENCE =  61 : PATTERN B = (1)
CUBE = 125
DIFFERENCE =  91 : PATTERN B = (1)
CUBE = 216
DIFFERENCE = 127 : PATTERN B = (7)
CUBE = 343
DIFFERENCE = 169 : PATTERN B = (9)
CUBE = 512
DIFFERENCE = 217 : PATTERN B = (7)
CUBE = 729
DIFFERENCE = 271 : PATTERN B = (1)
CUBE = 1000

Now we move on to pattern (C) this pattern is made by dividing the actual COUNT position,in relation to where the cube numbers are from the start onwards! NOTE: the resulting numbers are always WHOLE numbers! the end numbers are made up of 1,4,9,6,5,6,9,4,1,0 in brackets ()


1 st Group PATTERN C

CUBE =    1 : COUNT =  1 : EQUALS   1 : PATTERN C = (1)
CUBE =    8 : COUNT =  2 : EQUALS   4 : PATTERN C = (4)
CUBE =   27 : COUNT =  3 : EQUALS   9 : PATTERN C = (9)
CUBE =   64 : COUNT =  4 : EQUALS  16 : PATTERN C = (6)
CUBE =  125 : COUNT =  5 : EQUALS  25 : PATTERN C = (5)
CUBE =  216 : COUNT =  6 : EQUALS  36 : PATTERN C = (6)
CUBE =  343 : COUNT =  7 : EQUALS  49 : PATTERN C = (9)
CUBE =  512 : COUNT =  8 : EQUALS  64 : PATTERN C = (4)
CUBE =  729 : COUNT =  9 : EQUALS  81 : PATTERN C = (1)
CUBE = 1000 : COUNT = 10 : EQUALS 100 : PATTERN C = (0)




Now lets see if the number ( 6858900 ) the sum of the two numbers is a cube number or not! I will put the number in the group where the number is closest matched,which happens to be the tenth place within the 19 th Group below,and replace the true cube number ( 6859000 ) with the number ( 6858900 ) and see what happens to the patterns A,B,C!


19 th Group

CUBE = 5929741 :   PATTERN A =   (1) : COUNT = 181 : EQUALS 32761 : PATTERN C = (1)
DIFFERENCE = 98827 :  PATTERN B = (7)
CUBE = 6028568 :   PATTERN A =   (8) : COUNT = 182 : EQUALS 33124 : PATTERN C = (4)
DIFFERENCE = 99919 :  PATTERN B = (9)
CUBE = 6128487 :   PATTERN A =  2(7) : COUNT = 183 : EQUALS 33489 : PATTERN C = (9)
DIFFERENCE = 101017 : PATTERN B = (7)
CUBE = 6229504 :   PATTERN A =  6(4) : COUNT = 184 : EQUALS 33856 : PATTERN C = (6)
DIFFERENCE = 102121 : PATTERN B = (1)
CUBE = 6331625 :   PATTERN A = 12(5) : COUNT = 185 : EQUALS 34225 : PATTERN C = (5)
DIFFERENCE = 103231 : PATTERN B = (1)
CUBE = 6434856 :   PATTERN A = 21(6) : COUNT = 186 : EQUALS 34596 : PATTERN C = (6)
DIFFERENCE = 104347 : PATTERN B = (7)
CUBE = 6539203 :   PATTERN A = 34(3) : COUNT = 187 : EQUALS 34969 : PATTERN C = (9)
DIFFERENCE = 105469 : PATTERN B = (9)
CUBE = 6644672 :   PATTERN A = 51(2) : COUNT = 188 : EQUALS 35344 : PATTERN C = (4)
DIFFERENCE = 106597 : PATTERN B = (7)
CUBE = 6751269 :   PATTERN A = 72(9) : COUNT = 189 : EQUALS 35721 : PATTERN C = (1)

Below is the tenth place!

Below is the true cube number,and shows all three Patterns working as normal!

DIFFERENCE = 107731 : PATTERN B = (1)
CUBE = 6859000 : PATTERN A = (0) : COUNT = 190 : EQUALS 36100 : PATTERN C = (0)

Below is the new number,you will notice the number works with only two of the patterns,but is way out with the third,i.e. not a WHOLE number! and ending with the wrong number in the () this proves its not a cube number!

DIFFERENCE = 107631 : PATTERN B = (1)
CUBE = 6858900 : PATTERN A = (0) : COUNT = 190 : EQUALS 36099.473 : PATTERN C = (int(9)? or (3)? both are wrong!



MORE INFORMATION BELOW ON HOW MY DEMONSTRATION WORKS!

As I have already said,all cube numbers are made the same way using what I would like to call a MATHEMATICAL ENGINE!
the process is from the first single cube number (1) onwards.
The most important numbers are the end numbers in brackets () the reason they are the most important is because they are the end of the process for making cube numbers! they are if you like the TRUTH numbers! its a bit like the numbers go round and round mathematically,and eventually make a whole cube number!
I have put forward three patterns A,B,C that show how these TRUTH numbers are present in any group of ten cube numbers,I am again giving the 1st and 19th Groups as examples!
The size of the cube numbers never changes how these TRUTH numbers are made! another way of looking at what im saying is that the cube numbers in the 1st Group are the same as in the 19th Group by the way they are made, and both are the same as any other group! size does not affect the TRUTH number results!
So if these two groups are the same,then trying to find if any two cube numbers make a third ,within the first group! is the same as trying with any other group,then after that test! we only need to test two groups against each other,because the maximum test is two from one group,or one from each group!
Now if we want to find out if a group of any ten numbers are cube numbers,the ten numbers must pass the three tests,they must conform to the patterns A,B,C and at the same time if we want to test if any single number is a cube number,this can be done by finding if it fits in with the nine cube numbers that are related to it,as part of its own group!.




