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Thread: Dimensions, sets, and Fermat's Theorem

  1. #11
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by grapes View Post
    Values of c that do not equal a+b

    Values of a,b,c that satisfy the triangle inequality

    In other words, pretty much almost all the possibilities, all the ones that you have ignored in your "attempt" at a proof of Fermat's Last Theorem
    values of a,b,c that satisfy the triangle equality are Pythagorean triples.
    values of a,b,c are included in the binomial theorem for case n = 2 that do NOT satisfy the triangle equality.
    That covers them all for a and b on disjoint number lines (vectors).

    c = (a+b) = d on the same number line are simply different names for the same integer.
    (the same integer is just partitioned in the case a + b, but there are still just as many elements.)

    Let a = 6



    That's why one uses vectors.

    (You really are very slow.... you keep repeating the same dumb objections. Find something more useful to do.... )
    Last edited by BuleriaChk; 09-24-2016 at 04:22 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
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  2. #12
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by BuleriaChk View Post
    Quote Originally Posted by grapes View Post
    Values of c that do not equal a+b

    Values of a,b,c that satisfy the triangle inequality

    In other words, pretty much almost all the possibilities, all the ones that you have ignored in your "attempt" at a proof of Fermat's Last Theorem
    values of a,b,c that satisfy the triangle equality are Pythagorean triples.
    Triangle inequality, you mean.
    values of a,b,c are included in the binomial theorem for case n = 2 that do NOT satisfy the triangle equality.
    O, you do mean triangle equality? What are you talking about?
    That covers them all for a and b on disjoint number lines (vectors).
    You're just confused.
    c = (a+b) = d on the same number line are simply different names for the same integer.
    (the same integer is just partitioned in the case a + b, but there are still just as many elements.)

    Let a = 6



    That's why one uses vectors.
    6,3,3 doesn't satisfy the triangle inequality, 6,4,2 doesn't satisfy the triangle inequality, 6,5,1 doesn't satisfy the triangle inequality

    For n>2, your proof ignores all numbers that satisfy the triangle inequality.
    (You really are very slow.... you keep repeating the same dumb objections. Find something more useful to do.... )

  3. #13
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by grapes View Post
    Triangle inequality, you mean.

    O, you do mean triangle equality? What are you talking about?

    You're just confused.

    6,3,3 doesn't satisfy the triangle inequality, 6,4,2 doesn't satisfy the triangle inequality, 6,5,1 doesn't satisfy the triangle inequality

    For n>2, your proof ignores all numbers that satisfy the triangle inequality.
    All your number combinations are not in the relation for Fermat's formula, and are irrelevant, triangle equality or not

    Triangle equality refers to n = 2 (Pythagorean triple), which is irrelevant to Fermat's theorem...

    It does NOT refer to anything for n >2 (i.e., the Binomial Theorem. Study it, you might learn something)

    Anyway, this discussion is pointless, since you are now bringing up repetitive, irrelevant points, and you insist on misinterpreting me. Unless there is something interesting or relevant, I won't be responding to you from here on out.

    You have joined the echo chamber once again...
    Last edited by BuleriaChk; 09-24-2016 at 10:59 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.

  4. #14
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by BuleriaChk View Post
    All your number combinations are not in the relation for Fermat's formula, and are irrelevant, triangle equality or not

    Triangle equality refers to n = 2 (Pythagorean triple), which is irrelevant to Fermat's theorem...

    It does NOT refer to anything for n >2 (i.e., the Binomial Theorem. Study it, you might learn something)

    Anyway, this discussion is pointless, since you are now bringing up repetitive, irrelevant points, and you insist on misinterpreting me. Unless there is something interesting or relevant, I won't be responding to you from here on out.

    You have joined the echo chamber once again...
    This comes from the one that keeps bringing this up when we've pointed out repeatingly that you have not demonstrated any relation. It exists only in your head but not in reality.

