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Thread: Fermat's Theorem, Relativity, Quantum Field Theory

  1. #21
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Quote Originally Posted by grapes View Post
    My point exactly. The binomial theorem is valid for nonintegers as well, so when you use it to "show" that cannot equal , you are "showing" it for all numbers, not just integers.

    Should've been your first clue that your approach was bogus.
    That just doesn't make any sense whatever. The ONLY issue with integers (as opposed to real numbers) is for the case n=2 where the interaction terms can be dismissed with complex numbers.

    You seem to be saying the Binomial Theorem is invalid (or if it is, it is invalid because my application applies to positive real numbers, not only positive integers and is my proof is invalid because I apply it to integers) . Good luck with that one....

    The issue has to do with symmetry in two dimensions, which is why the relativistic circle applies with radius of one. You are still trying to argue from a one (or no) dimensional perspective, which is like looking into the universe with a flashlight (i.e., line-of sight) As long as you can't admit there is a discipline called trigonometry, there is no point in going on, since you keep trying to revert to a one-dimensional argument, where everything can be handled in simple arithmetic. However, even in one dimension, integers or not, there will always be interaction terms in the Binomial Expansion for positive integers, so Fermat's Theorem is valid.

    In one dimension, if there are no interaction terms, one is relegated to counting and partitioning sets using the same metric; which only works for the case n=2 if imaginary interactions are addied and subtracted. Otherwise, the difference in degree of elements in the expansion cannot be eliminated.

    You keep coming up with purported kindergarten "objections", and you're missing the whole point of the analysis, especially since you never answer any of my responses directly but make up theories of your own.
    Last edited by BuleriaChk; 02-18-2017 at 02:46 PM.
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  2. #22
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Quote Originally Posted by BuleriaChk View Post
    That just doesn't make any sense whatever. The ONLY issue with integers (as opposed to real numbers) is for the case n=2 where the interaction terms can be dismissed with complex numbers.

    You seem to be saying the Binomial Theorem is invalid (or if it is, it is invalid because my application applies to positive real numbers, not only positive integers and is my proof is invalid because I apply it to integers) . Good luck with that one....

    The issue has to do with symmetry in two dimensions, which is why the relativistic circle applies with radius of one. You are still trying to argue from a one (or no) dimensional perspective, which is like looking into the universe with a flashlight (i.e., line-of sight) As long as you can't admit there is a discipline called trigonometry, there is no point in going on, since you keep trying to revert to a one-dimensional argument, where everything can be handled in simple arithmetic. However, even in one dimension, integers or not, there will always be interaction terms in the Binomial Expansion for positive integers, so Fermat's Theorem is valid.

    In one dimension, if there are no interaction terms, one is relegated to counting and partitioning sets using the same metric; which only works for the case n=2. Otherwise, the difference in degree of elements in the expansion cannot be eliminated.

    You keep coming up with purported kindergarten "objections", and you're missing the whole point of the analysis, especially since you never answer any of my responses directly but make up theories of your own.
    Make up theories? What? Like you do constantaly and never atually consider some basic mathematical objections?

  3. #23
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Quote Originally Posted by BuleriaChk View Post
    That just doesn't make any sense whatever. The ONLY issue with integers (as opposed to real numbers) is for the case n=2 where the interaction terms can be dismissed with complex numbers.

    You seem to be saying the Binomial Theorem is invalid (or if it is, it is invalid because my application applies to positive real numbers, not only positive integers and is my proof is invalid because I apply it to integers) . Good luck with that one....
    You've said that before. I have never said the binomial theorem is invalid, or even came close to saying the binomial theorem is invalid.

    Your proof is invalid, for the reasons I've given.

    You've blinded yourself.
    The issue has to do with symmetry in two dimensions, which is why the relativistic circle applies with radius of one. You are still trying to argue from a one (or no) dimensional perspective, which is like looking into the universe with a flashlight (i.e., line-of sight) As long as you can't admit there is a discipline called trigonometry, there is no point in going on, since you keep trying to revert to a one-dimensional argument, where everything can be handled in simple arithmetic. However, even in one dimension, integers or not, there will always be interaction terms in the Binomial Expansion for positive integers, so Fermat's Theorem is valid.