PATTERN (A) = 1,8,7,4,5,6,3,2,9,0 in brackets ()

PATTERN (B) = 7,9,7,1,1,7,9,7,1 in brackets ()

PATTERN (C) = 1,4,9,6,5,6,9,4,1,0 in brackets ()


1 st Group

CUBE = 1:    PATTERN A = (1) : COUNT = 1  : EQUALS   1 : PATTERN C = (1)
DIFFERENCE = 7   : PATTERN B = (7)
CUBE = 8:    PATTERN A = (8) : COUNT = 2  : EQUALS   4 : PATTERN C = (4)
DIFFERENCE = 19  : PATTERN B = (9)
CUBE = 27:   PATTERN A = (7) : COUNT = 3  : EQUALS   9 : PATTERN C = (9)
DIFFERENCE = 37  : PATTERN B = (7)
CUBE = 64:   PATTERN A = (4) : COUNT = 4  : EQUALS  16 : PATTERN C = (6)
DIFFERENCE = 61  : PATTERN B = (1)
CUBE = 125:  PATTERN A = (5) : COUNT = 5  : EQUALS  25 : PATTERN C = (5)
DIFFERENCE = 91  : PATTERN B = (1)
CUBE = 216:  PATTERN A = (6) : COUNT = 6  : EQUALS  36 : PATTERN C = (6)
DIFFERENCE = 127 : PATTERN B = (7)
CUBE = 343:  PATTERN A = (3) : COUNT = 7  : EQUALS  49 : PATTERN C = (9)
DIFFERENCE = 169 : PATTERN B = (9)
CUBE = 512:  PATTERN A = (2) : COUNT = 8  : EQUALS  64 : PATTERN C = (4)
DIFFERENCE = 217 : PATTERN B = (7)
CUBE = 729:  PATTERN A = (9) : COUNT = 9  : EQUALS  81 : PATTERN C = (1)
DIFFERENCE = 271 : PATTERN B = (1)
CUBE = 1000: PATTERN A = (0) : COUNT = 10 : EQUALS 100 : PATTERN C = (0)




19 th Group

CUBE = 5929741 : PATTERN A = (1) : COUNT = 181 : EQUALS 32761 : PATTERN C = (1)
DIFFERENCE = 98827   : PATTERN B = (7)
CUBE = 6028568 : PATTERN A = (8) : COUNT = 182 : EQUALS 33124 : PATTERN C = (4)
DIFFERENCE = 99919   : PATTERN B = (9)
CUBE = 6128487 : PATTERN A = (7) : COUNT = 183 : EQUALS 33489 : PATTERN C = (9)
DIFFERENCE = 101017  : PATTERN B = (7)
CUBE = 6229504 : PATTERN A = (4) : COUNT = 184 : EQUALS 33856 : PATTERN C = (6)
DIFFERENCE = 102121  : PATTERN B = (1)
CUBE = 6331625 : PATTERN A = (5) : COUNT = 185 : EQUALS 34225 : PATTERN C = (5)
DIFFERENCE = 103231  : PATTERN B = (1)
CUBE = 6434856 : PATTERN A = (6) : COUNT = 186 : EQUALS 34596 : PATTERN C = (6)
DIFFERENCE = 104347  : PATTERN B = (7)
CUBE = 6539203 : PATTERN A = (3) : COUNT = 187 : EQUALS 34969 : PATTERN C = (9)
DIFFERENCE = 105469  : PATTERN B = (9)
CUBE = 6644672 : PATTERN A = (2) : COUNT = 188 : EQUALS 35344 : PATTERN C = (4)
DIFFERENCE = 106597  : PATTERN B = (7)
CUBE = 6751269 : PATTERN A = (9) : COUNT = 189 : EQUALS 35721 : PATTERN C = (1)
DIFFERENCE = 107731  : PATTERN B = (1)
CUBE = 6859000 : PATTERN A = (0) : COUNT = 190 : EQUALS 36100 : PATTERN C = (0)



THE ABOVE EXAMPLES CLEARLY SHOW ALL CUBE NUMBERS ARE MADE THE SAME WAY! SO THE METHODS USED TO SEE IF ANY TWO NUMBERS MAKE A CUBE NUMBER, WILL ALWAYS BE THE SAME! REGARDLESS OF THE SIZE OF THE NUMBERS.

Please forward any comments to me at my E-MAIL address below,or to any Forum this (FLT) Demonstration is on!

  TONYSWWW@YAHOO.CO.UK

A.R.B

 

#2 2006-12-14 06:11:49

Ricky
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Re: FLT DEMONSTRATION By Anthony.R.Brown

The first pattern certainly is interesting.  However, you haven't proved that it always happens.  I certainly believe it does, but you have to prove it before you can use it.  Same thing with your difference pattern.  However, you never even show how this means that the sum of two cubes can not be equal to a third.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#3 2006-12-14 06:27:11

Devantè
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Re: FLT DEMONSTRATION By Anthony.R.Brown

This is completely off-topic, but how did Anthony.R.Brown post so much in capital letters? When I try that, it gets censored into small-case. Even when I try to type at least 7 words in capitals.

