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In fact:
(z_xya)^a = x^a + y^a
(z_xya)^a = x^a + y^a
Yea that's just a minor detail...DeadWolfe said:Although I believe x, y and z must also be integers.
An alternative proof, especially a comparitvely simple one would however be something of note, most mathemaricians believe there is no 'simple' proof of FLT.DeadWolfe said:lol
That's what Fermat's Last Theroem states. It's already accepted to be true. This person just has an aternate proof.
Although I believe x, y and z must also be integers.
I figured this much. I only said it in an attempt to get him to detail the proof a little more (and apparently it worked).matt grime said:Probably because he cannot for legal reasons redistribute now the journal has it.
Not that anyone here of sufficiently high standing in the math community is going to bother looking at it. A perhaps sad state of affairs but as it's a proof by picture by the sound of it I doubt it is at all rigorous, and probably badly written by maths standards - we have a notoriously short attention span for such things - after all if it's so easy why is the proof so disguised sort of attitude. See sci.math responses to james harris
I've read parts of your multiple posts. When I see things like "How can z exist as both square root of z^2 and nth root of z^n? Answer: It cannot." I tend to stop reading, and I suspect you won't find many mathematicians who will take you seriously with statements like this.jcfdillon said:So I understand that most mathematicians will obey statistical probability and ignore my work, and also will have extreme hostility toward it.
matt grime said:There are seemingly non-trivial solutions in real x,y,z that your paper says can't exist, and in fact x=1 is not a trivial solution in any sense.
I think solutions using 1 in certain cases of any problem considering "infinite" or open-ended ranges, will be considered trivial solutions. The following definition of "trivial" is quoted from MathWorld:
Trivial
Related to or being the mathematically most simple case. More generally, the word "trivial" is used to describe any result which requires little or no effort to derive or prove. The word originates from the Latin trivium, which was the lower division of the seven liberal arts in medieval universities (cf. quadrivium).
According to the Nobel Prize-winning physicist Richard Feynman (Feynman 1997), mathematicians designate any theorem as "trivial" once a proof has been obtained--no matter how difficult the theorem was to prove in the first place. There are therefore exactly two types of true mathematical propositions: trivial ones, and those which have not yet been proven.
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Your method has been shown to be wrong since it apparently implies, for instance that there is no cube root of 16 in the real numbers, which is obviously nonsense.
I did not say there is no cube root of 16 in the real numbers. But why not test specific real numbers using my method, such as x = SQRT(2) and y = pi, (both held as constants, estimated to some specific degree of accuracy) and allow the exponent n to range through the positive reals in the given equation of FLT. Then z must vary inversely, decreasing as the exponent n increases, and the same dichotomy arises as detailed above.
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So not only is there a stastical knee-jerk reaction to your work, but said reaction seemingly turns out to be justified.
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I am just saying I dont blame people for not wanting to learn about my simple paper, when they know that so many FLT proof attempts have failed, and the one accepted proof is fiendishly complex. So my paper is akin to blasphemy.
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You may claim that x and y are integers (and z) but if you don't actually use any property of them being integer then that supposition is unnecessary.
This is hilarious. You act as though I dont know that a number can be square root of its square and also nth root of its nth power. The problem is not with individual terms like z^n, but with its existence in the given equation of FLT, which is more complex.shmoe said:I've read parts of your multiple posts. When I see things like "How can z exist as both square root of z^2 and nth root of z^n? Answer: It cannot." I tend to stop reading, and I suspect you won't find many mathematicians who will take you seriously with statements like this.
I dont have a math degree, it was interesting to me that this relationship was there when x = y--, and it seemed relevant to my proof method when I noticed it in '96-- that was prior to going back to a diagram found in 1980 and simply averaging the thing! A simple step, but it had apparently never been done before, and best of all it led to a fully geometrically generalized approach for illustrating, diagramming and analyzing FLT.matt grime said:Your discorver two posts back that you don't claim is original? Well, some of us might have pointed out that it was flippin' obvious otherwise sqrt(2) is rational (integer even). If you didn't sport even that, then as shmoe says, we ain't going to bother taking you seriously.
jcfdillon said:I dont have a math degree,