    In one dimension, if there are no interaction terms, one is relegated to counting and partitioning sets using the same metric; which only works for the case n=2. Otherwise, the difference in degree of elements in the expansion cannot be eliminated.

    You keep coming up with purported kindergarten "objections", and you're missing the whole point of the analysis, especially since you never answer any of my responses directly but make up theories of your own.

  4. #24
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    The Binomial Expansion provides an expression that is true for all values of a,b,c and n positive integers for n > 2.

    In order to provide an expression that is true for all values of , we simply equate them so that for all values of c, and ask "what condition needs to be true so that the expression

    is valid (for all values of c)?

    That condition is rem(a,b,n) = 0, which can only be true if a=0 or b=0. Therefore, a or b cannot be a positive integer.

    QED

    In other words, suppose there was an integer d such that

    Then

    But we already know that is valid for all values of a,b and n.
    Therefore is valid only if rem(a,b,n) = 0 so that

    rem(a,b,n)=0 only iff a=0 or b=0, so a or b cannot be a positive integer. QED
    Last edited by BuleriaChk; 02-18-2017 at 02:50 PM.
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  5. #25
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Quote Originally Posted by BuleriaChk View Post
    The Binomial Expansion provides an expression that is true for all values of a,b,c and n positive integers for n > 2.
    Absolutely false.

    For instance it is not true when c=5, b=3, and a=4
    In order to provide an expression that is true for all values of , we simply equate them so that for all values of c, and ask "what condition needs to be true so that the expression
    Yes, you "simply equate them". That's what I've pointed out all along. By equating them, you are saying c=a+b.

    Essentially, you are ignoring every other possibility. Your "proof" is not a general proof, you're only treating the cases where c=a+b.

    Which is nonsense.
    is valid (for all values of c)?

    That condition is rem(a,b,n) = 0, which can only be true if a=0 or b=0. Therefore, a or b cannot be a positive integer.

    QED

    In other words, suppose there was an integer d such that

    Then

    But we already know that is valid for all values of a,b and n.
    Therefore is valid only if rem(a,b,n) = 0 so that

    rem(a,b,n)=0 only iff a=0 or b=0, so a or b cannot be a positive integer. QED
    That's because you are insisting that c=a+b

    The only way that c=a+b and is if a or b equal zero. That's all you've proved!

    Nowhere close to a proof of Fermat's Last Theorem

  6. #26
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Numbered annotations mine:
    Quote Originally Posted by grapes View Post
    Absolutely false.

    1.)For instance it is not true when c=5, b=3, and a=4

    2.)By equating them, you are saying c=a+b.

    3.)Essentially, you are ignoring every other possibility. Your "proof" is not a general proof, you're only treating the cases where c=a+b.

    4.)you are insisting that c=a+b

    5.)The only way that c=a+b and is if a or b equal zero. That's all you've proved!
    Each of the above numbered points are the same exact ones Grapes (and others) have made over and over. If we dug through all the numerous threads and posts on this topic, we would find these points made hundreds of times. Literally, hundreds.
    Yet, each time these points are made, BuleriaChk goes into a pure and elegant case of absolute Denial. There is no other term for it; It is Blind Denial.
    A scientist is able to accept a failure. A mistake. A scientist is able to admit fault. It is necessary, in science, to be able to throw an idea into the Circular file when it fails and to Move on, having learned from that mistake.
    Chuck and JasonMe both like to poke fun at my Signature line. Yet, the quotes stand. Yes, Science encourages the Free Exchange of ideas but this does not mean that science accepts the totality of any freely exchanged idea. Ideas are often wrong. Theories are often wrong. And the proper course is to correct what is wrong.
    These two do not accept the errors but instead, get Hostile. They get angry. They take it very personally and claim anyone that points out their error is attacking their intelligence or is oppressing their ideas. Clearly, their ideas are not oppressed; this forum is filled with their ideas in numerous posts and revisions. They are Not Oppressed at all.
    In science, ideas must be tested for Accuracy and Merit. Lacking those, the idea is Trashed. It is not a conspiracy; it is a necessary ingredient of building models that are accurate and reflect reality in a way we can observe and build more models upon.
    --Inter Arma Enim Silent Leges--
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  7. #27
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Quote Originally Posted by Neverfly View Post
    Numbered annotations mine:


    Each of the above numbered points are the same exact ones Grapes (and others) have made over and over. If we dug through all the numerous threads and posts on this topic, we would find these points made hundreds of times. Literally, hundreds.
    Yet, each time these points are made, BuleriaChk goes into a pure and elegant case of absolute Denial. There is no other term for it; It is Blind Denial.
    A scientist is able to accept a failure. A mistake. A scientist is able to admit fault. It is necessary, in science, to be able to throw an idea into the Circular file when it fails and to Move on, having learned from that mistake.
    Chuck and JasonMe both like to poke fun at my Signature line. Yet, the quotes stand. Yes, Science encourages the Free Exchange of ideas but this does not mean that science accepts the totality of any freely exchanged idea. Ideas are often wrong. Theories are often wrong. And the proper course is to correct what is wrong.
    These two do not accept the errors but instead, get Hostile. They get angry. They take it very personally and claim anyone that points out their error is attacking their intelligence or is oppressing their ideas. Clearly, their ideas are not oppressed; this forum is filled with their ideas in numerous posts and revisions. They are Not Oppressed at all.
    In science, ideas must be tested for Accuracy and Merit. Lacking those, the idea is Trashed. It is not a conspiracy; it is a necessary ingredient of building models that are accurate and reflect reality in a way we can observe and build more models upon.
    Neverfly is a pompous ass, without any understanding of the foundations of mathematics or physics.

    Grapes has no concept of independent variables, upon which my proof is based. If the concept of the independence of (a,b) is not accepted, then the Binomial Theorem is invalid, since it allows the existence of multiplicative pairs (e.g., ab).

    Unfortunately, both of these nincompoops are moderators.

    Blind denial of WHAT? That there is only one dimension (number line, line-of sight) etc. in the analysis and that Descartes is somehow irrelevant? None of you understand Descartes, linear algebra, or independent variables (neither did John Gabriel). If the Peanut Gallery would only learn high school pre-calculus, they would have a ground for understanding the concept of (relative) metrics, which is what relativity is all about - the "spacing" of integers on a number line.

    That said, I soldier on (sigh) -----

    Consider the equation c=a+b. If a and b are unique integers, then either a or b must be zero so c=a or c= b. If they are not unique integers, then the sum must be equal to another unique integer so that d = a +b. But that merely means c = d, a tautology.

    Consider the equation c = ab Again, if a and b are unique integers, the product must be equal to some integer d. But this means that c=d, another tautology. If a and b are not unique integers, then c = k(ab) for some positive integer k. But then c = (ka)b=a(kb) so that c = a'b or c=ab', where the integers have been scaled to a specific k, so the variables are not independent, and any expression involving k will not be general.

    Therefore, for generality, (a,b) must be characterized variables over the positive integers, with the scaling ratio true for all integers, so that for t and t' positive integers, (at,bt') are independent. This is a metric comparison, in which a and b are scaled, such that the scaling is true for all possible combinations. If we select at as the foundation metric, to be compared with bt', with as the initial standard for comparison of the metrics, then the relation:

    fpr at, bt' independent variables, so the scalar operation is true for all a,b,t and t'. The only way to ensure that that the equality applies for all a and b is to assume that at and bt' are independent.

    Solving this gives the equation , where



    Under what conditions can all the parameters be integers?



    For c
    to be an integer,


    must be an integer, where .