 

#4 2006-12-14 07:42:02

Zhylliolom
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Re: FLT DEMONSTRATION By Anthony.R.Brown

You are correct that this is a demonstration. However, nothing has been proven; you have only shown a particular pattern in the cube numbers. Also, you have only considered the case n = 3 (which was already proven hundreds of years ago, and indeed the theorem is proved for n = 4 and all regular primes if we don't consider Wiles' approach. Then we only need to find a proof for all n that are irregular primes), which will not cover all possible n. Also, Fermat claimed to have a proof, not a mere demonstration.

 

#5 2006-12-14 07:51:04

mathsyperson
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Re: FLT DEMONSTRATION By Anthony.R.Brown

Fermat claimed that he had a proof, and then he later proved it for the specific case of n=4. If he'd already proved it generally, it wouldn't make much sense for him to do that, so it's speculated that he found a flaw with his general proof.

To Zhylliolom, what's the difference between regular and irregular primes? I've never heard of them before, but they sound interesting. Also, why would proving FLT for all primes mean that's proved for all n? Proving it for n=3 and n=5 doesn't mean that it's proved for n=15, does it? Or does it?

BY THE WAY, DEVANTÉ, I DON'T SEE WHAT YOUR PROBLEM IS. I CAN TYPE AS MUCH IN CAPITALS AS MUCH AS I WANT, AND CLEARLY SO CAN ANTHONY AND PROBABLY OTHER PEOPLE AS WELL.
(Not that I'm encouraging it or anything)


Why did the vector cross the road?
It wanted to be normal.
 

#6 2006-12-14 12:10:48

Zhylliolom
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Re: FLT DEMONSTRATION By Anthony.R.Brown

mathsyperson wrote:

To Zhylliolom, what's the difference between regular and irregular primes? I've never heard of them before, but they sound interesting. Also, why would proving FLT for all primes mean that's proved for all n? Proving it for n=3 and n=5 doesn't mean that it's proved for n=15, does it? Or does it?

A regular prime p does not divide the class number of the pth cyclotomic field. That's a pretty rugged definition, so we can give another meaning of a regular prime: a prime p is a regular prime if and only if it does not divide the numerator of the first p - 3 Bernoulli numbers (I wrote about such numbers in my zeta function thread long ago; recall that the Bernoulli numbers are the coefficients generated in the sequence



or also given by the contour integral



The first few Bernoulli numbers are 1, -1/2, 16/, -1/30...). It is conjectured that the regular primes are rather dense in the set of primes. Anyway, it would be interesting to find a proof of the Bernoulli requirement for regularity of a prime, I may work on that later. Proving the density of the regular primes among the primes would be a feat as well (the conjectured proportion is e-1/2, if you are interested. I find density relations among sets of numbers pretty interesting myself).

Proving Fermat's Last Theorem for all primes will indeed prove it for all n. From the Fundamental Theorem of Arithmetic, we know that any number other than 1 has a (unique) prime factorization. So for some arbitrary n, we can find a prime p such that m*p = n for some integer m. Then



becomes



or



So as you can see, we could substitute



and we would once again have a form of Fermat's Last Theorem:



Then essentially it is only necessary to solve the problem for prime values of n.

 

#7 2006-12-14 23:58:04

Anthony.R.Brown
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Re: FLT DEMONSTRATION By Anthony.R.Brown

Hi
Everyone

I Have shown that Xn3 + Yn3 does not Equal Zn3

The reason being all cube numbers are made the same! (look at the 3 Patterns ) as shown in my Demonstration!

The same applies for Xn>3 + Yn>3 The 3 Patterns are the same!

 

#8 2006-12-15 00:00:48

Anthony.R.Brown
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Re: FLT DEMONSTRATION By Anthony.R.Brown

The Beauty of the Method I have put forward to Show In a Demonstration, that all cube numbers are made the same! is just the opposite of what you are saying above.
When I started out to show in a Demonstration if Xn3 + Yn3 = Zn3,I knew it would always be impossible to include the Last test Group of Ten Cube Numbers! This is where the (it is not possible to have a 100% Proof for Any Infinite Math Problem!)
Comes from.
So no matter How large a Test number or Group is put forward, there will always be a larger one!
The only place we can try to prove a problem like this is by looking at the start (How Cube Numbers are Made).
It Can be (Proved!) from the First Ten Cube Numbers Onwards, For as far as (we!) can (Calculate!),That Every Group of Ten Cube Numbers, Repeats Itself as a Pattern!
It is very important to Note that when doing this, we are fully Aware of the Numbers we are Testing, i.e. We know we have started From the Smallest possible (Whole) Cube Number, and worked forward! (We Cannot Start From The End!!).
We are Aware of The Patterns A,B,C Continually being True, we are Aware of the Count Value! i.e. How far we have Gone forward.

