uncool

I truly don't know what distinction you are making here; the axioms are what give you the "material" or "structure" to "construct" things, and (usually more importantly) then provide a method to prove theorems about those things Constructing an uncountable set is easy. Use the axiom of infinity to get an inductive set; use the axiom schema of specification to get a set we think of as the positive numbers, and use the powerset axiom to get the powerset of the natural numbers. This powerset is uncountable, by Cantor's theorem.

By constructing it, using the axioms of set theory.

I said that for a specific reason: you keep trying to use a structure ("+*") that you fail to define, and then worse, you assume properties about it without proving those properties. That's not how a proof can work. You can define such a set. Basic set theory, axiom schema of specification. Prove it. Your attempted proofs have not been sufficient.

Do you know why that formula works, in a geometric sense? (I don't plan to simply give an answer at the moment; simply giving an answer is uninformative, and someone recently asked the same question in the Homework Help section)

There are countably many triplets of integers (a, b, c) such that a^2 + b^2 = 2c^2. There's even a method to find them. Do you know how to find a triplet (a, b, c) such that a^2 + b^2 = c^2?

I realize why you want to be able to use it, but you have to demonstrate that it makes sense in the first place. To be honest, I think you are attempting to write a proof where you don't have an understanding of the underlying question, or of how a proof works.

Then you're adding a restriction to the original problem (eta: as I have told you before). The question is whether the real numbers have some ordering that is well-ordered - no other restrictions. You are adding the restriction that addition and multiplication must be compatible with the ordering. The well-ordering principle does not require compatibility.

"As posed", the question does not assume the "niceties". It's asking about using this as a definition; when talking about definitions, "niceties" cannot be assumed unless explicitly stated.

That's not how proofs work. You have made the claim that it can't have a least element; it's your job to demonstrate it, not for me to provide a method to find a minimum. You can try a proof by contradiction, to show that any method will result in a problem, but you don't get to look at my method when your statement should be true independent of which method I choose. Alternatively: I choose 1/2 to be the minimum of S. Prove that that's impossible.

And I keep asking you: why is your statement true for "this set"? Why can't {x: a <* x <* b} have a least element? You're relying on "clearly" for a statement that isn't at all clear.

This isn't exactly true; the actual definition involves lots of "there exists" and "for all"s. If the function is nice enough, then yes, this "pseudo-proof" tells you that the second derivative must be equal to the expression given; however, the second derivative does not necessarily exist.

What do you think is the reasoning behind those equations? In other words, I'm not talking about whether they work, but why they work - what train of thought led to their discovery. There is some interesting geometric reasoning that immediately gives the answer for a^2 + b^2 = 2c^2.

Expand on that "clearly". Why can't {x: a <* x <* b} have a least element? As I have asked you before: What step in your proof fails if you try to apply it to the natural numbers instead of the reals?

This integral isn't meant to be evaluated in a straightforward manner. The solution is a trick.

Spydragon: do you understand the reasoning behind the equations you gave for a^2 + b^2 = c^2?

That is exactly the opposite of correct reasoning; it reminds me of the "25 dollars" riddle.

Any shape made by an angled cross section of a circular cylinder can also be made by an angled cross section of a cone, with the exception of the degenerate case (a pair of parallel lines). It will be an ellipse.

While you are correct that there is only one obvious symmetry, it turns out that the equations of the cone and plane result in another symmetry. The conic section (whenever the plane is at a shallower angle than the cone itself) is an ellipse.

This description does work universally; it takes a bit of algebra to demonstrate it, but it does work. Ellipses can be seen as cross sections through both cones and cylinders..

The statement that the function is discontinuous is calculus. Additionally, this reverses the usual order of definitions. In calculus, continuity is defined in terms of limits, not the other way around. So this begs the question: how do you know the function is discontinuous?

taeto - the function I described does exist; it's not hard to construct using the axioms. In nonstandard analysis (specifically, the version using Internal Set Theory), it isn't standard, and nonstandard analysis defines limits and derivatives for standard functions (as I understand it).

Dasnulium: I don't think you have ever answered this question. In the system you favor, what would the limit of f(x) be in this case?

I think you have either used standard calculus or assumed your conclusion without realizing it in your paper, by assuming that "r" (which you should more explicitly define than "the two sets of terms as a ratio") can always be expressed in the form "+- b epsilon^2 +- c epsilon^3 +-.../+- a epsilon". More generally: let's say we have the function f(x) = 0 if x is neither infinitesimal nor 0, and 1 if x is 0 or infinitesimal (in other words, if x is smaller than any rational number). What is the limit as x approaches 0 of f(x)? (More on this after an answer)

...the math is what you provided. You asked how we should understand the equation. What math are you asking for?

You should understand it by realizing that the force is proportional to the square of the distance only if acceleration is being held constant, which is generally an unnatural assumption. Mass being held constant is a somewhat natural idea; acceleration being held constant is not.