    , so that .

    , so that .

    Then if is an integer, either or so that the equation means that or .

    If then since no initial metric exists for comparison. If then and the metrics are the same there is no multiplicative scaling, so that the multiplicative product does not exist. (there is no multiplicative relation between a and b))
    _______________________________________
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  8. #28
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Quote Originally Posted by BuleriaChk View Post
    Neverfly is a pompous ass, without any understanding of the foundations of mathematics or physics.
    Great counter argument from the guy that thinks the constant "c" is a variable and that the Time Dilation equation can be applied to a photon.
    Quote Originally Posted by BuleriaChk View Post
    Blind denial of WHAT? That there is only one dimension (number line, line-of sight) etc. in the analysis and that Descartes is somehow irrelevant? None of you understand Descartes, linear algebra, or independent variables (neither did John Gabriel). If the Peanut Gallery would only learn high school pre-calculus, they would have a ground for understanding the concept of (relative) metrics, which is what relativity is all about - the "spacing" of integers on a number line.
    Shifting the goal posts: Red Herring. Focus on this:
    Quote Originally Posted by grapes View Post
    Absolutely false.

    1.)For instance it is not true when c=5, b=3, and a=4

    2.)By equating them, you are saying c=a+b.

    3.)Essentially, you are ignoring every other possibility. Your "proof" is not a general proof, you're only treating the cases where c=a+b.

    4.)you are insisting that c=a+b

    5.)The only way that c=a+b and is if a or b equal zero. That's all you've proved!
    ...Instead of the Circus side show distraction.
    Quote Originally Posted by BuleriaChk View Post
    Consider the equation c=a+b. If a and b are unique integers, then either a or b must be zero so c=a or c= b. If they are not unique integers, then the sum must be equal to another unique integer so that d = a +b. But that merely means c = d, a tautology.
    Sigh... See above.

    Quote Originally Posted by BuleriaChk View Post
    Consider the equation c = ab Again, if a and b are unique integers, the product must be equal to some integer d. But this means that c=d, another tautology. If a and b are not unique integers, then c = k(ab) for some positive integer k. But then c = (ka)b=a(kb) so that c = a'b or c=ab', where the integers have been scaled to a specific k, so the variables are not independent, and any expression involving k will not be general.

    Therefore, for generality, (a,b) must be characterized variables over the positive integers, with the scaling ratio true for all integers, so that for t and t' positive integers, (at,bt') are independent. This is a metric comparison, in which a and b are scaled, such that the scaling is true for all possible combinations. If we select at as the foundation metric, to be compared with bt', with as the initial standard for comparison of the metrics, then the relation:

    fpr at, bt' independent variables, so the scalar operation is true for all a,b,t and t'. The only way to ensure that that the equality applies for all a and b is to assume that at and bt' are independent.

    Solving this gives the equation , where



    Under what conditions can all the parameters be integers?



    For c
    to be an integer,


    must be an integer, where .

    , so that .

    , so that .

    Then if is an integer, either or so that the equation means that or .

    If then since no initial metric exists for comparison. If then and the metrics are the same there is no multiplicative scaling, so that the multiplicative product does not exist. (there is no multiplicative relation between a and b))
    That was a lot of song and dance to say, c=a+b and a must =0 and b must also =0 but uhhh:
    Quote Originally Posted by grapes View Post
    1.)For instance it is not true when c=5, b=3, and a=4

    2.)By equating them, you are saying c=a+b.

    3.)Essentially, you are ignoring every other possibility. Your "proof" is not a general proof, you're only treating the cases where c=a+b.

    4.)you are insisting that c=a+b

    5.)The only way that c=a+b and is if a or b equal zero. That's all you've proved!
    It is all right there. It has been there, pointed out hundreds of times and each time, you throw on blinders and stumble forward stubbornly trying to repeat the mistake again, again, again and again.
    --Inter Arma Enim Silent Leges--
    “Science needs the light of free expression to flourish. It depends on the fearless questioning of authority, and the open exchange of ideas.” ― Neil deGrasse Tyson

    "When photons interact with electrons, they are interacting with the charge around a "bare" mass, and thus the interaction is electromagnetic, hence light. This light slows the photon down." - BuleriaChk

  9. #29
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    The problem is that neither of you have any idea what it means for an integer to be an invariant (i.e., unique).