Below is The Truth Table for All Groups of Ten Cube Numbers Starting from the first Ten Onwards.
**********************************************************************
CUBE = (n01) last number must = 1) : PATTERN(A) = (a01(1) PATTERN(C) = (c01) : COUNT = (d01)

PATTERN(B) = (b01)
CUBE = (n02) last number must = 8) : PATTERN(A) = (a02(8) PATTERN(C) = (c02) : COUNT = (d02)

PATTERN(B) = (b02)
CUBE = (n03) last number must = 7) : PATTERN(A) = (a03(7) PATTERN(C) = (c03) : COUNT = (d03)

PATTERN(B) = (b03)
CUBE = (n04) last number must = 4) : PATTERN(A) = (a04(4) PATTERN(C) = (c04) : COUNT = (d04)

PATTERN(B) = (b04)
CUBE = (n05) last number must = 5) : PATTERN(A) = (a05(5) PATTERN(C) = (c05) : COUNT = (d05)

PATTERN(B) = (b05)
CUBE = (n06) last number must = 6) : PATTERN(A) = (a06(6) PATTERN(C) = (c06) : COUNT = (d06)

PATTERN(B) = (b06)
CUBE = (n07) last number must = 3) : PATTERN(A) = (a07(3) PATTERN(C) = (c07) : COUNT = (d07)

PATTERN(B) = (b07)
CUBE = (n08) last number must = 2) : PATTERN(A) = (a08(2) PATTERN(C) = (c08) : COUNT = (d08)

PATTERN(B) = (b08)
CUBE = (n09) last number must = 9) : PATTERN(A) = (a09(9) PATTERN(C) = (c09) : COUNT = (d09)

PATTERN(B) = (b09)
CUBE = (n10) last number must = 0) : PATTERN(A) = (a10(0) PATTERN(C) = (c10) : COUNT = (d10)
**********************************************************************
In The Truth Table Above,
The Cube Numbers (n01) to (n10) Are Any Ten Related Cube Numbers (First Ten Onwards).
The Pattern (A) Numbers (a01(1) to (a10(0) Are the End Cube Numbers, And will always be the Same Numbers.
The Pattern (B) Numbers (b01) to (b09) Are the Difference Numbers Working Backwards i.e. (n10)-(n09) Then (n09)-(n08) Etc.
The Pattern (C) Numbers (c01) to (c10) Are the Numbers Produced, From Dividing The Cube numbers (n01) to (n10),
by The Count Numbers (d01) to (d10) (How far we have Gone, from the Start!) i.e. (n01)/(d01) Then (n02)/(d02) Etc.

Below are the Three Constant Pattern Numbers, for All Groups of Ten Cube Numbers!
PATTERN (A) = 1,8,7,4,5,6,3,2,9,0.
PATTERN (B) = 7,9,7,1,1,7,9,7,1.
PATTERN (C) = 1,4,9,6,5,6,9,4,1,0.

So Within the Above Truth Table we Can Place Any Group of Ten Cube Numbers, From the First Ten Onwards, And Prove All Cube Numbers Are Made The Same!
It is Impossible to Place The Largest Possible Group Of Ten Cube Numbers, Because They Can Never be Found! But then that is the Same as Any other Infinity Problem!

So Now we Have a Sound Demonstration, That All Cube Numbers Are Made the Same Way!
The Only way my Demonstration Can Be Proved Wrong! is if Someone Can put Forward Ten Numbers, That pass the Truth Table Tests Above, And the Results Are! None of The Numbers Put Forward Are Cube Numbers! I am Still Waiting!! ARB.

 

#9 2006-12-15 00:05:31

Anthony.R.Brown
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Re: FLT DEMONSTRATION By Anthony.R.Brown

p.s Fermat never claimed he had a Proof!!

I HAVE DISCOVERED A TRULY MARVELLOUS DEMONSTRATION [ OF THIS GENERAL THEOREM ] WHICH THIS MARGIN IS TOO NARROW TO CONTAIN.

 

#10 2006-12-19 05:48:40

Ricky
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Re: FLT DEMONSTRATION By Anthony.R.Brown

"I have a truly marvelous proof of this proposition which this margin is too narrow to contain." (Original Latin: "Cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.")

That is what he wrote.  Emphasis is mine.

Assuming that your demonstration is proper (which I can't judge because I can't follow it), you have only done so for the n=3 case.  Now you need n=4, 5, 6, 7, ... or you could just stick to all the primes.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#11 2007-01-03 05:07:43

Toast
Real Member

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Re: FLT DEMONSTRATION By Anthony.R.Brown

0_o trying to understand. Is this applicable to the real world?

 

#12 2007-01-03 07:54:02

Ricky
Moderator

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Re: FLT DEMONSTRATION By Anthony.R.Brown

The larger theorem, the one which Wiles proved which also proved FLT as a special case, has many applications, especially in that of Number Theory.  As for FLT by itself, I know of none.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#13 2007-05-17 22:30:54

Anthony.R.Brown
Banned

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Re: FLT DEMONSTRATION By Anthony.R.Brown

Below is my Fermats Last Theorem QBASIC Computer program!

To Run the program copy/paste and then save the file as FERMATS.BAS or FERMATS.TXT
then open the program in the QBASIC program's window to Run.

