wtf

2 hours ago, studiot said:

 

So what is the additive inverse of the complex number z = (a +ib)  , given that a or b or both could themselves be either positive or negative ?

 

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.

11 minutes ago, studiot said:

So what is the additive inverse of the complex number z = (a +ib)  , given that a or b or both could themselves be either positive or negative ?

-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

 

10 hours ago, studiot said:

Actually there is another 'explanation'.

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? 

1 hour ago, ahmet said:

technical means , the methods used in one, of branch of science, profession or art. 

 

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.

32 minutes ago, ahmet said:

x2=−1 ( the existence of −1−−−√

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?

2 hours ago, wtf said:

yN=c1c1+c2+c3…..+cN−1y1+c2c1+c2+c3…..+cN−1y2+…..+cN−1c1+c2+c3…..+cN−1yN−1

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

On 3/12/2021 at 11:47 PM, Anamitra Palit said:

Σi(ciαi)=0

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? 

On 3/12/2021 at 11:47 PM, Anamitra Palit said:

Now each alpha_i=e+y_i belongs to V-W.

 

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.

On 3/12/2021 at 11:47 PM, Anamitra Palit said:

Σi=Ni=1ci=0

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.

 

On 3/12/2021 at 11:47 PM, Anamitra Palit said:

yN=c1c1+c2+c3…..+cN−1y1+c2c1+c2+c3…..+cN−1y2+…..+cN−1c1+c2+c3…..+cN−1yN−1

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.

 

1 hour ago, John Cuthber said:

Root (1)  = -1

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.

2 minutes ago, studiot said:

I'm sure there are plenty more but that will do.

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.

9 hours ago, studiot said:

As a californian, don't you have the verb to foal for horses ?

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.

On 2/12/2021 at 2:09 AM, studiot said:

But I didn't include the information that 5 of these sheep about to lamb.

 

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

34 minutes ago, IvoryEbony said:

& why is that? I mean, information is power. Would you share power with someone who may be irrational?

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.

 

12 hours ago, Tristan L said:

Since the reader knows that "witchcraft" wouldn't make much sense in the context, don't they come up with the thought to read again more carefully so as to check what actually stands written there?

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.

6 hours ago, Tristan L said:

lest you become conspiracy-theorists.

"I'm not a conspiracy theorist. I'm a conspiracy analyst." -- Gore Vidal

 

6 hours ago, Tristan L said:

However, all that has little to do with the topic at hand,

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.

44 minutes ago, studiot said:

I support Wikipedia with a few dollars a year and thought I would like to support the Standford project with a few dollars more.

 

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)

8 hours ago, studiot said:

I suppose that depends upon which presentation you download did you download the HTML version ?

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. 

1 hour ago, studiot said:

Actually page 45 of the stanford link you provided

(I have already thanked you for the link, paid my $10 and downloaded the pdf )

says rather more, particularly on page 45 et seq.

As always the devil is in the detail.

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. 

4 hours ago, Tristan L said:

23 hours ago, wtf said:

I suspect the OP's questions relate to this fact, that second order logic does not have a completeness theorem.

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? 

 

8 hours ago, joigus said:

Maybe @wtf has something interesting to say about it.

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."

 

13 hours ago, Tristan L said:

To me, axiom-systems seem to basically be ownships (properties). For instance, the group-axiom-system is basically the ownship of being an ordered pair (G,∗) such that G is a set and ∗ is a function from G×G to G such that ∗ is associative and has an identity element and each member of G has an inverse element with regard to ∗ .

Ok, that's a funny way of putting it but I know what you mean. 

 

 

13 hours ago, Tristan L said:

Just as the axiom-system itself is an ownship, so are what are called “propositions in the language/speech of the system” actually properties. For instance, when we say: “The proposition that the sum of the inner angles of a triangle is always 180° follows from the Euclidean axioms“, we actually mean that for every structure E, if E has the Euclidean axiom-system as a property, then E has the property that every triangle in it has an inner-angle-sum of 180°.

 

 

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.

 

 

13 hours ago, Tristan L said:

Some axiom-systems, such as the group-axiom-system and the field-axiom-system, are had by several structures of which not all are isomorphic to each other. In other words, such axiom-systems have at least two models which aren’t isomorphic to each other. Let’s call these axiom-systems “not-characterizing”. Others, such as the Dedekind-Peano-axiom-system (called “DP” henceforth) , are only had by one structure up to isomorphy – they have a model, and all their models are isomorphic to each other. Let’s call these axiom-systems “characterizing”. The only model of DP up to isomorphy (and indeed up to unique isomorphy), th.i. the only entity, up to (unique) isomorphy, which has DP as a property, is the structure of the natural numbers. The German mathematician Richard Dedekind showed this in 1888 with his Theorem of Isomorphy.

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.

