wtf

ps -- I don't think my previous explanations were very good. I found a much better page explaining this matter. If you google "why can't you distinguish i from -i" you get hundreds of totally irrelevant hits no matter how you alter or rephrase the question. Took me a while to find this. https://math.stackexchange.com/questions/177594/how-to-tell-i-from-i/177601#177601 The right answer is that there's an automorphism of [math]C[/math] that takes [math]i[/math] to [math]-i[/math]; namely, complex conjugation. In other words the difference between the two amounts to a relabeling with no change of meaning. By comparison, there is no automorphism of the reals that takes 1 to -1. They are algebraically different.

-a -ib. Why are you asking such an elementary question whose answer you perfectly well know? If z is a complex number, -z is its reflection through the origin. Perhaps you'll find this helpful. There are no positive or negative complex numbers because it's not possible to put a total order on the complex numbers that is compatible with their addition and multiplication. https://math.stackexchange.com/questions/788164/positive-and-negative-complex-numbers Or https://en.wikipedia.org/wiki/Complex_number#Ordering

I'm afraid I didn't see where you said what it is. There are no positive or negative numbers in the complex numbers, that's why you can't unambiguously use the sqrt sign convention that works in the real numbers.

OP never came back? After crossposting this to two different discussion forums?

Yes. The notation [math]\sqrt{-1}[/math] is often used casually, but it's imprecise in the branch of science, profession, or art of mathematics.

Me being the picky type, let me point out that [math]\sqrt{-1}[/math] is not good notation and is technically not correct. In the case of a nonnegative real number [math]x[/math], we can define [math]\sqrt x[/math] as the positive of the two values whose square is [math]x[/math]. However in the complex numbers there is no concept of positive or negative. That is, we can't algebraically distinguish between [math]i[/math] and [math]-i[/math]. So we define [math]i[/math] as a complex number such that [math]i^2 = -1[/math]. We pick one of the two possible values and call it [math]i[/math], and we call the other one [math]-i[/math]. The notation [math]\sqrt{-1}[/math] is technically inaccurate regardless of its ubiquity. That's why when people want to use that notation and are also being precise, they'll say, "Let [math]\sqrt{-1}[/math] be a square root of -1, rather than "the" square root of -1. The use of the word "a" signals that the writer understands the point and is using the notation [math]\sqrt{-1}[/math] anyway.

pps -- I looked at your handle and found this. You most definitely do know how to do math markup. Can you please do us a favor and mark up this linear algebra post?

ps -- I follow everything up to this line. If you can please put in proper parens and show exactly how you got this I'd find it very helpful. Also please note that the [math]y_i[/math]'s are presumably taken to be all distinct from each other, else you can't be sure they contain a linearly independent subset. And also note that when ask us to consider the equation you need to specify that at least one of the [math]c_i[/math]'s is not zero. There's no reason they couldn't all be. When you say, "We cannot have all c_i=0 [individually]in an exclusive manner since that would make the space N dimensional," that is not true. With all the c_i's equal to zero, that equation is true in any vector space no matter what the dimensions of the space and its subspaces. I think you are confusing this with the definition of linear independence, which says that if [math]\displaystyle \sum c_i v_i = 0[/math] implies that all the c_i's must be zero, then the v_i's are linearly independent. But just because you have [math]\displaystyle \sum c_i v_i = 0[/math], that doesn't mean all the c_i's can't all happen to be 0 regardless of whether the v_i's are linearly independent. Finally, you keep using the notation Σi=Ni=1 ... which I imagine is supposed to mean [math]\displaystyle \sum_{i= 1}^N[/math] but is incredibly confusing in context. Can you please fix that throughout?