If you dont have a copy of the QBASIC program,it can be download for Free from

http://members.lycos.co.uk/brisray/qbasic/qind.htm   


A.R.B

Code:

REM ***************************************************************************
DIM SHARED START AS STRING
DIM SHARED QUIT AS STRING
DIM SHARED RUNCUBESTEST AS STRING
DIM SHARED RUNMAKECUBES AS STRING
DIM SHARED CUBEVAL0 AS SINGLE
DIM SHARED CUBEVAL1 AS SINGLE
DIM SHARED CUBEVAL8 AS SINGLE
DIM SHARED CUBEVAL27 AS SINGLE
DIM SHARED CUBEVAL64 AS SINGLE
DIM SHARED CUBEVAL125 AS SINGLE
DIM SHARED CUBEVAL216 AS SINGLE
DIM SHARED CUBEVAL343 AS SINGLE
DIM SHARED CUBEVAL512 AS SINGLE
DIM SHARED CUBEVAL729 AS SINGLE
DIM SHARED DELAYN AS DOUBLE
DIM SHARED COUNTN AS DOUBLE
DIM SHARED TIMESN AS DOUBLE
DIM SHARED MAKECUBEN AS DOUBLE
DIM SHARED CUBECOUNTN AS DOUBLE
DIM SHARED SLOWSPRINT AS DOUBLE
DIM SHARED MAXCUBEN AS DOUBLE
DIM SHARED RNDRNDNUM AS DOUBLE
DIM SHARED RDXX AS DOUBLE
DIM SHARED RDYY AS DOUBLE
DIM SHARED RDZZ AS DOUBLE
DIM SHARED TOTALXXYY AS DOUBLE
DIM SHARED DIFFXXYYZZ AS DOUBLE
DIM SHARED X1 AS DOUBLE
DIM SHARED Y1 AS DOUBLE
DIM SHARED Z1 AS DOUBLE
DIM SHARED XX AS DOUBLE
DIM SHARED YY AS DOUBLE
DIM SHARED ZZ AS DOUBLE
CLS
PRINT
PRINT TAB(14); "*******************************************************"
PRINT
PRINT TAB(95); "    { FERMATS LAST THEOREM } DEMONSTRATION PROGRAM "
PRINT
PRINT TAB(95); "            BY,Anthony.R.Brown V.01/01/1998 "
PRINT
PRINT TAB(14); "*******************************************************"
PRINT
PRINT "  THE PROBLEM IS AS FOLLOWS! "
PRINT "  are there any whole numbers e.g (x,y,z) cube numbers "
PRINT "   where x3 + y3 = z3 NOTICE = (Xn,Yn,Zn) n Must be greater than (2) "
PRINT "  an example that does not work is given below "
PRINT "   x = 64 cube y = 64 cube Z = 125 cube X + Y = 128 (+ 3) > Z "
PRINT "  if you could use zero?? then the answer would be, "
PRINT "   x = 0 y = 0 z = 0 simple!  X + Y = Z "
PRINT
INPUT " PRESS (ENTER) TO RUN PROGRAM "; START
CLS
PRINT : PRINT
INPUT " ENTER (Y) TO RUN CUBE NUMBER TEST PROGRAM! "; RUNCUBESTEST
CLS
IF RUNCUBESTEST = "Y" THEN GOTO CUBETESTLB
IF RUNCUBESTEST = "y" THEN GOTO CUBETESTLB
CLS
PRINT : PRINT
INPUT " ENTER (Y) TO RUN MAKE CUBE NUMBER PROGRAM! "; RUNMAKECUBES
CLS
IF RUNMAKECUBES = "Y" THEN GOTO MAKECUBESLB
IF RUNMAKECUBES = "y" THEN GOTO MAKECUBESLB
RUN: REM NO INPUT!
REM***************************************************************************
CUBETESTLB:
REM***************************************************************************
CLS
PRINT : PRINT
PRINT "------------------------------------------------------------------------------"
PRINT "                    { FERMATS! CUBE NUMBER TEST PROGRAM! } "
PRINT "------------------------------------------------------------------------------"
PRINT : PRINT
PRINT " ENTER THE AMOUNT OF TIMES TO RUN CUBE NUMBER TEST! "
PRINT
INPUT ; TIMESN
IF TIMESN < 1 THEN GOTO RUNAGAINLB
PRINT : PRINT
PRINT " ENTER A NUMBER TO SLOW DOWN CUBE NUMBER TEST! SCREEN PRINT! E.G  9999 "
PRINT
INPUT ; SLOWSPRINT
IF SLOWSPRINT < 1 THEN GOTO RUNAGAINLB
CLS
REM---------------------------------------------------------------------------
CUBESTESTLOOPLB:
REM---------------------------------------------------------------------------
COUNTN = COUNTN + 1
REM---------------------------------------------------------------------------
REM BELOW RANDOM NUMBER ROUTINE FOR! RDXX,RDYY,RDZZ
REM---------------------------------------------------------------------------
GOSUB CUBESSTART
IF RDXX >= 1000000 THEN GOSUB CUBESSTART: REM FOR MORE LOWER NUMBERS!
RANDOMIZE TIMER
RDXX = INT(RND * MAXCUBEN) + 1
GOSUB CUBESSTART
IF RDYY >= 1000000 THEN GOSUB CUBESSTART: REM FOR MORE LOWER NUMBERS!   
RANDOMIZE TIMER
RDYY = INT(RND * MAXCUBEN) + 1
GOSUB CUBESSTART
IF RDZZ >= 1000000 THEN GOSUB CUBESSTART: REM FOR MORE LOWER NUMBERS!   
RANDOMIZE TIMER
RDZZ = INT(RND * MAXCUBEN) + 1
GOTO CUBESDONE
REM---------------------------------------------------------------------------
CUBESSTART:
REM---------------------------------------------------------------------------
RANDOMIZE TIMER
RNDRNDNUM = INT(RND * 23) + 1
IF RNDRNDNUM = 1 THEN MAXCUBEN = 10
IF RNDRNDNUM = 2 THEN MAXCUBEN = 100
IF RNDRNDNUM = 3 THEN MAXCUBEN = 1000
IF RNDRNDNUM = 4 THEN MAXCUBEN = 10000
IF RNDRNDNUM = 5 THEN MAXCUBEN = 100000
IF RNDRNDNUM = 6 THEN MAXCUBEN = 1000000
IF RNDRNDNUM = 7 THEN MAXCUBEN = 10000000
IF RNDRNDNUM = 8 THEN MAXCUBEN = 100000000
IF RNDRNDNUM = 9 THEN MAXCUBEN = 1000000000
IF RNDRNDNUM = 10 THEN MAXCUBEN = INT(RND * 1D+20) + 1
IF RNDRNDNUM = 11 THEN MAXCUBEN = INT(RND * 1D+30) + 1
IF RNDRNDNUM = 12 THEN MAXCUBEN = INT(RND * 1D+40) + 1
IF RNDRNDNUM = 13 THEN MAXCUBEN = INT(RND * 1D+50) + 1
REM BELOW NO 1D+60 ?!
IF RNDRNDNUM = 14 THEN MAXCUBEN = INT(RND * 9.999999999999999D+59) + 1
IF RNDRNDNUM = 15 THEN MAXCUBEN = INT(RND * 1D+70) + 1
IF RNDRNDNUM = 16 THEN MAXCUBEN = INT(RND * 1D+80) + 1
