taeto

We would get here in a finite amount of time from any previous point in time, there is nothing problematic about that

Who? Romeo D. Matshaba, the author?

I answer the OP's question, only to get a downvote. Not that I care a lot. But seriously, in a forum is it really supposed to be better rewarded not to answer questions? Possibly not as familiar as you are. But the transfer principle immediately allows me to ascertain that the limit p in (2) is identically equal to 0. I did not say that (2) does not make sense. What I said was that it only makes sense if p is 0.

It is de Broglie, not De Broglie. You say that you understand the explanation in the paper. Do you know how the de Broglie wavelength of a particle is defined at all? These are correct. But if you check again with the paper, it says something completely different.

Correct. Gravity is a concept. The gravitational constant is a quantity, and it is not measured in meters.

I am not sure that saying that all the mistakes are just "innocent typos" gives evidence in favor of the article likely having been reviewed. And we can throw in that the author's name is duplicate. No journal would allow that. The first line of the abstract states that we can "view gravity as the de Broglie wavelength of quantum mechanics". A few problems here. Gravity is not a physical quantity, and certainly not a length. And QM does not possess a de Broglie wavelength. The second to third line of the abstract is just nonsense as well. But you want equations. Equation (1) is nonsense. The LHS is a number, and the RHS is either undefined or a binary operator, which is anyone's guess so long as it is not defined. Thereafter, the quaternions do not have the properties that are claimed here. There is no "familiar Clifford space" either. The concept that is familiar is Clifford algebra. Equation (2) makes no sense unless p is identically zero. In the subsequent displayed equation, the "p_i" is not even defined. In the next displayed equation, there is no referent to the symbol "n". Right after that, there is division by zero; N=1/p. After that everything is mostly word salad and nosensical expressions. But tell me one single thing, somewhere in this paper, that you are convinced that it is actually correct, please?

It is unlikely that the version that you link to has been peer reviewed, seeing how it is full of mistakes of all kinds.

I suspect you are supposed to see that G is abelian. That certainly helps for showing normality.

Where are you stuck, showing that H is a subgroup or showing that H is normal?

You never have to show this. It is an axiom that if X is a any set, then P(X) is a set.

So you are only interested in the congruency class modulo 9?

How often does it happen that you actually have a value like that? I am not a violent person. But I suggest that if you come across a landlord who tells you that the rent per month in US$ amounts to 9517452……, then you are absolutely in the right to punch them on the nose, using up to medium strength.

Yes, thank you. Should not try this late evening after an afternoon of lectures. There is an \(x \in X\) which is not an element of \(Y.\) Which is the meaning of \(Y\) is a proper subset of \(X\).

No, because any function from \(X=\{1,2\}\) to \(Y=\{1\}\) maps both \(1\) and \(2\) to \(1\), so is not an injection. The condition is that if \(a\) and \(b\) are different, then \(f(a)\) and \(f(b)\) are also different. You already got a perfectly precise from wtf anyway, as I just noticed. And your reply was that you "cannot work with it". You are obviously not serious, so I am out of here.

We certainly can. Such as: "A set \(X\) is infinite if there exists a proper subset \(Y\) of \(X\) and an injective function \(f : X \to Y\)." Which means that if \(a\) is an element of \(X\), then \(f(a)\) is an element of \(Y\), and if \(a\) and \(b\) are two different elements of \(X\), then \(f(a)\) and \(f(b)\) are two different elements of \(Y\). And there is at least one \(y \in Y\) such that \(y\) is not an element of \(X\). As an example, the set \(X = \mathbb{Z}\) of all integers is infinite by this definition, because the set \(Y=2\mathbb{Z}\) of all even numbers is a proper subset of \(X\), and \(f(n)=2n\) defines an injective function from \(X\) to \(Y\).

Because some sets are finite, and some are not finite. In a context of using mathematical language, I would be justified to say that "infinity" is just a symbol \(\infty\), which is a kind of letter. And letters are neither sets nor are they infinite. It is used to denote limits and improper integrals, among other things. In the particular contexts when \(\infty\) is used as the name of an actual set, it is not required to be an infinite set. It is grammatically correct, if it means that each of the natural numbers is finite. The properties of each natural number do not depend on how many there are of them. I think that was not the context in which the question was asked. I believe the question concerned sets in general.

