User talk:JRSpriggs/Archive 6

This is the file "User talk:JRSpriggs/Archive 6" which archives User talk:JRSpriggs.

Healthcare

What purpose does healthcare serve in society? Is it a commodity? Or a public service?--WaltCip (talk) 14:24, 16 March 2017 (UTC)

Your question is ill defined. 'Society' is not a living thing and thus has no purposes.

Healthcare is best considered a service, like barbering. One who voluntarily receives and benefits from the service should be responsible for ensuring that his provider is paid.

I assume that your question was prompted by reading User:JRSpriggs#Guaranteed failure: Obamacare and Trumpcare. Obamacare and Trumpcare both assume that every person has a right that someone else pay for his healthcare (at least in part or as a last resort). That is an unjustified and indeed unworkable assumption.

Communicable diseases are a special case since one person may benefit from protecting another from the disease. JRSpriggs (talk) 20:26, 16 March 2017 (UTC)

Soapbox

I don't know if you have seen WP:NOTSOAPBOX. All the best: Rich Farmbrough, 21:02, 21 March 2017 (UTC).

This was asked and answered at MFD.--WaltCip (talk) 15:17, 22 March 2017 (UTC)

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Lord Roem ~ (talk) 21:34, 7 May 2017 (UTC)

oops

oops (now did what I intended to do, which is remove the 2 double spaces in the article)

Appreciate your proud user page :) Ethanbas ^(talk) 03:32, 21 June 2017 (UTC)

I am glad that you did not intend to replace blanks by "Latino". So it was some kind of mistake rather than deliberate vandalism. Good. JRSpriggs (talk) 03:53, 21 June 2017 (UTC)

Seeking support for proposed Wikiproject Quantum Mechanics

Hi, thanks for all your contributions! I'm reaching out to members of the community who might be interested in a Wikiproject dedicated to QM. The goal is to create articles which can be read and understood by laypersons but that also thoroughly present the technical details of the subject. As it stands now, too many QM articles feature ledes filled with jargon and lack introduction or overview sections.

I hope you'll support the proposal and contribute as a member when the time comes.

Thanks

Informata ob Iniquitatum (talk) 22:17, 12 September 2017 (UTC)

Here's the proposal page. Forgot that last time :-| Informata ob Iniquitatum (talk) 03:02, 13 September 2017 (UTC)

Thank you for the invitation. However, I am not an expert in quantum mechanics. So I doubt that I could help you in this project. JRSpriggs (talk) 04:09, 13 September 2017 (UTC)

Re: Friedmann Density, CMBR Hz Planck's Constant c^2

((((x GHz) * (3.71295774e-28 (m^2))) / Planck's constant) * (1 kg)) = 299792458^2

((((160.390033967 GHz) * (3.71295774e-28 (m^2))) / Planck's constant) * (1 kg)) = c^2

((((CMBR Hz) * (Friedmann Mass Density (m^2))) / Planck's constant) * (1 kg)) = c^2

1/(160.390033967 GHz * 3.71295774e-28 kg / Planck's constant / c^2 )^0.5 = 1 m^2/s (square meter per second)

https://en.wikipedia.org/wiki/Viscosity#Kinematic_viscosity

The kinematic viscosity "momentum diffusivity" is the ratio of the dynamic viscosity μ to the density of the fluid ρ. It is usually denoted by the Greek letter nu (ν) and has units m^2/s — Preceding unsigned comment added by 2602:306:833C:5980:EF:7508:2AC:611B (talk) 01:28, 20 September 2017 (UTC)

Why are you addressing this to me? What is your point? Please explain in words rather than in out-of-context equations!

Do you want me to do something, if so what? Did I do something wrong, if so what? JRSpriggs (talk) 14:38, 20 September 2017 (UTC)

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Comment at Gender Dysphoria

Hello, JRSpriggs. I hardly know how to respond to your comment at the Talk page discussion about uncertainties regarding incidence or prevalence of Gender dysphoria. The best possible face to put on it was that your comment was ignorant and tone deaf. Beyond the fact that the article Talk page is for discussions about how to improve the article and not a forum for expounding your personal opinions on Dysphoria, to claim as you did that the reason that the number of transgender individuals has "radically increased" (the discussion never said that) is because it is "now considered acceptable or even fashionable" to be transgender, is offensive. Your edit summary just doubles down on this. Transgender individuals suffer from societal discrimination of various types, culminating in high rates of rape, assault, and murder, and as a result have an attempted suicide rate approaching 50%.

If this is a misinterpretation of your thoughts about this, you're welcome to correct me below. If you have thoughts on how to improve the Gender dysphoria article, that would be an appropriate subject to add at the article talk page, but please keep your general opinions about the article topic to yourself. Cordially, Mathglot (talk) 06:01, 25 December 2017 (UTC)

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The Arbitration Committee has authorised discretionary sanctions to be used for pages regarding all edits about, and all pages related to, any gender-related dispute or controversy and people associated with the same, all broadly construed, a topic which you have edited. The Committee's decision is here.

Discretionary sanctions is a system of conduct regulation designed to minimize disruption to controversial topics. This means uninvolved administrators can impose sanctions for edits relating to the topic that do not adhere to the purpose of Wikipedia, our standards of behavior, or relevant policies. Administrators may impose sanctions such as editing restrictions, bans, or blocks. This message is to notify you that sanctions are authorised for the topic you are editing. Before continuing to edit this topic, please familiarise yourself with the discretionary sanctions system. Don't hesitate to contact me or another editor if you have any questions.

EvergreenFir (talk) 06:20, 25 December 2017 (UTC)

Dear Mathglot, I am sorry that I offended you and others. I came to the talk page because a question was raised at the math project talk page about how to resolve an apparent inconsistency in the data. I see no way to resolve the contradiction except two, either (1) reject one or more of the assumptions or (2) that the number of people identified as transgender has increased by a large amount in a short time. It could have increased quickly if there was a large change in the incentives to "come out" versus "stay in the closet". That is all I was trying to say. I was not aware of the high rate of suicide or that discrimination is continuing even now (as opposed to recently, like maybe two years ago). Again, you have my apology. JRSpriggs (talk) 20:48, 25 December 2017 (UTC)

Actually, the strong reactions of Mathglot and EvergreenFir are evidence supporting the notion that society's view of transgender people has changed a lot very recently. JRSpriggs (talk) 21:10, 25 December 2017 (UTC)

Hi, JRSpriggs: Thanks. You probably were just trying to say something mathematically valid, that came out wrong. I see from your tone that no ill will was intended. Appreciate your comment, we're good. (I wouldn't make any assumptions about societal change, though, on the basis of two data points, but your sentiment is appreciated, nonetheless. ) Happy editing! Mathglot (talk) 10:40, 28 December 2017 (UTC)

shutdown

This edit actually removes important information, which appears to have consensus on talk - specifically the "first shutdown" material. Additionally the claim that "Liberals have criticized the Democrats decision to withdraw the shutdown without a firm assurance and no guarantee on DACA" is misleading. Some did, some did not. As stated the claim is incorrect and POV.Volunteer Marek (talk) 07:40, 24 January 2018 (UTC)

