uncool

Tom - not quite; substitution isn't quite an axiom. However, the converse that you're stating actually itself is an axiom - the axiom of well-definedness, isn't it?

=Uncool-

It's been a while, Victor, and I'm ready to try this again.

Complet text of the proof

 

 

All wholes numbers in the next are given in the prime base [math]n[/math].

 

Here are the known positions, repeatedly discussed on the mathematical forums. Thus, let

 

(1°) [math]A^n+B^n = C^n[/math], where prime [math]n > 2[/math] and [math]A, B, C[/math] have no common factors.

Let for the moment the number [math]ABC[/math] is not divided by [math]n[/math]. Then:

Looks alright to me so far.

 

(2°) [math]C^n-B^n=(C-B)P[/math], where [math]C-B=a'^n, P=a''^n[/math],

[math]C^n-A^n=(C-A)Q[/math], where [math]C-A=b'^n, Q=b''^n[/math],

[math]A^n+B^n=(A+B)R[/math], where [math]A+B=c'^n, R=c''^n[/math],

[this follows from the simple lemma: if the numbers [math]A[/math] and [math]B[/math] have no common factors and the number [math]A+B[/math] is not divided by [math]n[/math], then the numbers [math]A+B[/math] and [math]R = \frac{A^n+B^n}{A+B}[/math] have no common factors (since [math]R=(A+B)^2T+nAB)[/math]];

Ahem. A = 1, B = 2, n = 3. Thank you.

 

 

 

(3°) the numbers [math]a', a'', b', b'', c', c''[/math] have no common factors;

OK, let's get to work on this slowly. When you say no common factors, do you mean for all 6 numbers in total? or pairwise?

 

(4°) [math]A=a'a'', B=b'b'', C=c'c''[/math].

Sure.

 

(5°) Important equalities:

[math]c'^n - b'^n = 2a'a'' - (C-B)[/math],

[math]c'^n - a'^n = 2b'b'' - (C-A)[/math],

[math]a'^n + b'^n = 2c'c'' - (A+B)[/math], or

 

(5°a) [math](A+B) - b'^n = 2a'a'' - a'^n [/math],

[math](A+B)- a'^n = 2b'b'' - b'^n [/math],

[math]a'^n + b'^n = 2c'c'' - (A+B)[/math],

from here we find the numbers [math]a''[/math] and [math]b''[/math]:

Could you please show where you got this math? I have no idea where you got it from.

(6°) [math]a'' = \frac{(A+B) - b'^n + a'^n}{2a'}[/math],

[math]b'' = \frac{(A+B) - a'^n + b'^n}{2b'}[/math].

Assuming what you said in 5 is correct, then sure.

At last, we find two important formulas, лежащие in the base of the proof of the FLT:

(7a°) [math]a'' - b'' = \frac{(b'-a')(A+B) + (a'+b')(a'^n-b'^n)}{2a'b'}[/math],

(7b°) [math]a'' + b'' = \frac{(b'+a')(A+B) - (a'-b')(a'^n-b'^n)}{2a'b'}[/math].

Sure, works for me, at the moment.

That is my analysis so far. I've found an error and a few points I'd like you to expand upon. I will continue after you have done so.

=Uncool-

Ooo, that looks like a nice proof...will take me a little while to understand it, as I don't know the Sylow theorems...

O MY JESUS YOUR JOKING. I thought he meant something else, anything else! I didnt actually read the posts seeing as I didnt know what FLT was...but omg Fermats Last Theorem? Proved, so easily, When The Best mathematicians in the field of modular forms would have trouble UNDERSTANDING Andrew Walters 200 page proof. Get some Humility.

This is why he is asking and correcting...it is possible, though unlikely, that he's found it.

Side note: Andrew Wiles.

There is nothing wrong in asking whether this will work...

=Uncool-

Excuse me last time, please!

 

Splendid joke.

 

Well known that:

If in the equality

1°. a^n+b^n=c^n the numbers a, b, c have no common divisors, then the number

2°. a+b-c=un^2.

It is obvious also, that if abc does not divided by n, then the number

3°. a^3n+b^3n-c^3n=Un^2.

From here it follows a simple proof of the Fermat’s Last Theorem

(at least, for case when abc does not divided by n):

 

Let us use the identity for A+B=C:

4°. A^3+B^3-C^3=3ABC.