    I don't think either one of you can have passed a pre-calculus course in analytic geometry; your perspective is of high school arithmetic, not even to the level of trigonometry; of counting sticks, and not understanding the concept of frequency (consider t and t' to be periods), much less the concept of sampling (in space or time, which involves the concept of metrics and orthogonal coordinates), signal processing and analysis, all of which are involved in my analysis in one way or the other.

    I can't believe that either Grapes or Neverfly have a technical education much above grade school level, since their questions miss the point - my comments are not "Red Herrings", they go to the root of what Einstein was concerned about in the metric tensor, but instead of discussing the metric tensor, these guys prattle on about beginning concepts based on counting (as integers) - like John Gabriel (but these guys have o idea why he was really wrong), ignoring the concepts of independence, metric scaling, and other more slightly advanced ideas.

    (Note: Given any three positive integers a,b, and c, I can always choose them so that and and by relabeling them accordingly). Then I can investigate the consequences of c=a+b as applies to the Binomial Expansion. The relativistic analysis provides a formal way of doing this for all combinations of such integers by defining a metric relationship in terms of which they can be compared. (at,bt')

    The proof by Binomial Theorem assumes this metric relation implicitly.

    But if the Binomial Theorem is true (i.e., valid for all positive integers a,b,c and n), why does it not prove Fermat's Theorem (n>2)? Flail around all you want; you have given me NOTHING that shows any intelligent reason that my proof is invalid. (As I'm sure Fermat understood, but didn't have space to write it down..)



    It is just that STR (and Quantum Field Theory) nails the coffin shut... except for wannabe "mathematicians" who can only accept counting widgets as the Foundation of Mathematics....
    Last edited by BuleriaChk; 02-18-2017 at 11:48 PM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

  10. #30
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    Default Re: Fermat's Theorem, Relativity, Quantum Field Theory

    Quote Originally Posted by BuleriaChk View Post
    The problem is that neither of you have any idea what it means for an integer to be an invariant (i.e., unique).

    I don't think either one of you can have passed a pre-calculus course in analytic geometry; your perspective is of high school arithmetic, not even to the level of trigonometry; of counting sticks, and not understanding the concept of frequency (consider t and t' to be periods), much less the concept of sampling (in space or time, which involves the concept of metrics and orthogonal coordinates), signal processing and analysis, all of which are involved in my analysis in one way or the other.

    I can't believe that either Grapes or Neverfly have a technical education much above grade school level, since their questions miss the point - my comments are not "Red Herrings", they go to the root of what Einstein was concerned about in the metric tensor, but instead of discussing the metric tensor, these guys prattle on about beginning concepts based on counting (as integers) - like John Gabriel (but these guys have o idea why he was really wrong), ignoring the concepts of independence, metric scaling, and other more slightly advanced ideas.

    But if the Binomial Theorem is true (i.e., valid for all positive integers a,b,c and n), why does it not prove Fermat's Theorem (n>2)?
    The binomial theorem is true.

    You think it proves Fermat's Last Theorem.
    Flail around all you want you have given me NOTHING that shows any intelligent reason that my proof is invalid.
    You cannot understand it because you've blinded yourself.

    It's what keeps you from understanding mathematics. Mathematics cannot be subjugated by simple bullheadedness. You run off a list of what you think are high level mathematics, but you really don't understand any of it. You could, but you have to admit that weakness.
    (As I'm sure Fermat understood, but didn't have space to write it down..)

    It is just that STR (and Quantum Field Theory) nails the coffin shut... except for wannabe "mathematicians" who can only accept counting widgets as the Foundation of Mathematics....

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