And btw, my removal challenges the "Liberals have criticized the Democrats decision to withdraw the shutdown without a firm assurance and no guarantee on DACA" text, so, per discretionary sanctions, it should not be restored without consensus.Volunteer Marek (talk) 07:41, 24 January 2018 (UTC)

And removing the other info is also a DS violation. I see that you have been notified here of the sanctions]. Please self-revert.Volunteer Marek (talk) 07:45, 24 January 2018 (UTC)

JRSpriggs edits were entirely appropriate and VM continues to violate the page DS with creative interpretations of the Discretionary Sanctions that he seems to interpret as an allowable "challenge" for his reverts and as edit warring reverts for anyone else. It creates a very hostile atmosphere to always interpret other editors removals and additions as DS violations. VM, you need to stop bludgeoning, badgering and hounding other editors with specious claims of violations. --DHeyward (talk) 08:14, 24 January 2018 (UTC)

No. I challenged the "Liberals have blah blah blah" text. He restored it. It's clear cut DS violation. There's nothing "creative" about this "interpretation". It's pretty straight forward. It's restoring challenged content.

Likewise, in your self-revert you implicitly acknowledged you were violating a DS sanction by repeatedly changing the text about the first shutdown. You then snuck back later, after full protection expired and reinstated your edit (which has been challenged) with a misleading edit summary ("update" - it wasn't an update, it was a revert).

I am not hounding anyone. AFAIK JRSpriggs has not been heavily involved in American Politics disputes, which is why I've brought this up on his talk page rather than file a WP:AE report. As far as you're concerned, DHeyward, tell you what - you start observing the page restrictions and stop trying to start edit wars, and I'll stop "claiming" that you're committing violations.Volunteer Marek (talk) 08:23, 24 January 2018 (UTC)

@DHeyward and Volunteer Marek: If you believe the DS have been violated by any editor(s) you should open a report at WP:AE, otherwise I'm going to have to ask you both to stop casting aspersions about editor behavior or intentions. — Coffee // have a ☕️ // beans // 08:27, 24 January 2018 (UTC)

As I say above, since JRSpriggs hasn't been involved in a lot of AP2 disputes, I'm assuming good faith and assuming that he wasn't aware of the "consensus required" provision (it can be sort of hard to miss unless you're looking out for it). Hence, my request that he simply self-revert - rather going to WP:AE.Volunteer Marek (talk) 08:35, 24 January 2018 (UTC)

To Volunteer Marek: I agree with DHeyward. I changed "Liberals" to "Some liberals". That is as far as I will go. JRSpriggs (talk) 10:24, 25 January 2018 (UTC)

JRSpriggs - this isn't about tweaking the wording. It's about you violating a discretionary sanction, despite prior notification. And then second notification right here. You restored challenged content, violating the "consensus required" provision. I gave you more than 24 hours to address this but if you refuse to self-revert then I will have no choice but to file a WP:AE report.Volunteer Marek (talk) 13:09, 25 January 2018 (UTC)

And you're ignoring the "first shutdown" text which you removed. Please restore that as well.Volunteer Marek (talk) 13:36, 25 January 2018 (UTC)

One last time - can you please self-revert and restore the sentence on it being the first general shutdown? [1]. Volunteer Marek (talk) 03:45, 26 January 2018 (UTC)

As you yourself said in another context, the lead is supposed to summarize the article so what is not in the body of the article should not be mentioned in the lead. What you want added to the lead, you should first place in the body of the article. I suggest that you could add a section at the end of the article on "history of shutdowns" and mention it there. But putting it in the lead attaches excessive importance to a historical accident.

In any case, I do not see why I should do your work for you. If you want to fix the article, fix it yourself. JRSpriggs (talk) 03:59, 26 January 2018 (UTC)

thankxs

Our article Goodstein's theorem claims, that this theorem - stating that every Goodstein sequence eventually terminates at 0 - is unprovable in Peano first order arithmetic. So (using Goedel numbers - I guess), Goodstein's theorem is expressible in the language of Peano first order arithmetic, so (as I guess) for every function F expressible in the language of Peano first order arithmetic, also the statement S(F) - stating that for every x the sequence (x, F(x), F(F(x)), F(F(F(x))), ...) terminates at 0 - is expressible in the language of Peano first order arithmetic, am I right? If I am, then I wonder how this statement - S(F) - is expressible in the language of Peano first order arithmetic. For example, let F be the function F(x)=2x, so how can the language of Peano first order arithmetic express the (false) statement S(F) - stating that for every x the sequence (x, 2(x), 2(2(x)), 2(2(2(x))), ...) terminates at 0? HOTmag (talk) 12:30, 25 February 2018 (UTC)

Using modular arithmetic and the Chinese remainder theorem, one can encode statements about finite sequences of natural numbers (of arbitrary length) by statements using only a fixed finite number of variables over the natural numbers. Then one just has to say that a_k has a certain relationship to a_k+1 for k in a certain range and constrain the first and last values of a. JRSpriggs (talk) 14:13, 25 February 2018 (UTC)

First off, would that need the use of Goedel numbers?

Anyway, if the required arithmetical statement is not too long (e.g. less than one hundred signs - I don't want to bother you too much), would you like to concretely construct it, e.g. for the example I've provided above, i.e. with the function F(x)=2x (or whatever F is), and with the statement S(F) - stating that for every x the sequence (x, 2(x), 2(2(x)), 2(2(2(x))), ...) terminates at 0? HOTmag (talk) 15:14, 25 February 2018 (UTC)

The technique I am describing is logically prior to Godel numbers, that is, it is a lower-level technique without which one cannot even begin to talk (in the formal language of PA) about Godel numbers or primitive recursive functions. However there are some similarities.

The formal language of Peano arithmetic (PA) uses only: variables over natural numbers, universal and existential quantifiers over those variables, equality, logical operations (and, or, implies, not, equivalence), the constant zero, the constant one, addition, and multiplication.

Define: x < y to mean there exists z such that x+1+z = y.

Define: z = x mod y to mean z < y and there exists w such that z+(w·y) = x.

It can be shown that there are arbitrarily long arithmetic sequences c+k·d for k = 0..n of natural numbers which are all relatively prime to each other.

Then the finite sequence a₀, a₁, ..., a_n can be encoded by choosing A, c, d for which a_k = A mod (c+k·d) for k = 0..n.

So the false formula which you want as an example would say: for any x, there are n, A, c, d such x = a₀ and 0 = a_n and for any k < n: a_k·2 = a_k+1.