I saw your similar post on another site where it didn't get any traction. May I suggest a couple of minor notational changes that will improve clarity? Since [math]e \in V \setminus W[/math], I'd call it [math]v[/math]. Likewise I'd call the [math]y_i[/math]'s [math]w_i[/math]. These minor changes would decrease the cognitive burden on the reader; and (if you don't mind my saying) your exposition here and on the other site are already a little convoluted and the reader can use all the help they can get. This notation's hard to figure out. You did convince me on the other site that [math]\displaystyle \sum_{i = 1}^N c_i = 0[/math], but here I don't know what you're trying to say. Are there some parens missing? I think so but who can be sure? I don't mean to be making only picky stylistic complaints, but I did make an honest attempt to work through your exposition on the other site and gave up due to lack of clarity. I suspect others might have done the same. You will get better responses if you clarify the exposition and notation. At least here you are no longer trying to divide through by [math]c_N[/math], which for all we know might be 0. So there is some incremental improvement over the version on the other site. Also where do you use the fact that N >> n? Your argument (whatever it is) seems like it would go through (or not) just as well for N = n + 1. In fact if you could give a concrete example with, say, [math]V = \mathbb R^3[/math] and [math]W = \mathbb R^2[/math] you might either figure out where you're making a mistake, or at the very least others would see what you're trying to do. Also do you happen to know Mathjax markup? For example if I write "e^{i \pi} + 1 = 0" between two "math" tags on this site, I get this beautifully rendered [math]e^{i \pi} + 1 = 0[/math] It's a small learning curve at first but greatly improves readability. As someone with terrible handwriting I wish they'd had this when I rode a dinosaur to school.

I don't believe this is true. sqrt(1) = 1 by definition, assuming by Root(1) you mean [math]\sqrt{1}[/math]. The square root of a positive real number is the positive of the two numbers whose square is the number. So if someone asks, find [math]x[/math] such that [math]x^2 = 1[/math], the answer is {1, -1}. But if someone asks what is [math]\sqrt{1}[/math], the answer is 1. There is no solution to the question in the title. What is true is that [math]- \sqrt{1} = -1[/math].

I happened to run across this terrific talk called The Secret Life of Quarks, well worth watching. https://www.youtube.com/watch?v=H_PmmMkGyx0 Like others, I'd always heard there are 3 quarks inside the proton. Turns out it's not really true. 3 is the number of quarks minus the number of antiquarks. But it's not a matter of counting and subtracting. Rather, you integrate something called the quark density function, and when you do, you get the answer of 3. The actual number of quarks and antiquarks depends on the scale at which you look. So there could be millions, zillions, whatever. I'm fuzzy on that part. But it's not 3 as in the counting number 3. 3 is what you get by integrating the quark density function, and it's much more complicated than just subtracting one integer from another. I found this article too, which I didn't read much of but that bears on the matter. https://en.wikipedia.org/wiki/Parton_(particle_physics) I'm curious about getting an amateur-level understanding of this answer myself. I'd always heard 3, but that's apparently a tremendous simplification that's not literally true. The other really interesting thing about all this (probably old hat to the pros in here), which I also learned recently from Youtube, is that mass comes from the binding energy among the quarks inside the nucleus. It takes a huge amount of energy to pull quarks apart, which is why you never see them in isolation. By Einstein's famous [math]E = mc^2[/math], that energy turns into mass. That's actually where mass comes from. Now the question is, why does the binding energy of the quarks in the nucleus bend spacetime? A Nobel awaits whoever figures that one out.

Thanks for the info. As a suburbanite I'm just getting over the shock of learning that cheeseburgers are made of chopped up dead cows. I had no idea.

The nearest horse is far from where I live. But my point was that horses foal (if you say so); they don't horse. Whereas evidently, lambs lamb. Which I didn't know. Now I'm gonna take it on the lam.

As a child of the suburbs I did not know that lamb could be verbed.

Are you saying that if RH was proven the math community would keep it secret? Did they change the fonts on this site so that the text is so light that I can no longer read it? Probably the same people who are covering up the proof of RH.

Nevermind. @joigus already linked the Motl reference that I was about to post. But there's no one-page proof of RH, I'm sure of that. And if RH had been solved we'd have heard about it.

No, they come to the conclusion that you're a little off. At least I did. Likewise brook, which has a different connotation in standard English, meaning "to stand for or tolerate." As in, "He brooks no difference of opinion." You give the impression of playing games with your own internal language, which detracts from whatever you're trying to say.