IF RNDRNDNUM = 17 THEN MAXCUBEN = INT(RND * 1D+90) + 1
IF RNDRNDNUM = 18 THEN MAXCUBEN = INT(RND * 1D+101) + 1: REM MAX = 1D300+
REM BELOW EXTRA LOW NUMBERS! FOR BETTER RANDOMNESS! BELOW 1D WHEN * 3!
IF RNDRNDNUM = 19 THEN MAXCUBEN = 10
IF RNDRNDNUM = 20 THEN MAXCUBEN = 100
IF RNDRNDNUM = 21 THEN MAXCUBEN = 1000
IF RNDRNDNUM = 22 THEN MAXCUBEN = 10000
IF RNDRNDNUM = 23 THEN MAXCUBEN = 100000
RETURN
REM---------------------------------------------------------------------------
CUBESDONE: REM BELOW MAKE XX,YY,ZZ CUBES!
REM---------------------------------------------------------------------------
X1 = RDXX
XX = X1 * X1 * X1
Y1 = RDYY
YY = Y1 * Y1 * Y1
Z1 = RDZZ
ZZ = Z1 * Z1 * Z1
TOTALXXYY = XX + YY
DIFFXXYYZZ = ZZ - TOTALXXYY
CLS
PRINT
PRINT "------------------------------------------------------------------------------"
PRINT " BELOW ( Xn1 ),( Yn1 ),( Zn1 ) SINGLE START NUMBERS! "
PRINT "------------------------------------------------------------------------------"
PRINT " ( Xn1 )                =  "; X1
PRINT " ( Yn1 )                =  "; Y1
PRINT " ( Zn1 )                =  "; Z1
PRINT "------------------------------------------------------------------------------"
PRINT " BELOW ( Xn3 ),( Yn3 ),( Zn3 ) CUBE NUMBERS! "
PRINT "------------------------------------------------------------------------------"
PRINT " ( Xn3 )                =  "; XX
PRINT " ( Yn3 )                =  "; YY
PRINT " ( Zn3 )                =  "; ZZ
PRINT "------------------------------------------------------------------------------"
PRINT " BELOW IF ( Xn3 + Yn3 ) = ( Zn3 ) THEN FERMATS THEOREM IS PROVED WRONG! "
PRINT "------------------------------------------------------------------------------"
PRINT " ( Xn3 + Yn3 )          =  "; TOTALXXYY
PRINT "------------------------------------------------------------------------------"
PRINT "    ( Zn3 )             =  "; ZZ
PRINT "------------------------------------------------------------------------------"
PRINT " COUNT ="; COUNTN; " ( Zn3 ) - ( Xn3 + Yn3 ) DIFFERENCE = "; DIFFXXYYZZ
PRINT "------------------------------------------------------------------------------"
FOR DELAYN = 1 TO SLOWSPRINT
NEXT DELAYN
IF TOTALXXYY = ZZ AND TOTALXXYY <> 0 THEN GOTO PROVEDWRONGLB
IF COUNTN = TIMESN THEN GOTO RUNAGAINLB
GOTO CUBESTESTLOOPLB
REM---------------------------------------------------------------------------
PROVEDWRONGLB:
REM---------------------------------------------------------------------------
PRINT " WOW! HAVE YOU PROVED FERMATS LAST THEOREM IS WRONG! ? "
GOTO RUNAGAINLB
REM***************************************************************************
MAKECUBESLB:
REM***************************************************************************
CLS
PRINT : PRINT
PRINT "------------------------------------------------------------------------------"
PRINT "                    { FERMATS! MAKE CUBE NUMBER PROGRAM! } "
PRINT "------------------------------------------------------------------------------"
PRINT : PRINT
PRINT " ENTER HOW MANY CUBE NUMBERS! YOU WANT TO MAKE! "
PRINT
INPUT ; MAKECUBEN
IF MAKECUBEN < 1 THEN GOTO RUNAGAINLB
PRINT : PRINT
PRINT " ENTER A NUMBER TO SLOW DOWN CUBE NUMBER! SCREEN PRINT! E.G  9999 "
PRINT
INPUT ; SLOWSPRINT
IF SLOWSPRINT < 1 THEN GOTO RUNAGAINLB
REM---------------------------------------------------------------------------
CLS
XX = 1
COUNTN = 1
CUBECOUNTN = 1
REM ABOVE PRINTS FIRST CUBEVAL1 AND NOT CUBEVAL0! AS = ZERO VALUES!
CUBEVAL0 = 0
CUBEVAL1 = 1
CUBEVAL8 = 8
CUBEVAL27 = 27
CUBEVAL64 = 64
CUBEVAL125 = 125
CUBEVAL216 = 216
CUBEVAL343 = 343
CUBEVAL512 = 512
CUBEVAL729 = 729
REM---------------------------------------------------------------------------
CUBESMAKELOOPLB:
REM---------------------------------------------------------------------------
IF CUBECOUNTN = 0 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 0 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 0 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL0;
IF CUBECOUNTN = 0 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 1 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 1 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 1 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL1;
IF CUBECOUNTN = 1 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 2 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 2 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 2 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL8;
IF CUBECOUNTN = 2 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 3 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 3 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 3 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL27;
IF CUBECOUNTN = 3 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 4 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 4 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 4 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL64;