OK? JRSpriggs (talk) 07:30, 26 February 2018 (UTC)

A genius trick (following the Chinese remainder theorem)! Thank you so much. Some questions:

1. Who invented that trick?

2. Are there mathematical disciplines making a significant use of that trick?

3. So, for every function F (no matter whether or not F is arithmetically expressible), and for every x, the sequence (x, F(x), F(F(x)), F(F(F(x))), ...) terminates at 0, if and only if:

$\forall x\exists (n,A,c,d)((x=A{\bmod {c}})\land (0=A{\bmod {(}}c+nd))\land \forall k((k<n)\to F(A{\bmod {(}}c+kd))=A{\bmod {(}}c+kd+d)))$.

Correct?

4. Are there mathematical disciplines making a significant use of the explicit formula mentioned above?

HOTmag (talk) 09:09, 26 February 2018 (UTC)

I do not know who invented it. I learned it so long ago that I have forgotten when or how. Perhaps it was in elementary school or junior high school.

It is used implicitly virtually throughout mathematical logic. All serious mathematical logicians use it like breathing, without thinking about it.

Notice that F must be arithmetic or it cannot be put into the formula and still have the formula in the language of PA.

And, of course, it is not limited to sequences which end in zero. For example, a slight change in your formula will make it into a definition of 2ⁿ. JRSpriggs (talk) 10:21, 26 February 2018 (UTC)

Elementary school? Junior high school? You're kidding...

I still wonder, who - the first mathematician to use this technique - was. Anyways, I guess it had never been used before Goedel invented his techniques for encoding formulas in natural numbers.

1. Do you think this technique is mentioned somewhere in Wikipedia? I'm eager to read more about its historical background.

2. Is it the case that for every function F and for every natural number x, the arithmetical formula mentioned above is true, if and only if there's a finite number n such that F₀(F₁(F₂(...F_n-1(F_n(x))...)))= 0 ?

HOTmag (talk) 10:38, 26 February 2018 (UTC)

According to the article on the Chinese remainder theorem, "The theorem was first discovered in the 3rd century AD by the Chinese mathematician Sunzi in Sunzi Suanjing.".

And frankly, this technique is a fairly obvious application of that theorem. So it was probably developed as soon as the need for it arose, i.e. right after Peano arithmetic was defined (circa 1889). JRSpriggs (talk) 01:54, 27 February 2018 (UTC)

By the way, it is easy to determine possible values for c and d. We can let d be the product of all the prime numbers less than or equal to n. Then if M is an upper bound on the values of the members of the sequence, we can let c = ceiling (M/d) · d + 1 . JRSpriggs (talk) 02:19, 27 February 2018 (UTC)

All right, c and d can easily be determined, but what about n, I asked you about? Must it be finite (so that F₀(F₁(F₂(...F_n-1(F_n(x))...)))= 0), if the long arithmetical formula ("For all x there exist n, A, c, d" etc.) is true? HOTmag (talk) 12:36, 27 February 2018 (UTC)

Yes. If the formula you gave above using A, c, d to determine a_k is true, then the sequence of a₀=x, a₁=F(x), ..., 0=a_n=F(...F(F(x))...) will exist of course. And conversely, if such an n exists, then we can determine d, c, and A (which last uses the Chinese remainder theorem) so that the formula is true. JRSpriggs (talk) 18:46, 27 February 2018 (UTC)

I still suspect, that if x is infinite (being a non-standard natural number), then any appropriate sequence may turn out to be infinite (at least in some cases) - even if the long arithmetical formula ("For all x there exist n, A, c, d" etc.) is true. In other words, I suspect that the Chinese remainder theorem may turn out to be false for some non-standard natural numbers. Correct? HOTmag (talk) 19:52, 27 February 2018 (UTC)

If I were you, I would not be worried about non-standard models. They are a waste of time for the most part.

Just how "non-standard" is your model? If it obeys all the axioms of Peano arithmetic including the axiom schema of induction, then it must also obey all the theorems as well, including the Chinese remainder theorem. But here I am assuming that all the sentences in your theory are "standard" sentences, i.e. they contain a truly finite number of symbols.

Although there are versions of logic which deal with infinite formulas (see infinitary logic). I am not aware of any which use sentences which might have Godel numbers which are non-standard. I doubt that it would be possible to say whether such sentences are true or false of a particular model or give them any meaning at all. JRSpriggs (talk) 00:33, 28 February 2018 (UTC)

Regarding your questions: My theory obeys all the axioms of Peano arithmetic, including the axiom schema of induction. I don't use Goedel numbers at all, so all of the sentences in my theory are finite. However, infinite (non-standard) natural numbers - are not excluded from the domain the variables in my theory run over (even though all sentences - which may talk about infinite natural numbers as well - are finite), so I'm still asking about what if the domain of the function F contains also infinite (non-standard) natural numbers: Are you sure there must still be a finite appropriate sequence then, assuming that the long arithmetical formula ("For all x there exist n, A, c, d" etc.) is true? HOTmag (talk) 04:51, 28 February 2018 (UTC)

Before I answer your new question, let me correct what I said before. The Chinese remainder theorem is not actually a single theorem (in first-order PA) rather it is a theorem schema because it cannot even be stated for an arbitrary number of moduli without using the trick I explained above (which would be circular reasoning). So there is a Chinese remainder theorem for two moduli, a Chinese remainder theorem for three moduli, a Chinese remainder theorem for four moduli, etc.. However in second order arithmetic, it could be proved as a single theorem. Thus in a non-standard model, you would not have to have the Chinese remainder theorem hold for n moduli if n is a fake natural number rather than an actual natural number.

To answer your new question: If that formula is true for n and n is fake, then the sequence a₀, a₁, ..., a_n it describes will have length n and thus not be actually finite. JRSpriggs (talk) 09:27, 28 February 2018 (UTC)

As for the Chinese remainder theorem, yes I had already noticed it couldn't be proven in first order PA because of the reason you're indicating now, so I'd wondered why you'd claimed the opposite.

