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. Ahem. A = 1, B = 2, n = 3. Thank you. 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? Sure. Could you please show where you got this math? I have no idea where you got it from. Assuming what you said in 5 is correct, then sure. 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...

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-

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

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.

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

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-

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-

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-

...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-

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-

Heh...sorry if I was talking condescendingly, as it's clear you do know this. I'm not sure if there's a nicer way, though. I thought you meant something slightly different, in which you had an identity and in which every element had a right inverse. However, we can look at yours this way: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-