IF CUBECOUNTN = 4 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 5 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 5 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 5 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL125;
IF CUBECOUNTN = 5 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 6 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 6 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 6 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL216;
IF CUBECOUNTN = 6 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 7 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 7 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 7 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL343;
IF CUBECOUNTN = 7 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 8 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 8 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 8 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL512;
IF CUBECOUNTN = 8 THEN PRINT " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 9 THEN PRINT "CUBE = ";
IF CUBECOUNTN = 9 THEN PRINT USING "######################"; XX;
IF CUBECOUNTN = 9 THEN PRINT ; " : PATTERN = "; : PRINT USING "####"; CUBEVAL729;
IF CUBECOUNTN = 9 THEN PRINT " : COUNT = "; COUNTN
REM
REM---------------------------------------------------------------------------
SF$ = "FERMFILE.TXT"
OPEN SF$ FOR APPEND AS #1
REM
IF CUBECOUNTN = 0 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 0 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 0 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL0;
IF CUBECOUNTN = 0 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 1 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 1 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 1 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL1;
IF CUBECOUNTN = 1 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 2 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 2 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 2 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL8;
IF CUBECOUNTN = 2 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 3 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 3 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 3 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL27;
IF CUBECOUNTN = 3 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 4 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 4 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 4 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL64;
IF CUBECOUNTN = 4 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 5 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 5 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 5 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL125;
IF CUBECOUNTN = 5 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 6 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 6 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 6 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL216;
IF CUBECOUNTN = 6 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 7 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 7 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 7 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL343;
IF CUBECOUNTN = 7 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 8 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 8 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 8 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL512;
IF CUBECOUNTN = 8 THEN PRINT #1, " : COUNT = "; COUNTN
REM
IF CUBECOUNTN = 9 THEN PRINT #1, "CUBE = ";
IF CUBECOUNTN = 9 THEN PRINT #1, USING "######################"; XX;
IF CUBECOUNTN = 9 THEN PRINT #1, ; " : PATTERN = "; : PRINT #1, USING "####"; CUBEVAL729;
IF CUBECOUNTN = 9 THEN PRINT #1, " : COUNT = "; COUNTN
REM
CLOSE
REM---------------------------------------------------------------------------
FOR DELAYN = 1 TO SLOWSPRINT
NEXT DELAYN
IF COUNTN = MAKECUBEN THEN PRINT : REM SPACE BETWEEN ABOVE!
IF COUNTN = MAKECUBEN THEN PRINT " NOTICE! CUBE NUMBERS HAVE BEEN SAVED TO FERMFILE.TXT TO LOOK AT!"
IF COUNTN = MAKECUBEN THEN GOTO RUNAGAINLB
CUBECOUNTN = CUBECOUNTN + 1
IF CUBECOUNTN = 10 THEN CUBECOUNTN = 0
COUNTN = COUNTN + 1
X1 = COUNTN
XX = X1 * X1 * X1
GOTO CUBESMAKELOOPLB
REM***************************************************************************
RUNAGAINLB:
REM***************************************************************************
IF COUNTN <= 0 THEN PRINT
IF COUNTN <= 0 THEN PRINT
IF COUNTN <= 0 THEN PRINT " NO INPUT? OR WRONG INPUT? TRY AGAIN! "
PRINT
INPUT " PRESS (ENTER) TO RUN AGAIN OR (QU) TO QUIT! "; QUIT
IF QUIT = "QU" THEN END
IF QUIT = "qu" THEN END
RUN
END
REM ***************************************************************************
 