Anyways, now I understand from your answer that the reason of the unprovability of Goodstein's theorem in first order PA (even though it's expressible in first order PA by the technique you've taught me), is not because there are some infinite natural numbers for which the appropriate Goodstein sequence needs infinite number of stages to terminate at zero, but rather because there are some infinite natural numbers for which no appropriate Goodstein sequence - even an infinite one - can terminate at zero. HOTmag (talk) 11:16, 28 February 2018 (UTC)

If I understand you rightly, the reason you are looking at non-standard models is: If Goodstein's theorem is unprovable in first order PA, then first order PA plus the negation of Goodstein's theorem is consistent. If it is consistent, then it has a model C. But since Goodstein's theorem is true of the standard model, that model C must be non-standard. Then as you said, in C there are some fake numbers whose sequences never terminate even after a fake number of steps. Now you want to turn that around and see how to construct such a C and thus verify that Goodstein's theorem is unprovable in first order PA. Right? JRSpriggs (talk) 06:35, 1 March 2018 (UTC)

I think the most promising way to do this (given that I have never studied it and have no references on how to do it), would be to start with the standard model and take an ultrapower using a non-principal ultrafilter over the natural numbers. This should give a non-standard model of first order PA into which the standard natural numbers are elementarily embedded. Then we need to make a cut in the fake (i.e. non-standard) natural numbers and discard the elements above the cut. There are two constraints we need to satisfy in choosing the cut: (1) the part below the cut must still be a model of PA; and (2) there must be a fake number x below the cut whose Goodstein sequence does not end until n is above the cut. These are very difficult conditions to satisfy with my current knowledge. JRSpriggs (talk) 08:40, 2 March 2018 (UTC)

I am guessing here, so please correct me if I am wrong. We can define the primitive recursive functions using the Chinese remainder theorem technique. But proving that the result f(x) of applying unary primitive recursive function f to x exists requires using j nested mathematical inductions upto x where j is the number of primitive recursions done in defining f. This suggests that the Ackerman function#Table of values may be the key to defining the cut because the Ackerman function A(m,n) is primitive recursive in n for each fixed m but dominates all primitive recursive functions when m is allowed to vary.

So pick and fix a fake natural number x and generate the ω-sequence A(m,x) for actual natural numbers m. Then put the cut just after that sequence. So a fake y will be in the reduced model if and only if there is an actual m such that y<A(m,x). JRSpriggs (talk) 22:15, 2 March 2018 (UTC)

Let me add a numerical example. Consider the sequence < 1, 2, 4, 8 > which is the beginning of the list of values of 2ⁿ. The best (smallest) arithmetic sequence of relatively prime natural numbers larger than the given sequence is < 5, 7, 9, 11 > which are the values of 5 + k·2 for k in 0..3. d = 2 is the only prime number less than n = 3. And c = d·n - 1 = 5 differs by one from a multiple of any prime which might divide the difference of two members of the arithmetic sequence. Grinding through the Chinese remainder theorem calculation, we get A = 2956. It satisfies: 1 = 2956 mod 5; 2 = 2956 mod 7; 4 = 2956 mod 9; and 8 = 2956 mod 11. The reason I did not choose larger numbers or a longer sequence is that A grows very rapidly as the numbers in the sequence increase and especially as the length of the sequence increases. But it is not hard to see that such d, c, and A must always exist. JRSpriggs (talk) 00:27, 15 March 2018 (UTC)

For the sequence < 1 2 4 8 16 >, the best arithmetic sequence of moduli is < 7 11 15 19 23 > with c = 7 and d = 4. Then it turns out that A = 40174 which is surprisingly small since it appeared that it could be as much as half a million. The product of the moduli is 504735. JRSpriggs (talk) 03:09, 16 March 2018 (UTC)

For < 1 2 4 8 16 32 >, the best moduli are < 7 13 19 25 31 37 > whose product is 49579075. I am not going to work out A because the calculation would be too tedious. JRSpriggs (talk) 02:30, 18 March 2018 (UTC)

1	2	4	8	16	32	64	128	256	512
2
2	3
3	4	5
5	7	9	11
7	11	15	19	23
7	13	19	25	31	37
7	19	31	43	55	67	79
11	29	47	65	83	101	119	137
17	47	77	107	137	167	197	227	257
11	71	131	191	251	311	371	431	491	551

n	c	d	A	Π
0	2	--	1	2
1	2	1	5	6
2	3	1	34	60
3	5	2	2956	3465
4	7	4	40174	504735
5	7	6	7175183	49579075
6	7	12	?	51611487235
7	11	18	?	113189163812705
8	17	30	?	?
9	11	60	?	?

JRSpriggs (talk) 01:16, 23 March 2018 (UTC)

ln(2)

Dear JRSpriggs,

Okay, so I've deleted the extra integral in integral representations, but would you mind if I put this integral in series representations?

Thank you for your time, BetaHolds.

@BetaHolds (talk · contribs): I have doubts about the value of those formulas because they 'represent' the natural logarithm of 2 in terms of π which is an even more transcendental number. Perhaps they would be more appropriate in one of the articles dealing with π.

However, in the interest of peace between us, I am willing to allow them to remain in the article provided that I am convinced that they are true. That is why I asked you to justify them. If you can prove one of them is true or give a link to a reliable source for it, then I will allow it to remain. JRSpriggs (talk) 21:47, 23 April 2018 (UTC)

I have now figured it out for myself. Your formulas are correct. So I will allow them to stay. In particular, you may revert your removal of the integral, if you wish to do so. JRSpriggs (talk) 17:55, 24 April 2018 (UTC)

Axiom of regularity and consistency of ZF without it

Hello, why did you revert my edit that clarified how the question of consistency of ZF without regularity is relevant to proving that there is no set of all sets? --Alexey Muranov (talk) 09:01, 18 July 2018 (UTC)

Currently the phrase "However, if the ZF axioms without regularity were already inconsistent, then adding regularity would not make them consistent." is out of place, because consistency of ZF was not a subject of the paragraph, and in the whole section only inconsistency of Naive set theory was mentioned. Moreover it is in any case obvious that adding axioms to an inconsistent theory will not make it consistent. So, now we have an obvious remark without any context that could justify it. I will restore my edit that is in line with the first part of the paragraph (discussion of a proof that there is no set of all sets). Removing the phrase altogether would also be fine IMO, but you opposed it previously. Otherwise the whole paragraph needs to be rewritten to make the obvious remarks on the question of consistency of ZF somehow relevant. --Alexey Muranov (talk) 09:21, 18 July 2018 (UTC)

My original edit summary explained why I put the sentence back in. It said "restore important (even if obvious to some) sentence". The fact that it was obvious to you and to me does not imply that it will be obvious to all readers. I expect that some readers of this section will be thinking something like << This section is about how to eliminate the Russell paradox. Apparently, we can just say that it is not allowed in our set theory. The axiom of regularity is how we say that. >>. Although I was not the author of that sentence, I appreciated that it was there to counter-act that mistaken thinking. That is why I put it back. Your attempt to change what I did (thinking that you know better what I want than I do) does not disabuse the reader of this error. JRSpriggs (talk) 17:17, 19 July 2018 (UTC)

I am sorry, but I think that what you expect about what the reader will think is not a good enough justification to add a remark that is irrelevant to the context (and thus is likely to confuse readers who will not think what you expect they will think). It confused me a bit, and I started to look how it could be relevant to what was said before. Russell's paradox appears in the Naive set theory, not in ZF, as is clearly stated in this section. It is even explained what remains of it in ZF (a theorem on the absence of the "universal set"). If ZF without regularity is inconsistent, then everything is a theorem, as I explained in my edit, and there is not much do discuss. --Alexey Muranov (talk) 22:25, 19 July 2018 (UTC)

It is now in a separate paragraph and explained more clearly, so please leave it alone. JRSpriggs (talk) 03:14, 20 July 2018 (UTC)

Dear JRSpriggs, to me the added paragraph still looks out of place: the possibility of inconsistency of ZF was not a topic of this section until that point and it seems to be irrelevant to the rest. It is already stated that (literal) Russell's paradox does not happen in ZF, so even if ZF is inconsistent, it is not "because" of Russell's paradox, end of story. It also seems that calling consequences "undesirable" is mathematically meaningless, unless there is some context that explains what is "desired." (Models with "undesirable properties" in the first paragraph make more sense to me: excluding "pathological" models means adding consequences.)