"I'm not a conspiracy theorist. I'm a conspiracy analyst." -- Gore Vidal Yes, so how did you get from linguistic imperialism to the fine points of the second-order completeness theorem for Henkin models? Thanks for explaining witcraft. Most English-speaking readers probably took that as witchcraft, which made no sense in context.

Most praiseworthy, and you put my freeloading self to shame. I should mention though that according to Google, Stanford University's endowment is 27.7 billion USD. Now 27.7B and ten dollars 😉 Also I went to Cal, and therefore am required to consider Stanford my mortal enemy, at least one day a year when they play football in a rivalry that goes back to 1892. https://en.wikipedia.org/wiki/Big_Game_(American_football)

You paid ten dollars to download a pdf? When you click on the article it comes up in your browser for free. You don't need to download anything. I don't even see a way to download a pdf, let alone pay for one. I am confused. Anyway thanks for the link. I did go back and read section 9 of the SEP article and did not understand a word of it. It seems to be fairly advanced mathematical logic. But there does seem to be a completeness theorem for "general" models, which is apparently a technical term that lets you get a completeness theorem but has some drawbacks.

The SEP article doesn't have page numbers. Can you please point me to what you're referring to? https://plato.stanford.edu/entries/logic-higher-order/ Early on it says, "The situation changed somewhat when Henkin proved the Completeness Theorem for second-order logic with respect to so-called general models (§9). Now it became possible to use second-order logic in the same way as first order logic, if only one keeps in mind that the semantics is based on general models." Whatever a "general model" is. Maybe I'll read it later.

Yes, it does. I'm not sure what you mean by this. I linked and quoted a SEP article saying there is no completeness theorem for second-order logic. Do you have a reference to the contrary? I admit I'm no specialist on this subject. Or is "yes it does" referring to what your question relates to?

I'll take a run at this. I think the key issue is that there is no completeness theorem for second-order logic. In general the OP's post doesn't mention the key distinctions between first and second order logic. OP should give this a read, perhaps there is some insight to be had. https://plato.stanford.edu/entries/logic-higher-order/ In particular, note section 5.2: "We shall now see that there is no hope of a Completeness Theorem for second-order logic." Ok, that's a funny way of putting it but I know what you mean. Ok. But you are conflating syntax (axiom systems) and semantics (models). That is, the 180 degree theorem follows syntactically from the axioms of Euclidean geometry. That's a purely syntactic fact. And it's also the case that the theorem is true in any model of those axioms. That's a syntactic fact. Two different things. In first-order logic, a theorem is provable (syntax) if and only if it's true in every model of the axioms (semantics). This is Gödel's completeness theorem. But beware, as the SEP article I linked above indicates, there is no completeness theorem for second-order logic. I confess to not knowing much about the fine points of second-order logic, but I suspect the OP's questions relate to this fact, that second order logic does not have a completeness theorem. The terminology is that an axiom system with a unique model up to isomorphism is categorical. And one that has non-isomorphic models is non-categorical. OP, please read this. https://en.wikipedia.org/wiki/Categorical_theory Now the first-order theory of Peano arithmetic most definitely does have nonstandard models. For example the hyperintegers of the hyperreal numbers are a nonstandard model of PA. They include the usual finite natural numbers 0, 1, 2, 3, ... as well as all the infinite ones. However the second-order theory of PA is categorical, and perhaps that's what you are referring to as Dedekind's theorem. So this is an example of where you need to distinguish between first and second order theories. There ARE non-categorical models of first-order PA; but none of second-order PA. This I believe is the crux of your concern, if I'm understanding your post. Yes, but note that the real numbers axiomatized as an infinite, complete ordered field is categorical. That's because completeness is a second-order property. It quantifies over subsets of the real numbers, not just individuals. That is, the least upper bound property says that all nonempty subsets of the reals that are bounded above have a least upper bound. Since we have quantified over subsets, that's a second-order property, and then it turns out that the second-order reals are categorical. This is where my knowledge ends. Since there's no completeness theorem for second-order logic, there must be some axiom system in which something is a theorem that's not true in all models, or true in all models but not a theorem. I don't know any specific examples or anything more about it. EDIT -- this isn't right, see below You're asking if there are undecidable statements in the categorical theory of second-order PA. This I do not know. A related question of interest is whether there are "natural" statements of arithematic that are undecidable (presumably in first-order PA). Harvey Friedman has been searching for examples of such. Here's an overview. https://plus.maths.org/content/picking-holes-mathematics Well that's what I know about it, hope something in here was helpful. The key is that first-order theories are never categorical and second-order theories sometimes are. But there's no completeness theorem for second-order theories, leading to all the aspects of this that I don't know, and to the question you're asking. ps -- Aha, a clue. I was perusing the Wiki article on categorical theories at https://en.wikipedia.org/wiki/Categorical_theory and it says: Every categorical theory is complete. However, the converse does not hold So this tells us that IF a theory is categorical (has only one model up to isomorphism) THEN every theorem provable from the axioms is true in every model of those axioms. But if the converse fails, then there is an axiomatic system with a statement that is true in all models of the axioms, but that is not provable from the axioms. Well, "today I learned!" pps -- The Wiki footnote leads to Carl Mummert's answer to this math.SE question: https://math.stackexchange.com/questions/933466/difference-between-completeness-and-categoricity/933632#933632 Mummert gives a concrete example of a comple, noncategorical theory. This doesn't seem to be the same thing as what we are looking for, a categorical, non-complete theory. There may well be clues on this page though. Reading a little more of Mummert's response, he's using "complete" in a different sense, that every statement is either provable or its negation is. That's not the same kind of completeness as in Gödel's completeness theorem, which says that a statement is provable if and only if it's true in every model. These two subtly different meanings of complete are yet another confusing aspect of all of this. So I was wrong about what it means for a theory to be categorical but not complete. Nevermind that part. These are deep waters and I got in a little over my head. From the Wiki article on completeness: https://en.wikipedia.org/wiki/Complete_theory