#14 2007-05-20 11:01:55

Sekky
Full Member

Offline

Re: FLT DEMONSTRATION By Anthony.R.Brown

Anthony.R.Brown wrote:

(it is not possible to have a 100% Proof for Any Infinite Math Problem!)

Stop making things up

http://en.wikipedia.org/wiki/Proof_by_induction

 

#15 2007-05-24 04:54:54

mathsyperson
Moderator

Offline

Re: FLT DEMONSTRATION By Anthony.R.Brown

I've attempted to tidy this topic up a bit.

The jist of the deleted stuff was Anthony trading insults with a few other people, and nothing much really being achieved other than general discontent and animosity.

I think one of the main reasons why everyone argues with Anthony is that his posting style is a little hard to understand. The demonstration in the first post might or might not be valid, but Anthony doesn't make clear what he's actually doing.

Maybe if you post again, trying to explain the demonstration more clearly, then we can progress.
I can't see anything 'wrong' with what you've got so far (although that doesn't mean there isn't anything), but I don't think it constitutes a full proof at the moment.

Please keep discussions mature without the use of needless insults, and PLEASE don't bring up the 0.999... thing again. We've got like seventeen bajillion topics on that already.


Why did the vector cross the road?
It wanted to be normal.
 

#16 2007-05-24 07:42:32

Sekky
Full Member

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Re: FLT DEMONSTRATION By Anthony.R.Brown

Ok, let's keep it pointed. He says the following:

"(it is not possible to have a 100% Proof for Any Infinite Math Problem!)"

Which isn't true, by the simple concept of

http://en.wikipedia.org/wiki/Proof_by_induction

Any quarrels?

 

#17 2007-05-24 10:03:01

Ricky
Moderator

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Re: FLT DEMONSTRATION By Anthony.R.Brown

To make Sekky's point a little more clear, once you accept Zermelo Frankel set theory, you can prove there exists a set (specially, the natural numbers) in which the principle of mathematical induction is valid.  This is how the natural numbers are always defined, ever since Peano.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#18 2007-05-24 21:53:31

Anthony.R.Brown
Banned

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Re: FLT DEMONSTRATION By Anthony.R.Brown

THE PROBLEM IS! FOR SOMETHING TO HAVE A 100% PROOF! IT HAS TO BE PROVED!!................

We all know the Sequence 1,2,3,4,5,6,7,8,9.....etc... but We cant prove what the Last Number is!?

All Infinite Math problems have the same Problem!!....................................................................

A.R.B

Last edited by Anthony.R.Brown (2007-05-24 21:54:06)

 

#19 2007-05-24 23:27:13

Sekky
Full Member

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Re: FLT DEMONSTRATION By Anthony.R.Brown

Anthony.R.Brown wrote:

THE PROBLEM IS! FOR SOMETHING TO HAVE A 100% PROOF! IT HAS TO BE PROVED!!................

We all know the Sequence 1,2,3,4,5,6,7,8,9.....etc... but We cant prove what the Last Number is!?

All Infinite Math problems have the same Problem!!....................................................................

A.R.B

No, they don't, at all, you're making things up. There IS not last number, but properties can apply for each and every number up to infinity if the property is induced.

Just read the induction page and stop typing in caps, and stop partitioning your sentences with exclamation marks.

 

#20 2007-05-24 23:34:25

Ricky
Moderator

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Re: FLT DEMONSTRATION By Anthony.R.Brown

THE PROBLEM IS! FOR SOMETHING TO HAVE A 100% PROOF! IT HAS TO BE PROVED!!................

There is a proof, as I have stated.  I don't have time to type it up, and you wouldn't understand it anyways.  Heck, it takes me a while to read through it myself.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#21 2007-05-24 23:56:59

Identity
Power Member

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Re: FLT DEMONSTRATION By Anthony.R.Brown

I heard it was something like 150 pages long.

 

#22 2007-05-25 00:02:20

Sekky
Full Member

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Re: FLT DEMONSTRATION By Anthony.R.Brown

Identity wrote:

I heard it was something like 150 pages long.

The general proof is 150 pages long, yes.

I believe Ricky was referring to just a proof for n=3

Last edited by Sekky (2007-05-25 00:03:02)

 

#23 2007-05-25 00:18:18

Ricky
Moderator

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Re: FLT DEMONSTRATION By Anthony.R.Brown

No, I was referring to proving that the principle of mathematical induction is valid on the natural numbers.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#24 2007-05-25 00:19:31

Sekky
Full Member

Offline

Re: FLT DEMONSTRATION By Anthony.R.Brown

Ricky wrote:

No, I was referring to proving that the principle of mathematical induction is valid on the natural numbers.

I thought that was a peano axiom.

Although I suppose it makes sense to deduce it, but it doesn't mean it can't be axiom.

 

#25 2007-05-25 22:31:37

Anthony.R.Brown
Banned

Offline

Re: FLT DEMONSTRATION By Anthony.R.Brown

Any so called Proof by induction! is still only Guess work! whether it is 150 pages or 150 million pages! size don't matter if you can't see the actual end of something 100%.....

a good example is " I state that in the sequence 1,2,3,4,5,6,7,8,9...for as far as we can see the numbers will always end in an Odd Number!


but if I could see a bit further! " I state that in the sequence 1,2,3,4,5,6,7,8,9,10...for as far as we can see the numbers will always end in an even Number!


Both examples above are by induction!........but neither is 100% correct or a Proof!.............

 

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