You wrote: "if ZF without regularity is extended by adding regularity to get ZF, then any contradiction (such as Russell's paradox) which followed from the original theory [...]". The mention of Russell's paradox in parentheses is rather confusing if not wrong: there is no literal Russell's paradox in ZF.

To make that added remark somehow on topic, I can only reword it thus: "There is no Russell's paradox in ZF with or without regularity, but if it were there without regularity, we would not be able to avoid it by adding regularity back." To me this qualifies as a moderately funny mathematical joke.

In general, this added remark seems to be on the way to making a list of all possible silly ideas a reader can come up with and warning against them all. As I keep saying, the question of consistency of ZF was not even mentioned before, so a hypothetical reader who, instead of reading what is written, starts making strange hypotheses about fixing hypothetical inconsistencies in ZF can be honestly left on their own, IMO.

I suggest removing the sentence which was the original cause of dispute and the added paragraph, and instead I'll add a paragraph along the lines of what you suggest, but about the Naive set theory instead of ZF, because this will be at least on topic (Russell's paradox), though somewhat redundant IMO. --Alexey Muranov (talk) 10:50, 20 July 2018 (UTC)

I edited it, and then edited again, because I realised that trying to follow your line of thought I got confused myself again, and that it is better to be clear that this remark is an aside.

Even if we assume that someone will imagine the possibility of avoiding Russell's paradox by forbidding the set of all sets, that person would not need to add axioms for this: the set of all sets is already forbidden by Rassell's paradox!

IMO you are making several unjustified assumptions: (1) that the reader is likely to be confused in that particular way, (2) that you suggested explanation will likely address the root cause of their confusion (otherwise it would be better to avoid giving any explanations and to let the reader think on their own), (3) that that reader will not be confused about some more important points (otherwise why to turn their attention, and also the attention of those who were not initially confused, to that particular point?). --Alexey Muranov (talk) 13:36, 20 July 2018 (UTC)

1. We do not know that Russell's paradox does not occur in ZF, merely that the easy derivation of it in naive set theory fails in ZF.

2. I think that we can safely say that a contradiction would be an undesireable consequence of any theory.

Again, please leave my paragraph alone! It is meaningful and true and relevant. JRSpriggs (talk) 04:49, 21 July 2018 (UTC)

What do you mean by Russell's paradox? I meant a certain derivation of contradiction in Naive set theory, which does not work in ZF because axioms are not the same. However, it can be "turned around" (as mentioned in the section we are discussing), in which case it becomes a proof that ZF excludes the existence of "set of all sets" -- as well as Naive set theory does. (This by itself cannot be called a "paradox" IMO.) --Alexey Muranov (talk) 11:52, 21 July 2018 (UTC)

Please do not insert your comments in the middle of my comments. It makes it difficult to read, especially to tell who wrote what.

If someone found a way to prove in ZF without regularity that the class of all sets was a set, then one could use the axiom of separation to show that the "Russell set" existed and get the paradox that way. This seems very unlikely since if it were possible, it probably would already have been done. But one cannot prove (in ZF) that it is impossible, unless it is possible. JRSpriggs (talk) 04:08, 22 July 2018 (UTC)

Sorry, i still fail to see which hypothetical derivations of contradiction you call "Russell's paradox." If someone derived a contradiction in ZF and used it to prove the existence of Russell's set, from which a contradiction can be derived, could that be called Russell's paradox in ZF, in your opinion? I hope not, otherwise it would seem like you can call "Russell's paradox" any derivation of contradiction in ZF. --Alexey Muranov (talk) 08:08, 22 July 2018 (UTC)

I clarified in the preceding paragraph that ZF without the axiom of regularity already prohibits the set of all sets, so your added paragraph seems to be clearly redundant now, please take a look. I still do not see how it is relevant: your explanation seemed to be that you expect a substantial part of readers to be mistaken in some particular way, and you suppose that you paragraph would be helpful, but I do not see what makes you think so. I am also fuzzy about the way you use the term "Russell's paradox" in relation to ZF. --Alexey Muranov (talk) 08:29, 22 July 2018 (UTC)

You said "... Russell's paradox yields a proof that there is no "set of all sets" using the axiom schema of separation ...". If this is how you show that Russell's paradox cannot occur, then you are begging the question. JRSpriggs (talk) 07:09, 23 July 2018 (UTC)

Excuse me, but you seem to have ignored what i was asking in my last comments. Regarding yours "If this is how you show that Russell's paradox cannot occur, [...]", it seems to be "vacuously true," because its premise is apparently false, though I still do not see what exactly you mean by Russell's paradox, and I hesitate to comment on the truth value of sentences that i do not quite understand (like "this sentence is false")...

Could you please (1) clarify which exactly hypothetical derivations of contradiction in ZF you would call Russell's paradox (see my question with more details above), (2) confirm whether you agree that your added paragraph is now clearly redundant and somewhat off topic. --Alexey Muranov (talk) 18:27, 23 July 2018 (UTC)

I already said "If someone found a way to prove in ZF without regularity that the class of all sets was a set, then one could use the axiom of separation to show that the "Russell set" existed and get the paradox that way.". I hesitate to give a more precise answer because I think that classifying derivations of a contradiction as one kind or another is a fool's errand. That is why I put the reference to Russell's paradox in parentheses. I think that given a derivation of a contradiction, one could either manipulate it to make it a instance of Russell's paradox or manipulate it to make it not an instance. Here, by an instance, I mean that the first obviously absurd step (other than possibly the set of all sets) is the set of all sets not containing themselves.

No, I do not agree. JRSpriggs (talk) 04:57, 24 July 2018 (UTC)

(1) "[...] I think that classifying derivations of a contradiction as one kind or another is a fool's errand." -- isn't this what you need to accomplish to be understood when referring to "Russell's paradox" in ZF, especially since the literal Russell's paradox does not occur? So far your use of "Russell's paradox" in parentheses looks confusing to me. I would rather take care of this, than of a hypothetical confusion of a reader about something that was never claimed or implied.