 

 

13 hours ago, Tristan L said:

Now, there are two ways in which a ‘proposition’ P (in reality a property, see above) can be undecidable (neither provable nor disprovable) in an axiom-system AS . One way is that AS has several non-isomorphic models of which some have P and others don’t. For instance, being unendly (infinite) isn’t decidable from the field-axioms since there are finite (endly) fields as well as unendly ones. This underkind of undecidability is, I think, obviously not very interesting. The Continuum-Hypothesis (CH) is one example, for it’s true in some models of ZFC and false in others. Here, the lack of our knowledge stems from the fact (deedsake) that ZFC doesn’t contain all information about the set-universe to start with.

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. 

 

13 hours ago, Tristan L said:

The second way in which P  can be undecidable in AS is that it either holds in all models of AS or its negation holds in all models of AS , and yet we still can derive neither one from AS because our logical (witcrafty) tools are too weak. This reflects a true unableness on our part to get info out of AS which nevertheless is there. This is always the case if AS is characterizing (has only one model up to isomorphy). Such an axiom-system contains all info about its model, so undecidability in it means inability to get info which is nonetheless there. So while undecidableness in ZFC need not be interesting, undecidableness in DP always is. Since characterizing axiom-systems have just one model up to isomorphy, what we call “propositions in their speech” – in truth ownships – can actually be regarded as propositions, namely the propositions resulting from predicating those properties of the system’s unique model.

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

 

 

13 hours ago, Tristan L said:

Now, from Gödel’s Incompleteness Theorems, we know that there are undecidable propositions in DP, that is, not every statement about the naturals can be shown or disproven. However, my question is this: Is there an individual proposition about the naturals of which we know that we can neither prove it nor disprove it in DP?

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

Quote

In mathematical logic, a theory is complete if, for every closed formula in the theory's language, that formula or its negation is demonstrable. Recursively axiomatizable first-order theories that are consistent and rich enough to allow general mathematical reasoning to be formulated cannot be complete, as demonstrated by Gödel's first incompleteness theorem.

This sense of complete is distinct from the notion of a complete logic, which asserts that for every theory that can be formulated in the logic, all semantically valid statements are provable theorems (for an appropriate sense of "semantically valid"). Gödel's completeness theorem is about this latter kind of completeness.

 

3 hours ago, slomobile said:

That interaction was exactly what I was looking for.  I appreciate the further external references.  Non-Standard Analysis is still above my head, but I will be drinking it in over the next few weeks.  That was part of the plan, to observe a mathematical conversation I do not understand.  Then systematically research, and record all the foundational concepts I need to add to my repertoire to achieve some functional level of understanding.  It is a step change in my understanding.  In general, step changes are revelatory when examining complex dynamic systems, which is why I chose this strange method.

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.

 

 

3 hours ago, slomobile said:

In order to help me know when I have achieved "some functional level of understanding" I ask that you try to come up with some test for me.  It can be just a single question and answer pair that I should be able to answer after understanding what you both just presented.  Please send it to me via the message system on this forum with the word "TEST" in the subject line.  I will not have it opened until after I have done what feels like enough research.  At which point, I will have someone else open it and test me.

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/. 

 

 

3 hours ago, slomobile said:

How you came to this thread, was an interest in the mention of NSA, but what prompted you to open a thread titled "Notation Study"?

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.

 

 

3 hours ago, slomobile said:

It was obviously your choice, not someone else.

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.

 

 

3 hours ago, slomobile said:

Is your interest in this thread professional, idle speculation, academic, or other?

 

Math forum junkie going back to sci.math on Usenet.

 

 

3 hours ago, slomobile said:

Tell me about the difference between what you expected when you entered this thread, versus what you actually discovered here.

 

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.

 

3 hours ago, slomobile said:

Sorry it took me a while to respond to your excellent posts.  I mistyped my password and locked myself out of the forum for a few days.

Could be worse, you might have  been one of those bitcoin holders hodlers who forgot their wallet password and lost millions.

 

  

5 hours ago, Col Not Colin said:

Definition:  A Language is a set of symbols appropriate for the structure under consideration.

  A first order Language must possess the following symbols:   Connectives, Variables, Commas, Paranthesis.     

Connectives must include the following symbols    "Implies" ;   "if and only if" ;   "Not"   ;   "And"  ;   "Or

If this is your interest, you'd be better served by a book on elementary mathematical logic, not a treatise on the hyperreals. 

 

17 minutes ago, Col Not Colin said:

The Hyperreals are non-Archimedian.  Yes, I agree but ... who said it was and does it matter?  It's interesting but not a problem.

The least upper bound property is the defining characteristic of the real numbers. It's important. The hyperreals are deficient in that respect.  

17 minutes ago, Col Not Colin said:

Anyway, +1 for your post.  Anyone who knows a little NSA can't be all that bad

 "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/