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You are not logged in. #1 20061214 01:50:26
FLT DEMONSTRATION By Anthony.R.BrownFERMATS LAST THEOREM DEMONSTRATION! BY, ANTHONY.R.BROWN (1998 SOLVED ) #2 20061214 06:11:49
Re: FLT DEMONSTRATION By Anthony.R.BrownThe 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 20061214 06:27:11
Re: FLT DEMONSTRATION By Anthony.R.BrownThis is completely offtopic, but how did Anthony.R.Brown post so much in capital letters? When I try that, it gets censored into smallcase. Even when I try to type at least 7 words in capitals. #4 20061214 07:42:02
Re: FLT DEMONSTRATION By Anthony.R.BrownYou 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 20061214 07:51:04
Re: FLT DEMONSTRATION By Anthony.R.BrownFermat 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. Why did the vector cross the road? It wanted to be normal. #6 20061214 12:10:48
Re: FLT DEMONSTRATION By Anthony.R.Brown
A regular prime p does not divide the class number of the p^{th} 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 20061214 23:58:04
Re: FLT DEMONSTRATION By Anthony.R.BrownHi #8 20061215 00:00:48
Re: FLT DEMONSTRATION By Anthony.R.BrownThe 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. #9 20061215 00:05:31
Re: FLT DEMONSTRATION By Anthony.R.Brownp.s Fermat never claimed he had a Proof!! #10 20061219 05:48:40
Re: FLT DEMONSTRATION By Anthony.R.Brown
That is what he wrote. Emphasis is mine. "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 20070103 05:07:43
Re: FLT DEMONSTRATION By Anthony.R.Brown0_o trying to understand. Is this applicable to the real world? #12 20070103 07:54:02
Re: FLT DEMONSTRATION By Anthony.R.BrownThe 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 20070517 22:30:54
Re: FLT DEMONSTRATION By Anthony.R.BrownBelow is my Fermats Last Theorem QBASIC Computer program! 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 20070520 11:01:55
Re: FLT DEMONSTRATION By Anthony.R.Brown
Stop making things up #15 20070524 04:54:54
Re: FLT DEMONSTRATION By Anthony.R.BrownI've attempted to tidy this topic up a bit. Why did the vector cross the road? It wanted to be normal. #16 20070524 07:42:32
Re: FLT DEMONSTRATION By Anthony.R.BrownOk, let's keep it pointed. He says the following: #17 20070524 10:03:01
Re: FLT DEMONSTRATION By Anthony.R.BrownTo 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 20070524 21:53:31
Re: FLT DEMONSTRATION By Anthony.R.BrownTHE PROBLEM IS! FOR SOMETHING TO HAVE A 100% PROOF! IT HAS TO BE PROVED!!................ Last edited by Anthony.R.Brown (20070524 21:54:06) #19 20070524 23:27:13
Re: FLT DEMONSTRATION By Anthony.R.Brown
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. #20 20070524 23:34:25
Re: FLT DEMONSTRATION By Anthony.R.Brown
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 20070524 23:56:59
Re: FLT DEMONSTRATION By Anthony.R.BrownI heard it was something like 150 pages long. #22 20070525 00:02:20
Re: FLT DEMONSTRATION By Anthony.R.Brown
The general proof is 150 pages long, yes. Last edited by Sekky (20070525 00:03:02) #23 20070525 00:18:18
Re: FLT DEMONSTRATION By Anthony.R.BrownNo, 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 20070525 00:19:31
Re: FLT DEMONSTRATION By Anthony.R.Brown
I thought that was a peano axiom. #25 20070525 22:31:37
Re: FLT DEMONSTRATION By Anthony.R.BrownAny 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%..... 