(2) Could you explain how it is relevant, given that ZF without regularity already prohibits the set of all set? (How is it relevant, an example of how one cannot construct a consistent theory if ZF without regularity is inconsistent? There are a lot of ways how one cannot do whatever you like.) --Alexey Muranov (talk) 20:17, 24 July 2018 (UTC)

It seems to me that you are conflating $\vdash \lnot ((R\in R)\iff \lnot (R\in R))$ with $\lnot \vdash ((R\in R)\iff \lnot (R\in R))$. JRSpriggs (talk) 21:30, 24 July 2018 (UTC)

I do not.

You do not do much (besides your very first reply) to explain how your added paragraph is relevant, and sometimes you avoid answering direct questions (my question in (2) just above, for example). Don't you agree that if someone claims that something is relevant and has to be included, the burden of justification should ultimately be on him? (I hardly see how it could be possible in principle to justify that something is irrelevant, if my efforts above do not count.) --Alexey Muranov (talk) 14:57, 25 July 2018 (UTC)

The first occurrence of the section on Russell's paradox in axiom of regularity was at [2]. It was clearly motivated by showing that the axiom of regularity cannot rescue ZF from contradictions, contrary to a common myth. You say "ZF without the axiom of regularity already prohibits such a universal set". But what do you mean by "prohibits"? Do you mean that one can prove that the universal set does not exist or do you mean that one cannot prove that it does exist? All that you justify is the former. My paragraph is clarifying that regularity does not help us with the latter, a point which you do not make in your paragraph. JRSpriggs (talk) 21:35, 26 July 2018 (UTC)

Thanks for the link to the original version, it makes much more sense IMO than the paragraph we are discussing. The current paragraph is included without any context and thus seemingly off-topic, but apparently it tries to address a "common misconception" (which was the motivation for the origin of this section and which was clearly stated in the initial version) without providing any evidence that such a misconception is common (I asked you about some such justification above) and without even mentioning it. If such a misconception exists, it is indeed interesting to mention it in the article providing some references (like what the original version did, mentioning Everything and More (book)), because it is the existence of the misconception that is worth discussion, not so much some trivial proof that it is a misconception. In the current version of the section, at least after my edit, it is clear that adding an axiom to be able to prove that there is no set of all sets is already useless (as it can already be proved), no need to look further.

IMO a solution would be to restore a big part of the original version of the section as a paragraph in the current section. I would even vote to revert the section to the initial version completely, as besides the discussion of the "common misconception" there is basically no relation between Regularity and Russell's paradox.

Note: I wrote the above under the assumption that the mentioned misconception is indeed common, which i know nothing about. If it is not, then it should be called "curious misconception," or "Regularity and Russell's paradox in popular culture," or something else, and in such case no explanation would be needed i think.

To be consistent, I used the word "prohibits" in the same sense in which it was used in the previous sentence (by someone else). --Alexey Muranov (talk) 07:57, 28 July 2018 (UTC)

Cardinality confusion

In regards to your revert edit in the cardinality article, their are two interpretations of one to one, one (each member of set) to one (for each or some members of set) or one to one for all of each set once only for each member. Additionally bijection is claimed by different authors as being determined by one to one, or by either one of two versions of a ≤ test and a ≥ test thus 'bi…'

Victor Kosko (talk) 00:07, 27 July 2018 (UTC)

Equivalents to Axiom of Choice

Whoops! You are absolutely right, thanks for catching that; it was meant to say "there is either an injection or a surjection from any nonempty A to B" but the equivalence of that to the first sentence is trivial and probably not worth adding separately. 2601:42:0:4C76:B8C1:CEBA:5ACF:59A2 (talk) 08:55, 9 September 2018 (UTC)

Multilinearity

@JRSpriggs: Hello! I noticed you made an edit to the article on tensors (which has since been re-worked by Purgy Purgatorio), where you removed "multilinear" because, as you said, all tensors are "multilinear": https://en.wikipedia.org/w/index.php?title=Tensor&type=revision&diff=860649481&oldid=860618369&diffmode=source

It seems to me that "multilinear" is an ambiguous term, due to the "multi" part of the word. I could be wrong about this, but it seems that, usually, "multi" implies "more than one", and sometimes it implies "an arbitrary whole number". In an article such as "Multilinear map", this ambiguity leads to apparent contradictions in the article.

If all tensors are unambiguously multilinear, then (geometric) scalars should be multilinear, and it would seem that "Multilinear map" should be reworked to clearly explain this general definition and include the trivial example of (geometric) scalars. If this "multilinear" terminology is actually ambiguous, then it seems that that should be explained both in the "Multilinear map" article and the "tensors" article.

Zeroparallax (talk) 05:45, 25 September 2018 (UTC)

Anime suggestion

Hey, I'm the guy that made the tiny edit in the natural deduction section of the propositional calculus article. I noticed that you enjoy watching anime, have you watched Death Note? Maybe this article will interest you since it applies math to a (pretty good) anime series: https://www.gwern.net/Death-Note-Anonymity

Yes, I watched Death Note. Thanks for the link. I read a large part of it. Above the foolish way Light went about his murders is the lack of morality in the quantity and quality of the victims he picked. JRSpriggs (talk) 21:07, 30 October 2018 (UTC)

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Thank you for contributing to MacCullagh ellipsoid

Cocorrector (talk) 11:14, 20 November 2018 (UTC)

natural logarithm

Hi,

Whoops. You are right! I did miss the section for x as an integer. But I must disagree that my submission is without encyclopedic merit. It is indeed less efficient than the prior example but that is because it more fundamental. The prior example is more efficient because it has been simplified - like how 2x + 5x = 7x. 7x is more 'efficient' but it loses the fact that it came from 2x and 5x. My equation demonstrates how the above equations work much more efficiently (IMO) than what is being shown on the page. But it's real value is when used in proofs where one wants to deconstruct the natural logarithm (for integers) into a more common and simpler (albeit less efficient) form. Examples would include writing the Stieltjes constants in terms of summations and many other applications in number theory. — Preceding unsigned comment added by 98.125.226.43 (talk) 04:18, 29 November 2018 (UTC)

Kinetic energy in fluid flow

Hi! I noticed your reversion re KE in fluid flow at Kinetic energy. Please give additional details re the apparent lack of sense of my addition to article on article talk page.--109.166.133.121 (talk) 15:31, 22 January 2019 (UTC)

sets

emPty sets.

