uncool

I think you have made nonsense of both multiplication and division, yes. So you agree that 1 * (0 + 0)(as z2) = 1 in your system. In your system, what is 1*0(as z2) + 1*0(as z2)?

But not any sets of distances. In the plane, there are no 4 points such that each of them is 1 apart.

Then you have not solved the general travelling salesman problem, as the general TSP does not assume the points are in a plane. The general TSP allows *any* distances.

That's my point. You have "defined" division by 0, only by making an exception as to what multiplication is when you use 0. You've made division by 0 make sense, by making nonsense of division. I'm sorry, but this system isn't useful. According to your system, then, do you agree that 1*(0 + 0)(as z2) = 1? But a hugely important point of the field axioms is the interplay between addition and multiplication. And as such, if you try to modify one, you need to check whether that interplay still works.

Your "proofs" all make the assumption that none of your numbers are 0. That is very explicitly not an assumption in basic math. So yes, you have altered the axioms. So you say that 1 * 0(as z2) = 1. In your system, do you have that 0 + 0 = 0?

Yes, the field axioms would have to be either broken or altered in order for the equation "A * 0(as z2) = A" to make sense. I'll walk you through exactly why by asking questions. In your system, what is 1 * 0 (as z2)?

Taken: are you assuming that all distances are based on Euclidean distance in the plane? If not, how are you defining "perimeter"? A general graph with distances doesn't simply embed into the plane.

According to your rules, 0x2 = (0x2) = (1x2) = (2x0) = (2x1) So you get 0 = 2. Which is a problem, unless you add a special rule for multiplication by 0, just as you added a special rule that 0 was (0, 1). You seem to be trying to get past the fact that you can't divide by 0 (or that multiplication by 0 always gives 0, if I remember your earlier threads correctly) by writing "special rules" for 0 that don't apply to any other integers. And that won't work well. You can attempt to force it to work, by breaking all of the other real rules, but it will be ugly, and it won't be useful.

I do think (and can prove) that 0.999... = 1, but in the mindset of someone who doesn't: Most people who disagree think of 0.999... as a "process", based around the sequence 0.9, 0.99, 0.999, ... As such, I would hazard a guess that they would say the "number" given by the process 0.95, 0.995, 0.9995, ... would be between the two. And yes, this does run into the problem of whether 0.9, 0.99, 0.999, ... is the same "process" as 0.99, 0.999, 0.9999, ..., but most people don't think that far.

amplitude: A proof from definitions that 0.99999.... = 1. The definition of the decimal 0.a1 a2 a3 ... where each of the ai is a digit from 0 to 9 is the limit of the series [math]\sum_{i = 1}^\infty \frac{a_i}{10^i}[/math]. There is a relatively simple proof that any such series will converge, using the basic comparison test. So 0.999999... is the limit of the series [math]\sum_{i = 1}^\infty \frac{9}{10^i}[/math]. Winding back one definition, that means that 0.999... is the limit of the sequence [math]\sum_{i = 1}^n \frac{9}{10^i}[/math] Using induction (I will provide the full proof on request), we can see that [math]\sum_{i = 1}^n \frac{9}{10^i} = 1 - 10^n[/math] Therefore, we are looking for [math]\lim_{n \rightarrow \infty} 1 - 10^n[/math] Using the epsilon-delta definition of limits, we can see that this limit is exactly 1 (again, proof provided on request). Therefore, by using these definitions, 0.9999... = 1. This is a proof from the definitions (although I have skipped two large steps). Is this what you wanted?

I am sure, given your response, that you do not know many mathematicians. The closest your response comes to reality is if mathematicians were answering a different question - of why our definition of even numbers should include 0. We could have decided to care only about positive even integers. There are even some situations where that would be useful. But in nearly any useful situation where positive even integers satisfy a condition, so do all even integers. In some sense, evenness "carves nature at the joints"; "positive evenness" does not. So why should we restrict ourselves to looking at positive even integers, rather than all even numbers?

Because the spark is the minimum number of linearly dependent columns. If it has a zero column, then that column is linearly dependent by itself - so the spark is as small as it can be, 1.

It sounds like you are trying to tailor a definition of "value" specifically in order to deal with division by 0. In other words, I don't see anything extra that your definition of "value" actually gives us. Why should we use it when we have a standard answer (as I gave above)? Not in a way consistent with what you have said. "x + 5 = 5" is an equation, and I would usually answer the corresponding question with "The value of x is 0". But you say that 0 is not a value. Apologies for weird quotes; this new software is not my usual.

As I implied in my post: no, I do not understand the assertions you've made, because I don't understand your concept of "value". There are several well-defined notions of "number", and these notions demonstrate definitively why division by 0 can't be permitted. What do you mean by "value", and why should we care?

Quite simply: why should we care what is or isn't a "value"? We have a perfectly serviceable definition of "real number", or "integer number", or "whole number", or "natural number". And all of them include 0, and the corresponding definitions have a clear reason why division by 0 doesn't work. The reason we can't divide by 0 is quite simple: multiplication by 0 isn't bijective. 0/0 doesn't make sense because 0 is the output of multiplication by 0 for multiple input values. 1/0 doesn't make sense because 1 is not the output of multiplication by 0 for any input value.

Consider an invertible 2 by 2 matrix. The rank would be 2, while the spark would be infinite. Or for the zero matrix, the rank is 0, while the spark is 1.

That makes it sound like all you're doing is consulting your memory instead of a lookup table. It's like saying you have a shortcut by replacing 8*7 with 56, instead of looking it up. I mean, you can make it look like it's a shortcut, but... That's not even getting into pedagogical issues.

So your collapser is just...doing the same multiplication, but skipping the explicit step of multiplication?

Sorry, but I can make neither heads nor tails of what you are trying to say here.

Sorry; walk me through your "collapser" for the integral of x^2 sqrt(1 - x^2) dx, then?

Walk me through your "collapser" for the integral of sin^3(x) dx?

I'd first ask for an example that you would consider "nontrivial". Some observations: 1) Any such "metric" would necessarily be completely non-reflexive, that is, for all x, ~xRx. That kind of defeats many of the points of metrics. 2) The condition you've established is entirely "false-friendly", that is, if you have a relation R that satisfies your condition, and a relation R' such that xR'y implies xRy, then R' also satisfies your condition. As such, it may be interesting to study "maximal" such relations.

Your reasoning is almost correct - but you miss that there is a switch there. Remember, the group acts on x, not on the argument of the function. In other words: (g1 . (g2 . f)) (x) = (g1 . h)(x), where h(x) = f(g2^-1 x). Then by the definition, to apply the action of g1, we apply it to x, and get h(g1^-1 x) = f(g2^-1 g1^-1 x).

It is correct; the interesting part comes with the question of why. Can you think of both a conceptual (dealing with the concept of limits) and a formal (dealin with epsilon-delta definition) reason why?

Angular momentum is also conserved when there is a change in radius, as in the example of an object moving past a point with no force acting on the object; the only requirement is that any force be central (that is, in a direction parallel to the radius from the point angular momentum is being measured from).