## Neverfly vs. BulariaChk FLT reference thread

Because BulariaChk edits so heavily and because I made a lengthy post referencing his previous comments:
I am making this thread to contain the bulk of direct quotes made by BulariaChk to prevent him from altering his posts to change the record that I recorded in the FLT thread, post 293 onward.

1.
Originally Posted by BuleriaChk
Originally Posted by mathnerd
I have no problem with sums, products, and powers. However, when you do a proof by contradiction, you show that an assumption leads to a false proposition, so conclude that the assumption must have been false. Here, you're trying to show that FLT being false implies . Where is the proof of that claim?

Let's focus on this proposition:

Where is the justification for this assertion?

Your argument structure is (is it not?):

2.
Originally Posted by BuleriaChk
Originally Posted by grapes
What are you trying to say with that not-equals sign?
Uh, lessee now...

That (a + b)2 = c2 is not the same as a + b = c, maybe?

(unless the 2's and parentheses are invisible....

Keep working on it; I'm outta here except for revising my first post maybe...
3.
Originally Posted by BuleriaChk
Assume

c = a + b is a positive integer.

Then cn = (a + b)n is a positive integer.

Then cn = (a + b)n = an + bn + rem(a,b) (by Binomial Expansion) is a positive integer.

If Fermat's theorem is false, cn = an + bn is true.

If Fermat's Theorem is false, then rem(a,b) must be equal to zero.

rem(a,b) is ALWAYS > 0 so cn an + bn is true, which validates Fermat's Theorem. That is, Fermat's theorem is true.

????

(I dunno, but I still think I may be ok here....)
4.
Originally Posted by BuleriaChk
1. if c (a + b) then cn (a + b)n

(a + b)n = an + bn + rem(a,b) by Binomial Theorem
then cn an + bn + rem(a,b) by Binomial Theorem
If Binomial Theorem is false, then the above equation is invalid

(I freely acknowledge my proof depends on the validity of the Binomial Theorem).
That is, my proof is correct iff the Binomial Theorem is correct
--------------------------------------------

If Binomial Theorem is valid:
cn = (a + b)n = an + bn only if rem(a,b) = 0 which means a or b or both are equal to zero.

My proof is invalid only if the Binomial theorem fails. Again, I freely acknowledge that my proof depends on the Binomial Theorem. (which you ignore).
5.
Originally Posted by BuleriaChk
Originally Posted by grapes
Utterly false

It's not a matter of belief.

You're just wrong.
Like John Gabriel (and unlike Fermat), you haven't a clue about Descartes......
(nor does the Naked Emperor, for that matter)

Others... see Posts #145, 147

Trying to prove Fermat's Theorem on a single positive integer number line is a fools' errand in the Religion of Mathematics (like Dedekind cuts to justify fractions, irrational, an real number extensions from integers)
- which was actually John's point, except he didn't know about independent variables and dimensions (i.e., Cartesian products)....

(I'm waiting for a phone call from Dr. Wiles....
6.
Originally Posted by BuleriaChk
Although, the more I think about it, the more I think that while maybe vector functions may relevant, another (easier) way to think about this is to imagine the terms in rem(a,b) to be functions on the two dimensional plane of integers:

e.g., rem2,3(a,b) = a2b3 envisioned as a continuous function z = x2y3 and then pick off the integer values on the positive quadrant of the x,y plane (i.e., x > 0, y > 0) -> (a > 0, b > 0).

In any case, all of the positive integer terms in rem(a,b) > (0,0), and
rem(a,b) = 0 iff:
a = 0 (cn = bn),
b = 0 (cn = an)
or
a = b = 0 (cn = 0)

(i.e., for n > 2, rem(a,b) is non-linear and always positive) ....
7.
Originally Posted by BuleriaChk
Although, the more I think about it, the more I think that while maybe vector functions may relevant, another (easier) way to think about this is to imagine the terms in rem(a,b) to be functions on the two dimensional plane of integers:

e.g., rem2,3(a,b) = a2b3 envisioned as a continuous function z = x2y3 and then pick off the integer values on the positive quadrant of the x,y plane (i.e., x > 0, y > 0) -> (a > 0, b > 0).

In any case, all of the positive integer terms in rem(a,b) > (0,0), and
rem(a,b) = 0 iff:
a = 0 (cn = bn),
b = 0 (cn = an)
or
a = b = 0 (cn = 0)

(i.e., for n > 2, rem(a,b) is non-linear and always positive) ....
8.
Originally Posted by BuleriaChk
Originally Posted by mathnerd
I think you're really confusing yourself over the restriction . If you do this, you're just spinning your wheels by restricting yourself to the trivial trio , which of course solves for any n. For FLT, we're interested in nontrivial .
Actually, I think the requirement of a metric might validate my proof. The equation c = a + b true for all integers IS an example of a metric (a distance relation between positive lengths), true for all a, b, and n, where c is the metric. So the proposed metric cn = an + bn cannot be a metric for independent powers of a and b in the integer space (an,bn). The proof shows that such a metric cannot exist (there is no cn satisfying the equality) by means of the binomial theorem in the orthogonal vector space (an, bn) in I+ I+ -> I+ .....

