Introduction to number theory lecture 13. The Chinese remainder theorem.

[@benjamindavid7371]

I think at 31:33 it's because the step where x = 1 mod 2^n implies x^2 = 1 mod 2^{n+1} works only for 2 as the modulo due to the binomial theorem.

[@labuclek2918]

@19:30 argument only works if both the sets are finite

[@area51xi]

This 34:46 min video took me several days to watch and it's still sinking in.

[@naveenjacob2440]

Great video!

[@AyushTripathi-j1z]

At 33:51, why don't we take the other solutions like -1,-3,--5,-7 as they also satisfy the equation?

[@olivergray8534]

We are only counting solutions modulo 8, so -1 (=7) has already been counted.

[@jimadams8385]

The consolations of mathematics. I would love Richard to produce new work on Verdier's thesis. Verdier died of AIDS a long time ago. The thesis purports to generate the whole of mathematics from an octagon diagram. Rather than going into the meaning of this it should be possible to generate diagrams for prime number-gons. This should describe new mathematics, but how are different diagrams related?

[@jimadams8385]

Sorry, this is not about the Chinese remainder theorem. I made some comments on Galois theory, with a wrong counterexample which amounts to refactorization, which could never be a counterexample. Schweinmachtbree states it is impossible to find any counterexample to Galois theory, because Galois theory has been proved by theorem provers. As having had a career in programming, I would state that a program is only as good as the programmer. Galois theory notes that the permutation of roots in a polynomial forms a group. The group theory is correct. What I claim is a gap in the literature, if not in theorem provers, is the proof that the solvability of polynomials corresponds bijectively with the presence of normal groups in a Jordan-Hoelder sequence. I have seen no proof of the existence of this model, which in category theory is called a representation. David Cox relates cube roots of unity to the solution in terms of linear combinations of them of roots of the cubic. The cubic is then represented by a quadratic. A similar procedure for the quartic yields a cubic in quadratic terms to give the roots. If the quintic is unsolvable, then the same method would show that the quintic cannot be represented in terms of a descending sequence of a quartic, cubic and quadratic. I have not done this calculation, but the question arises whether the Galois nonsolvability result can or cannot be obtained by these means, thus bypassing Galois theory entirely.

[@migarsormrapophis2755]

yeeeeeeee

[@jimadams8385]

I have solved FLT by elementary methods as many amateurs have claimed to do. It uses descent. The second method has been shown by schweinmachtbree to be wrong. The first method is unrefuted. Did you get the solution? I now understand galois theory. We must distinguish between rotation/reversal and noncommutative field extensions. The polynomial model is an initial object in a rotation/reversal category. The galois model is non abelian. Since the Jordan-Hoelder theorem operates on noncommutative groups and terminates for 5-permutations, the same must happen in the rotation/reversal case, since these groups are subobjects of nonabelian groups. However, if we look at the practical solution of the cubic, there are three equations for the roots, involving cube roots of unity. One is a linear term without roots of unity, the other two are complex conjugates on the coefficients. This might be related to galois theory. For the quartic, there seems to be no such relationship. I am adrift as to how galois theory may be said to operate here. Perhaps what we are saying here is galois theory has nothing practical to say on the solution of the cubic and quartic.

[@evgenykuznetsov7759]

What have you done, have solved a conjecture? OMG

[@jimadams8385]

@@evgenykuznetsov7759 Hello Evgeny. Beal's conjecture is divided into 16 components. There are two proofs of FLT using the binomial theorem. See entry 44 for one of the proofs (incorrect). I am humble enough to know I am not perfect. I just say what my reason tells me, irrespective of social constraints. The result is I am called a crank. The mathematics is the truth, not me. I am just a reporter of what I find. That is why I say I am a member of Rebel University. I say that there are deep imperfections in present day science, in mathematics clustering around conceptions of incomputability. There is new work on functors. Exponentiation is nonassociative, so almost never use categories. Universals remain. This is 'structure' theory. I have a theory of Topologic, unifying topology and logic. Insight theory is part of hyperintuition. See intuition - categories. The reason I thought for the defect in galois theory is it does not distinguish between commutative and noncommutative field extensions. The sextic is not solvable I now understand. The technique I give is just the product of two cubics, so is not solvable. I think what I am saying on galois theory is correct, now. There is a difference between commutative and noncommutative field extensions. The true theory should be simpler than galois theory, but have the same end result. It is a factorisation problem. Can anyone come up with an alternative to galois theory based on killing central terms, and/or not field extensions based on highly noncommutative algebra?

[@evgenykuznetsov7759]

OMG @@jimadams8385 , anyways care of yourself.

[@jimadams8385]

@@evgenykuznetsov7759 Thank you Evgeny. I go too far, and it is dangerous, but God I hope is with me. Take care of yourself too! The reaction is so bad I give up!

[@schweinmachtbree1013]

@@jimadams8385 Not all sextic polynomials can be solved by radicals (that is, have roots expressible using +, -, ×, ÷, square roots, cube roots, and fifth roots).