--Amanbir Singh Grewal,NHH (talk) 16:33, 21 March 2019 (UTC)

To Amanbir Singh Grewal: Please do not edit articles on subjects, such as set theory, which you do not understand. JRSpriggs (talk) 07:47, 22 March 2019 (UTC)

Following the technique you described in this thread, for encoding - in a first order arithmetical language - second order statements about sets of natural numbers:

Let $R(a,b)$ be a first order arithmetical binary relation. Can you use the technique mentioned above, to encode in a first order arithmetical language, the following second order statement:

"For every natural $x$, and for every set $S$ (whether finite or not) of natural numbers of which every number $y$ - both satisfies $R(x,y)$ - and either isn't greater than any number in $S$ or is an (additive) successor of a number in $S$, there exists a natural $z$ such that every $y\in S$ satisfies $R(y,z)$". 185.46.78.19 (talk) 15:14, 19 May 2019 (UTC)

What do you mean by "set $S$ (whether finite or not) of successors greater than $c$"?

There are an uncountable number of sets of natural numbers. So clearly, they cannot all be systematically encoded by the natural numbers (a countable set) using any encoding scheme whatever. JRSpriggs (talk) 00:27, 20 May 2019 (UTC)

S is a set of successors, if and only if both, S contains natural numbers only, and every number in S either isn't greater than any number in S or is an (additive) successor of a number in S. Hence, S contains (additive) successors only (if they are not equal to the smallest number in S). Please notice that the number of sets of successors is countable. 185.46.78.36 (talk) 07:31, 20 May 2019 (UTC)

Please do not alter your comments after I have replied to them since this creates a false impression of the meaning of my reply. If you no longer agree with what you wrote, you may strike out all or part of your comment with <s>strike out</s>.

In the context of a natural number n, the word "successor" means n+1. You appear to be using the word to mean something else. Please explain what it is. JRSpriggs (talk) 00:39, 21 May 2019 (UTC)

By "(additive) successor" of n, I really mean n+1 (That's why I added the word "additive", because I suspected that if I didn't add it you would think that by "successor" I meant something else. Surprisingly and unfortunately, even after I added the word "additive", you still thought that by "successor" I meant something else...)

Hence, [for every countable model of natural numbers] we can very easily order countably, all sets of successors, stage by stage; [e.g. for the standard model of natural numbers] as follows: For every natural number n, we build the n-th stage, by listing every set of successors which, either only contains natural numbers not larger than n, or only contains all natural numbers not smaller than n.185.46.78.55 (talk) 22:26, 21 May 2019 (UTC)

If I understand you correctly, you are talking about sets S of the form [a, b] = {n|a≤n≤b} or of the form [a, ∞) = {n|a≤n}. Is that right? I do not know why you used such a complicated description for such simple sets (just intervals within the natural numbers). JRSpriggs (talk) 03:08, 22 May 2019 (UTC)

Correct. A set is a set of successors if and only if it's an interval within the natural numbers.

As for your comment against (LOL) "the complicated description": Please notice that my original phrase ("set of successors") has less words than yours ("interval within the natural numbers") has, while the longer phrase I used following your question about what "set of successors" means ("A set is a set of successors, if and only if both, the set contains natural numbers only, and every number in the set either isn't greater than any number in the set or is an (additive) successor of a number in the set"), is the shortest verbal description, whereas your formal one (using the parameters a, b, along with the constant ∞) is not verbal. I wanted to describe it verbally. 185.46.78.36 (talk) 05:18, 22 May 2019 (UTC)

So your question is can

$\forall x\forall a\forall b(\forall y(a\leq y\leq b\to R(x,y))\to \exists z\forall y(a\leq y\leq b\to R(y,z)))\,\land \,\forall x\forall a(\forall y(a\leq y\to R(x,y))\to \exists z\forall y(a\leq y\to R(y,z)))$

be written as a first order formula (i.e. arithmetical formula)? Right? Well it already is. JRSpriggs (talk) 06:56, 22 May 2019 (UTC)

Please notice, that I've added some new additions in brackets, to my old comment about the countability of the number of sets of successors.

As for your question "Right?": No. My verbal second order description - that should be described in a first order arithmetical formula, is not equivalent to your first order arithmetical conjunction, because mine - that speaks about "every set $S$ (whether finite or not) of natural numbers" et cetera - doesn't exclude sets of the form {y|y is a finite natural number not smaller than a}, which are ignored (for every non-standard model of natural numbers) by your first order arithmetical conjunction - including by its second part that only speaks about sets of the form {y|y is a natural number not smaller than a}, unless your first order arithmetical formula assumes that every natural number is finite - but my verbal second order description assumes nothing with regard to that (Please notice that when I wrote "correct" in the beginning of my previous response, I was referring to the continuation: "A set is a set of successors if and only if it's an interval within the natural numbers"). 185.46.78.9 (talk) 08:01, 22 May 2019 (UTC)

You have continued to edit your old comments in spite of my request that you not do so. Now you are re-defining "natural number" to include things which are not natural numbers. I have no interest in non-standard models. So I am going to end this conversation. Please do not edit my talk page any more. If you do so anyway, I will not reply further but simply revert you. JRSpriggs (talk) 07:25, 23 May 2019 (UTC)

Friedmann equations

Hey, I was wondering why you undid my revision? The section does not state that rho or rho_c are being evaluated at the current time (and they don't have to be, generally omega changes with time). And even if rho and rho_c were evaluated today, then H0 should be defined and omega should be written omega0 instead (this is more consistent with the following equation on this page). P.espenshade (talk) 16:29, 13 June 2019 (UTC)

Makes sense. I changed Friedmann equations back. Sorry about that. JRSpriggs (talk) 00:06, 14 June 2019 (UTC)

"Practical Cryptography" listed at Redirects for discussion

An editor has asked for a discussion to address the redirect Practical Cryptography. Since you had some involvement with the Practical Cryptography redirect, you might want to participate in the redirect discussion if you wish to do so. LFaraone 13:51, 25 October 2019 (UTC)

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Methods of computing square roots

I've answered your principal objections by restoring/deleting sections of the article, as per your request on the talk page. I had no idea anyone would object, given several editors' comments about verbosity. I'm going to continue editing, sparingly. If you have any other suggestions, please forward them on the talk page. Thanks for your efforts. Sbalfour (talk) 16:11, 21 November 2019 (UTC)

Continuum Hypothesis, History

Is there a particular reason why my edit was reverted? I attempted to improve the clarity by rewording the confusing phrase “the unprovability of the nonexistence of an intermediate-sized set”. Phrases like “the A of the B of the C” are very hard for many readers to comprehend, so I reworded it to the form “the B of the C is A”. I would argue that my revision actually made the paragraph more clear, and, in addition, avoids the passive voice (“Paul Cohen proved the …” instead of “The … was proven by Paul Cohen”). I also made the revision to remove the incorrectly-used en dashes. By convention, en dashes are used mainly for numerical ranges, and em dashes would have been the appropriate punctuation mark (regardless, dashes shouldn’t’ve even been used at all, as there’s nothing intrusive or dramatic here). Anyway, I would like to hear the reasoning behind why you rejected my edit, and any suggestions you might have on how it can be improved. Thanks. — Preceding unsigned comment added by Chharvey (talk • contribs) 00:05, 18 February 2020 (UTC)

Your version of the last sentence of the History section is "In 1963, Paul Cohen proved the second half of the independence of the continuum hypothesis: the nonexistence of an intermediate-sized set could also not be proved.".