  5. #15
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by BuleriaChk View Post
    All your number combinations are not in the relation for Fermat's formula, and are irrelevant, triangle equality or not
    Those were your number combinations.
    Triangle equality refers to n = 2 (Pythagorean triple), which is irrelevant to Fermat's theorem...

    It does NOT refer to anything for n >2 (i.e., the Binomial Theorem. Study it, you might learn something)
    Inequality. I have no idea why you keep referring to triangle equality, even after being corrected.

    It is easy to show that any solution to cn=an+bn has to satisfy the triangle inequality. When you assume c=a+b, or even that a and b are legs of a right triangle with c the resultant, you are ignoring over 99.9999% of the possibilities.
    Anyway, this discussion is pointless, since you are now bringing up repetitive, irrelevant points, and you insist on misinterpreting me. Unless there is something interesting or relevant, I won't be responding to you from here on out.

    You have joined the echo chamber once again...

  6. #16
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    Default Re: Dimensions, sets, and Fermat's Theorem

    It is impossible to prove Fermat's theorem without two independent number lines representing a and b, because there is no consistent distinction between the partitions within the same set, e.g. for a = 24, , etc., etc., etc.....

    All the coefficients are simply ways of counting all the elements of the set represented by 24 - which is the single element (positive integer) a = 24 represented in the equation for Fermat's Theorem:

    for a,b,c and n positive integers, n > 2

    (If there is only one number line, there is no second element "b")... it is not that b = 0, but that b does not exist in the equation, it is only a meaningless symbol and all one has is

    .... with no relationship specified except equality.

    To the reader: read my ignore list before continuing.
    Last edited by BuleriaChk; 09-25-2016 at 12:51 PM.
    _______________________________________
    "Flamenco Chuck" Keyser
    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.

  7. #17
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    Default Re: Dimensions, sets, and Fermat's Theorem

    To the reader, don't bother reading his list. It is just him crying over that poeple are showing how much of a crank he is.

  8. #18
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by BuleriaChk View Post
    It is impossible to prove Fermat's theorem without two independent number lines representing a and b, because there is no consistent distinction between the partitions within the same set, e.g. for a = 24, , etc., etc., etc.....

    All the coefficients are simply ways of counting all the elements of the set represented by 24 - which is the single element (positive integer) a = 24 represented in the equation for Fermat's Theorem:

    for a,b,c and n positive integers, n > 2

    (If there is only one number line, there is no second element "b")... it is not that b = 0, but that b does not exist in the equation, it is only a meaningless symbol and all one has is

    .... with no relationship specified except equality.
    Number lines exist, there's an infinite number of them.


    To the reader: read my ignore list before continuing.

  9. #19
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by grapes View Post
    Number lines exist, there's an infinite number of them.
    Existence is in the mind. One chooses in the imagination which ones and how many of each to address for a particular problem.
    (One doesn't do experiments on number lines.... or does one? ....
    _______________________________________
    "Flamenco Chuck" Keyser
    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. #20
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    Default Re: Dimensions, sets, and Fermat's Theorem

    Quote Originally Posted by BuleriaChk View Post
    Existence is in the mind. One chooses in the imagination which ones and how many of each to address for a particular problem.
    (One doesn't do experiments on number lines.... or does one? ....
    Or one can simply use that imagination to have both a and b on the same number line, that refutes this post:
    Quote Originally Posted by BuleriaChk View Post
    It is impossible to prove Fermat's theorem without two independent number lines representing a and b, because there is no consistent distinction between the partitions within the same set, e.g. for a = 24, , etc., etc., etc.....

    All the coefficients are simply ways of counting all the elements of the set represented by 24 - which is the single element (positive integer) a = 24 represented in the equation for Fermat's Theorem:

    for a,b,c and n positive integers, n > 2

    (If there is only one number line, there is no second element "b")... it is not that b = 0, but that b does not exist in the equation, it is only a meaningless symbol and all one has is

    .... with no relationship specified except equality.

    To the reader: read my ignore list before continuing.

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