(i.e., there will always be interactions between powers of a and b in rem(a,b).

(If correct, it would have interesting consequences for QFT and GTR, especially many-body theory at a field point ..

Not relevant to Fermat's theorem ================

(for n = 2 the metric is still valid under multiplication, since a,b, and c are the same power in the "interaction" product 2ab; i.e., the product 2ab = 0 does not invalidate the metric for Pythagorean triples. Also, there is only one term in the expansion that is eliminated for PT's)

That is, even if 2ab = 0, (a2,b2) are independent (orthogonal) variables with metric c2 = a2 + b2 for Pythagorean triples .

Note that in the equation:

c2 = (a + b)2 = a2 + 2ab + b2 if the product ab is an inner product, a.b can vanish if a and b are orthogonal (Pythagorean triples), since 2a.b = (2ab)cos(/2) = 0 (a scalar in one dimension; the cross product creates a pseudo vector which represents the area)

This is because a and b can be considered linear scalars multiplying orthogonal vectors, where the resultant products result either in the scalar 2ab = 0 or 2(

=========================================
For the case n > 2, if each element of the nonzero terms in rem(a,b) a or b in higher order terms are considered variables over unique number lines (e.g. a2b3 = , then the resultants of different ordering of the terms will be different, so the resultant metric cannot be mapped into a unique number line)
9.
Originally Posted by BuleriaChk
In the context of "c = a + b", to be technically precise, I should have used direct sums and products to indicate the relationship between independent sets in {a}

{c} = {a} {b} for addition and
{c} = {a} {b} for multiplication

I thought the context would be clear, since I specified independent variables a and b, and clear even if I didn't originally. Subsequent posts of mine have emphasized issue in the context of vector and matrix transformations ....

Silly me....

See my new thread on Independent Variables

Proof of Fermat's Theorem --------------

Fermat's Theorem: for a,b,c,n positive integers, n > 2

a,b, and n are independent variables and c is the dependent (resultant) variable

Let + rem(a,b) By Binomial Theorem

only iff rem(a,b) \neq 0

rem(a,b) > 0, therefore rem(a,b) 0

Therefore, for a,b,c,n independent positive integers, n > 2

QED
================

a and b integers in the same set

The point is that if c = a + b with a,b, and c in the same set, then both sides of the equation refer to the same number. Similarly for c = ab, or any powers of the same. Therfore, "Fermat's Theorem" as stated if in the same set: is equivalent to which is obviously wrong, so "Fermat's Theorem" would be false.

Again, here "Fermat's Theorem" refers to the formula applied to elements within the same set.

However, the statement is true. So if the negative elements are omitted, the statement is again false, since

So the negative elements form an independent (i.e., disjoint) set w.r.t. the positive integers if, in addition, multiplication is excluded from the negative integers. (All elements are EITHER positive OR negative, but not both).

Then if a and b are different positive integers in the same set, the Binomial Theorem is valid, since they are individually independent subsets each containing a single element (invariants), and Fermat's Theorem is valid, since , and .
(if a = a there is no distinction between the subsets, then the Fermat's Theorem is again wrong, since then .

-------------------------------------------

If this result is then extended to the case of two independent variables |a| and |b| over the sets {a} and {b} with the operations above, then the Binomial Theorem can be applied for the proof... In this case, Fermat's Theorem (the real Fermat's Theorem) MUST refer to independent sets and, together with the Binomial Theorem with the appropriate operations, the theorem is proved.

QED

In Fermat's Theorem, a,b, and n are independent variables and c is the dependent (resultant) variable
in the equation.
10.
Originally Posted by BuleriaChk
(I'm not convinced this is an issue, since equality was what established the Fermat relationship to rem(a,b)

That said,

assume c (a+b)
cn (a + b)n = an + bn + rem(a,b)
cn an + bn (since rem(a,b) > 0) (Fermat's Theorem)

QED

(sigh)

cn = an + bn means Fermat's Theorem is false

BUT

suppose c < (a+b)
then cn < (a + b)n = an + bn + rem(a,b)

(I remember now, this is why I introduced complex numbers in my previous thread)

Then rem(a,b) must have negative terms for equality: cn = an + bn + rem(a,b) -|d|, where -|d| < 0 so that -d = ik|d| and d = -rem(a,b) to prove Fermat's Theorem false.

If k is odd, then -d = ik|d must be imaginary,so k must be even.
if k is even, then d = -|d|, d can be either positive or negative; the positive solutions can be rejected immediately.

So d must be a bivariate polynomial that includes negative terms that cancel those in rem(a,b) (term by term), so one has to look at the individual terms to see if anything is left after setting one of the powers in the product to be < 0.

This is what led me to my original thread (and the consideration of powers as vectors (independent variables) in each of the terms; exponential powers form the bases of terms in polynomials and these terns are bi-variate (a,b), so dot and cross products are relevant in the analysis).

That is, a and b independent variables (a,b) in R x R means that a is perpendicular to b geometrically; as is (ia,ib) in the complex plane.