I think that most readers will see this as "... Paul Cohen proved the second half of the independence of the continuum hypothesis, the nonexistence of an intermediate-sized set ...". Which is incorrect. And then they will see "... the nonexistence of an intermediate-sized set could also not be proved.", and think "What? Is it proved or not proved?".

Replacing the colon with ", that" might help, but I do not think that it would be clearer than the older version. JRSpriggs (talk) 19:07, 18 February 2020 (UTC)

Question at Reference desk/Mathematics

Can you help out with this question about GCH and cardinal exponentiation? Thanks.  --Lambiam 19:59, 6 June 2020 (UTC)

Thanks for the reply, and sorry for my late response now... Dan Gluck (talk) 07:03, 25 June 2020 (UTC)

Kinetic energy: formatting

Hello, thanks for looking over the text I edited at kinetic energy here. I've looked over the text again and I think it would be most consistent to write dx in this context. In particular, d would not be bold, which is consistent with the following text, and italic, in keeping with the usual convention on Wikipedia to write dx rather than dx. What do you think?

Eelworm (talk) 23:21, 8 October 2020 (UTC)

Ordinary variables are italicized. Vector-valued variables are bolded (see "F" in the same sentence). I would treat "dt" the same as "t" and "dx" the same as "x" rather than regarding "d" as an operator which deserves separate treatment. JRSpriggs (talk) 03:18, 10 October 2020 (UTC)

Thanks! No objection to that convention, but we should use one convention consistently. Are you going to change the rest of the text? Eelworm (talk) 08:00, 10 October 2020 (UTC)

No. I do not have the time or energy to do that. Revert me, if you must. JRSpriggs (talk) 19:44, 10 October 2020 (UTC)

Free vs bound variables in binary connectives

Could you explain why the definition of a bound variable in a formula with a binary connective would include those variables that are bound only on one side of the connective? What is the purpose of having variables be both free and bound? A reference would also be good. Thanks! Undsoweiter (talk) 23:36, 3 November 2020 (UTC)

From free and bound variables, "However, it could be confusing to use the same letter again elsewhere in some compound proposition. That is, free variables become bound, and then in a sense retire from being available as stand-in values for other values in the creation of formulae.". Once a variable has "retired", we do not want it to become active again. The only reason we define bound variables is to forbid the use of a rule of inference in a case where the substitution of a free variable would become bound. See first-order logic#Rules of inference. If "and" was used instead of "or" in the definition of bound, then one could side-step that restriction by simply combining the clause with another clause not containing the variable. JRSpriggs (talk) 01:17, 5 November 2020 (UTC)

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Nomination of Former Muslims United for deletion

A discussion is taking place as to whether the article Former Muslims United is suitable for inclusion in Wikipedia according to Wikipedia's policies and guidelines or whether it should be deleted.

The article will be discussed at Wikipedia:Articles for deletion/Former Muslims United (2nd nomination) until a consensus is reached, and anyone, including you, is welcome to contribute to the discussion. The nomination will explain the policies and guidelines which are of concern. The discussion focuses on high-quality evidence and our policies and guidelines.

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Hemiauchenia (talk) 05:54, 21 January 2021 (UTC)

A request to contribute to the discussion

Good morning,

please express your opinion in the thread "relativistic mass" in the discussion with DVdm:

"The legitimacy of removing entries by DVdm in a topic of relativistic mass"

thank you in advance, Best Regards, RodriguesVector.

To RodriguesVector (talk · contribs). Regarding User talk:DVdm#The legitimacy of removing entries by DVdm in a topic of relativistic mass, I have confidence in DVdm's understanding of and respect for the rules of Wikipedia. I suggest that you follow his advice. I regret that I do no have the time to give a more detailed response. JRSpriggs (talk) 05:00, 24 May 2021 (UTC)

@JRSpriggs. Thanks a lot for your time. I have looked at your entries and I can see that you solve problems quickly and efficiently.

Unfortunately, my dispute with the DVdm is escalating. There is a reasonable supposition that the DVdm editor is confusing its idea of Wikipedia with Wikipedia's principles. It looks as if he is only reading the headings of the rules of procedure without looking inside. Similarly, he judged my entry only by the title of the article, which he disregarded and considered to be "fringle theory".

If you had any more time, I'd like to ask you to intervene on my talk page.

Thank you for your time RodriguesVector (talk) 08:20, 24 May 2021 (UTC)

Absolute values in logarithms

Can you provide a standard reference where the unnecessary absolute value in real logarithms is used?.

Let me provide an example of how abs(x) can make a mess, consider the integral of tan(x) from 1 to pi, this integral does not converge in reals, but if you use the absolute value you find a real value. This is not correct as the integral is not real-valued. I argued this with the other guy and he just ignored me. Taking x>0 is the only thing needed to have real-valued logarithms. Putting absolute value is unnecessary and generates confusion when the logarithm must be extended to x<0 in order to have correct non-real results. — Preceding unsigned comment added by David phys davalos (talk • contribs) 20:09, 10 July 2021 (UTC)

Your counter-example is incorrect. See Fundamental theorem of calculus. In order to be able to apply it, you would need the tangent function to be defined and continuous in an interval including 1 and pi. However, tangent is not defined at pi/2 which is in that interval. JRSpriggs (talk) 06:13, 11 July 2021 (UTC)

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Talk:Ordinal notation major collapse

I collapsed your recent additions to the Talk page. Your search for fundamental sequences is not a contribution to the issue of making the Ordinal notation page better, since it is obviously OR. You should delete the material. I expect I will do so in a month or so anyway. See WP:TALK, in particular: "Delete. It is common to simply delete gibberish, test edits, harmful or prohibited material (as described above), and comments or discussion clearly about the article's subject (as opposed to the treatment of the subject in the article)." (emphasis added). 96.5.122.4 (talk) 18:23, 20 June 2022 (UTC)

Done. I hope you do not mind that I took out your surrounding text urging this action. JRSpriggs (talk) 20:24, 20 June 2022 (UTC)

I do not mind. The only positive value it would have is that perhaps a future editor might similarly get carried away without knowing any better. But it's a problem I've seen in many math Talk pages. It's a distraction when the editor knows his stuff, and it's a serious nuisance when he doesn't. 128.91.40.241 (talk) 17:40, 23 June 2022 (UTC)

Please register a user-id. See Wikipedia:Why create an account?. How can I be sure to whom I am talking when your IP address changes? JRSpriggs (talk) 20:22, 23 June 2022 (UTC)