 

But the number a^3n+b^3n-c^3n=3(a^3n)(b^3n)(c^3n) (cf. 4°) does not divided entirely by n^2 (cf. 3°)!!!

I do not see 2 as being true. 3 either. Are you assuming n is prime?

=Uncool-

You do? Isn't it just [math]\ln{x}\cdot e^\frac{1}{x}[/math]?

No - ln is the integral, not the derivative, of 1/x. And in any case, you did use the chain rule to do that.

The problem in itself is quite difficult, because it includes three different methods of differentiation - the derivative of an exponential function, the derivative of an inverse of a power function, and the derivative of a product. The three together make this problem rather annoying at the least...

=Uncool-

Can you prove the lemma? I'm not sure that it's true. In fact, if it were true, it would prove FLT all by itself, because:

If a^n + b^n - c^n = 0 = nabcd, then either n = 0 (not true), a, b, or c = 0 (assumed not true), so d = 0, but then n|d, which you don't allow.

Also, do you mean (c-a)^n + (c-b)^n - (a+b)^n?

How do you know that (c-a), (c-b), and (a+b) have no factors of n? If they do have enough factors of n, then they can be equal.

see the application of this equation at this link

 

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2

I looked, no sphere there at all. Please stop making plugs for your site.

;316624'']Ok, I still fail to see how thats the most important use of imaginary numbers. It looks more like the least important use.

Looks to me like someone's trying to get traffic to their site.

=Uncool-

It would usually be done wth mathematical induction - each time, you add 2n + 1, making the numbe ron the left side n^2 + 2n + 1 = (n+1)^2, and as you have a base case with 1 working, mathematical induction guarantees it for all n.

=Uncool-

Actually momonaco27, the inverse of Planck length would be (1/1.616E-35) and that would be: 6.188E+34, not 1.616E+34 as you formerly stated. Never-the-less, an interesting thought anyway.

Another thing is that the inverse of Planck length is 6.188E+34 m^(-1), not 6188E+34 m - define your inverses correctly.

=Uncool-

How is D equal to A^n + B^n - C^n?

Why can't A^n + B^n - C^n have an extra 0 at the end?

A - a = dn^(k+1) - 2a, not dn^(k+1), and so on. Perhaps you meant to write D + E?

Can you try to write this all out formally, starting from the beginning? It's a bit hard to read it all bunched up as you have it.

=Uncool-

I am basically a grad student in math, and what you have written makes no sense. How do you get frequencies from lines?

And P.S. Swansont is a mathematician...

=Uncool-

Alright, please give us a full explanation of everything, and I mean everything that went down for this. I want dates, times, places, reasons for the errors, reasons why the errors weren't caught, reasons why we ourselves can catch them, reasons why Russia didn't catch this, etc. We have already given basically all of this nd more. Now it's your turn.

=Uncool-

Clearly, all 3 have to be less than 1. Then, at least two of them must be less than 1/2, as otherwise we have 1 + (1/2)^y + (1/2)^z where y, z < 1. Let y, z < 1/2. That should start off the problem...

Are there any nonabelian groups of squarefree order where none of the primes divides any of the others minus 1? If not, is there a proof of this? I've been trying to do this, but the proof for 2 primes doesn't extend, as I can't find how to prove the group with order of the largest prime is normal.

=Uncool-

Here is the very first problem, ghostunitt, go back and review the definition of an equilateral triangle.

Actually, this is correct - the height of an equilateral triangle with 18 feet-long sides is about 15.6.

=Uncool-

You know, you really haven't said anything at all here, especially in mathematics. You've simply complained that people are saying that what they see in the math really is the universe. In that, you're once again wrong. They say that what they see in the math models the universe. There is a large difference, and you would do well to figure out what it is.

=Uncool-

Dear Uncool, I could be utterly and entirely wrong. It wouldn’t be the first time in my 76 years. If I am wrong, it won’t be for a long period of time. Then, I will know and, if your were correct, I will bow to your excellent reasoning. I may even come back and pull your toes. Until then, I’ll keep smiling and smelling the roses along the way.............TreborS

...and this is all I get for my second post. Thank you very much for showing that you have no care for content, unless you will reply to my second post (and basically, my first post as well, since you completely ignored the point of that post, too).