Ok YOU are the OP. My apologies to everyone for that confusion on my part. NSA is a bit of a niche area of study, it won't do you much good in general. It's an alternate model of the real numbers. You'd be better off studying the standard model first; that is, the real numbers as taught in high school, and their formalization in the undergrad math curriculum as in a course on Real Analysis. I'll leave it under the rock two meters due north of the old oak tree in the park. Make sure you're not followed. There are many math resources on the Web, you should just consult some of them at whatever level of math you feel comfortable with. If you want to see mathematicians discussing things, you should become a daily reader of https://math.stackexchange.com/. And if you want to see actual professional mathematicians talking among themselves, read https://mathoverflow.net/. I periodically check out this site to see if there are any new posts in the Math section. Your OP was new so I looked at it. Unless one is a determinist, in which case my looking at the thread was determined at the moment of the Big Bang. You can't discount that possibility and nobody knows for sure whether it's true. Math forum junkie going back to sci.math on Usenet. I haven't read the thread. I didn't understand your OP and didn't have an interest. Recently when the subject of NSA came up, I jumped in, because I have an interest in the subject and took the trouble to understand the basics some time ago. Could be worse, you might have been one of those bitcoin holders hodlers who forgot their wallet password and lost millions.

If this is your interest, you'd be better served by a book on elementary mathematical logic, not a treatise on the hyperreals. The least upper bound property is the defining characteristic of the real numbers. It's important. The hyperreals are deficient in that respect. "Can't be all bad." I'll try not to blush with false modesty LOL. I thought you were the OP but if not my apologies. Like I say I haven't followed this thread and should probably go back into my hidey hole. You drew me out mentioning the hyperreals, which I spent some time looking into a while back. I greatly recommend Terence Tao's brilliant blog post on nonprinciple ultrafilters as voting systems. That's the article that snapped all this into focus for me. https://terrytao.wordpress.com/2007/06/25/ultrafilters-nonstandard-analysis-and-epsilon-management/ See also https://terrytao.wordpress.com/2012/04/02/a-cheap-version-of-nonstandard-analysis/ https://terrytao.wordpress.com/2010/11/27/nonstandard-analysis-as-a-completion-of-standard-analysis/