uncool

The precise answer is: Spin is the part of angular momentum that does not come from position, where angular momentum is the conserved quantity associated with rotational invariance. I don't expect you to understand this as is; it takes quite a bit of study. In order to understand the above, the best thing to do is to learn classical mechanics (up through the Lagrangian and method of least action), then to learn quantum mechanics (including spin, but taking it as a "given"), and then to learn some quantum field theory.

If a positive integer n is not a square, then its square root is irrational. This includes all primes. My point was that it isn't as simple as looking at the definition - it does take some facts about the integers in the first place, as vovka implicitly points out.

Not the definition, no. It does need to be proven that given a prime p, then for any nonzero integers a, b that a^2/b^2 is not p.

Only if you require that the map be continuous. There are (highly discontinuous) bijections between R^m and R^n for any positive m and n.

It is a thought experiment that has associated calculations. Yes, those calculations have a non-calculation explanation, but there are still calculations. I don't understand your picture. What, exactly, is your "third event"? And remember, in the original ladder paradox, there are 4 relevant events: 1) Back of ladder passes entrance of barn 2) Front of ladder passes exit of barn 3) Entrance door closes 4) Exit door opens.

I'm going to put my marker in for "There is no infinitely rigid rod in relativistic mechanics" as the reason this "discrepancy" fails.

How, exactly, is this supposed to prove the Collatz conjecture? You have a "probabilistic" reason to think that most things get down to 1; that simply is not enough. The conjecture must be true not just "probabilistically", but for all integers.

You really think the people listed don't abuse process for their own gain? You don't think Microsoft, Apple, Facebook, and others have been found doing unethical things, not just once, but on a repeated basis? I'd suggest that many of them have abused process for their own gain - but (at least recently) in a way that many people here find amenable to their own goals (or not especially objectionable), especially in contrast to some other especially distasteful examples.

I did not "force" them to be equal. I literally picked 3 arbitrary numbers. That's the entire point of the distributive property. Also: you still have your equation wrong. It's not "a( b + c) = a*b + b*c"; it should be a*(b + c) = a*b + a * c. Further: the fact that you have to say "excluding zero" is a huge problem. It means that for literally everything you do in your system, you may have to say "Unless we ran into zero somewhere along the way". And since operations (namely, multiplication) with zero are the only difference between your system and the usual system of multiplication, that takes away any possible advantage your system could possibly have. If you don't want to talk distribution any more, that's fine; I'm done. I've given my "peer review" (although to be honest, no, I am not your peer), which is: your system does not add anything useful to the usual system of multiplication, and rather strongly takes something away.

Change the 1 to 5, the first 0 to 2, the second 0 to 3. 5*(2 + 3) = 25 = 10 + 15 = 5*2 + 5*3. Yes, it does work with numbers that aren't 0. "a ( b + c ) = (a + b) * c" is not the distributive property whatsoever.

You still haven't answered how distribution can possibly work, except by breaking the usual way distribution works. Which would make this useless to any normal use of multiplication.

And once again: it seems that you are choosing z1 or z2 in every case to simply match what happens for the original multiplication in order to check the axioms. Which means that your system doesn't add anything useful - in every case that matters, we'd just use basic multiplication.

You had it right the first time. Once again: the distributive property is a statement of equality, not of notation.

Once again: It's not that they simply are equivalent. It's that the distributive property holds only if they are equal. And in your system, they are not equal.

Again: the distributive property is not an act. It is a statement of equality. And the point is that the two are not equal. 1*(0 + 0) means "Do the operation 0 + 0 first, then perform the multiplication 1 * the result". The distributive property says that that must be equal to what you get when you add 1*0 and 1*0. And in your system, it isn't. Of course when you "distribute" 1*(0+0) you get 1*0 + 1*0. That's what distribution says. The point is that in your system, they're not equal, and therefore that the distributive property fails in your system.

The problem is that you are now not using PEMDAS. The distributive property isn't telling you how to do that multiplication. It's specifying what the result of that multiplication must be.

That's my point - it yields 2 products, and you seem to be willing to pick and choose which one is important for the axioms. Which...isn't useful. Especially because the one that is important for the axioms is always the one that yields the original multiplication.

OK, that gets into a major problem: It sounds like you're saying your system is "Multiplication gives two answers, and I pick and choose which answer seems to work". That's not how multiplication works, and breaks multiplication entirely. One important thing about multiplication is that it gives a single answer, period. Moreover, one of those answers is simply "Apply the usual multiplication". Which will always work. So your extra answer doesn't add anything.

1) I did. Remember when I asked you whether 1 * 0(as z2) = 1? And therefore that 1 * (0 + 0)(as z2) = 1? That's because I was following PEMDAS. 2) The distributive property of multiplication is the fact that the two must be equal. No, the distributive property is not an "act" of multiplication. Distribution of multiplication is an "act" that uses the distributive property. To be more precise with that inequality: 1 * (0 + 0)(as z2) = 1 * 0(as z2) = 1 =/= 2 = 1*0(as z2) + 1*0(as z2) The fact that the far left is not equal to the far right means that in your system, distributivity fails.

Distributivity isn't multiplication. It's the property that the two sides must be equal. And you have agreed that they are not.

I am following PEMDAS (note: the 5th operation is addition, so PEMDAS). To use all the relevant parentheses: 1*(0 + 0)(as z2) =/= (1*0(as z2)) + (1*0(as z2)). It fails distributivity because...that's what distributivity is. I just put in 1, 0, and 0 for the 3 variables. Distributivity in general says that: a*(b + c) = a*b + a*c. That inequality is a counterexample, setting a = 1, b = c = 0.

I just did. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2).

Good. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) That means you have broken distributivity. In other words, you have broken the exact way in which addition and multiplication interact.

I have no idea what you are trying to say here. Your method doesn't explain how to solve the travelling salesman problem for the graph with 4 equidistant vertices. It only works on a subclass (graphs that can be embedded into the Euclidean plane), and not even then - it requires extra information in the form of an embedding in the plane. As such, it isn't a solution to the TSP.

So in your system, the following is NOT true: 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2)