=Uncool-

In the real world, there is no such thing as a 3-body problem. Since there is no gravitational umbrella, there is only a 2-body problem: the subject body and the rest of the universe. Any subject body cannot react gravitationally with just another or two other bodies. It is constantly solving the 2-body problem between it’s own center-of-mass and the apparent center-of-mass of everything else. Nature just does it. Man tries to reduce it to a mathematical problem by isolating a few bodies from the n-bodies of the universe. ......TreborS

You are utterly and entirely wrong. The center of mass is used as an approximation - not as the actual event. If, say, there is a tiny object that is an extremely small distance away - such a small distance that it has the largest effect on the other object - then even though the center of mass hasn't changed much, the gravitational pull on the object has. The center of mass is not correct as is - it is simply an approximation.

If what you were saying were true, then the 3-body problem itself would reduce to a 2-body problem, and therefore be solvable - which it is not.

=Uncool-

The one problem with this is that it completely misunderstands how center of mass works. The attraction is neither directly between the center of masses, nor is it only related to that. This can easily be shown by using 2 objects of radius r at distance 2r. While the attraction between the center of masses is GM1M2/4r^2, at certain points the attraction is much, much greater, such as at the point of tangency.

This is the reason why the 3-body problem cannot be solved. Nature doesn't act as if it's a two-body problem - if it did, then the 3-body problem would be reducible to the 2-body problem and therefore be exactly solvable, which it is not. The argument entirely breaks down there.

=Uncool-

The 0th and 1st Isomorphism theorem works in these partial rings, that is:

0th: The kernel of any "partial ring" homomorphism is a partial normal ideal, that is, it is a normal subgroup under addition and is an ideal under multiplication. Any partial normal ideal N of R also has a homomorphism to R/N with kernel N.

The definition of a "partial ring" homomorphism is that it takes sums to sums and products to products.

1st: If you have an onto ring homomorphism from R to R' with kernel N, then R/N is isomorphic to R'.

I believe that this also implies the second and third isomorphism theorems, namely, H+N/N is isomorphic to H/H intersect N for any subring H and normal partial ideal N of R, and also that G/N1 = (G/N2)/(N1/N2) where N2 and N1 are normal partial ideals of G, and N2 is in N1.

=Uncool-

Ah, now you're using that the dimension of the row space is equal to the dimension of the column space. If A has a right inverse the column space has dimension n (A is nxn). And A has a left inverse iff it has dim(rowspaceA)=n.

 

That's a neater way to look at it. But showing that dim(rowspace)=dim(columnspace) (=rank A) is shown in my book by counting pivots. I hope there's a nicer way to look at it apart from counting pivots.

I forgot what I those things were called, but I meant RINGS. The set of nxn matrices form a ring. I`m pretty sure you cannot prove that if an element A of your ring has a right inverse then it will also have a left inverse. I dont think it's hard to conjure up a counterexample. That fact that it is true for matrices means you have to use some properties of matrices and that's why I dug into vectors and pivots and whatnot.

Let us look at the set of elements with a right inverse (they may have left inverse as well). This has a well-defined multiplication, is closed under multiplication, is associative, and has an identity. It therefore is a quasi-group. It also has a right inverse for every element, as defined - and therefore, it can be proven that they have a left inverse, that is equal to the right inverse.

=Uncool-

Recently found: 32 element ring (with no identity) such that the set of products is not a group.

Under addition, it is Z2^5. Let the basis be named a, b, c, d, and e.

Under multiplication: c, d, and e annihilate the ring (their product with anything is 0). a*a = c, a*b = b*a = d, b*b = e. This is distributive, associative (as all triple products are 0), etc., and c + d + e is not generated.

=Uncool-

well, 2 things, Zhareon:

1) There's a thing called kernel, etc. It measures the independence of each line of a matrix - that is, if you make the matrix into a set of vectors, it measures how many are idependent. It is simple to prove that the dimension of the horizontal kernel is equal to that of the vertical kernel - so that if the matrix has an inverse on the right, then its horizontal kernel has dimension 0, so the vertical kernel has dimension 0, so it has a left inverse (this is from a while back, so anyone with a more correct way of saying it is welcome.)

2) Actually, there is no group-like structure with right inverse but no left inverse unless you remove associativity.

=Uncool-