Can you make a finite set out of the minutes on a clock, since it does not seem to have an end, and it just moves around in cycles? I am on the safe side here, because I am just the one saying that there is no such thing as a particular concept of "infinity" in mathematics. No way i can get confused about something that is not there. But I see where strange is coming from as well, because apart from mathematics itself, it is not uncommon to refer to "infinity" as that aspect of math that deals with some kind of theory of infinite structures. It is difficult to set precise borders between theories that do and theories that don't. If you are only interested in arithmetic with natural numbers, then every mathematical object that you will ever consider will be a finite object, namely a natural number. And yet, the thing which is "arithmetic with natural numbers" is the prime example of something in real life which consists of infinitely many other things, namely all the infinitely many natural numbers. That makes it natural for observers of mathematical practice to suggest that "infinity" is an actual thing in mathematics. Just like the set \(\{ \ldots , -4,-3,-2,-1,0 \}\), which has end \(0\)?

I don't know of it being defined at all. Do you know what "well defined" means? Usually it is defined by its properties. You could decide to assign a value to it, but what the value is would not matter. That doesn't make any sense. If you mean in set theory, then the set of natural numbers exists. If you do not mean in set theory, then you should not address the notion of a "set". The set is infinite, and it does not contain as elements anything that is infinite. It contains only natural numbers, and they are all finite, by definition. What do you mean by "infinity"?

Look, notice how every time we discover a prehistoric fossil, we always tend to ask why that organism happened to develop its long neck, or its ability for flight, its huge size, massive fangs, or any other anatomical features. Usually the answer goes along the lines of saying that it just happened to be that adaptation which made its survival and procreation within its current environment most likely. Compare with a species trying to advance by transmitting radio waves and laser beams out into space, or listening for same from remote sources. What is that going to do? As a method for evolutionary progress it seems totally doomed. No advanced extraterrestrial life form would survive for long by using such a survival strategy. Surely they are just like trilobites, happily feeding along the bottom of their respective oceans.

I have no doubt that he can do just that. I just want to remark that one thing which seems confusing about your idea is that a common definition of "finite" says that a set is finite if and only if the number of elements in the set is a natural number. How do you explain to a person who is aware of this definition how one may think that some natural numbers are infinite?

To one answer that you got in that thread you replied: "I got some concrete answers this time pointing out my error." You were lying then, or you are lying now?

Maybe you could say that it is a kind of discontinuity. I.e. \(\lim_{n\to \infty} 2^n \neq 2^{\lim_{n\to \aleph_0} n}\). Expressed like that, it seems suspect that the LHS means the limit of a series of numbers, whereas the RHS appears to be a limit of sets, where the \(n\) has to be understood as something like \(n = \{0,1,2,\ldots,n-1\}.\) For the LHS we would usually write \(\lim_{n\to \infty} 2^n = \infty\) to mean that the series is divergent, i.e. it does not have a proper limit, but it increases without bound in some regular manner. But that is only a symbolic notation, there is no inherent meaning, no mathematical object to attach, to the \(\infty\) symbol that appears on the RHS of that expression. For the RHS, if we force the sequence \(n\) to be considered as a sequence of sets, then we have no choice other than \(\lim_{n\to \aleph_0} n = \aleph_0,\) and the RHS becomes \( 2^{\aleph_0} = c\), i.e. the continuum. You end up being forced to interpret the LHS as a limit of sets: \(\lim_{n\to \aleph_0} 2^n\) which is a limit of sets all elements of which are finite sets. Even in the limit, there is no way that you somehow get to include infinite sets as elements. Maybe more philosophically, we can also argue that even though we write \(\lim_{n\to \infty}\) and we think "\(n\) approaches infinity", this is entirely different from writing \(\lim_{x\to 0}\) and think "\(x\) approaches 0." In the latter case \(x\) actually "gets infinitely closer" to 0 in the limit, closer than distance \(\varepsilon\) for every \(\varepsilon > 0\). Whereas in the expression \(\lim_{n\to \infty}\) every \(n\) is still infinitely far away from "\(\infty\)". Perhaps this is what causes the discontinuity; that there always remains an infinite gap between \(n\) and its limit?

Just keep rubbing it in, please. I did admit you were right, so what more do you want? And anyway, this is the internet, and I can stick with whichever story I want.

No, I am sure that a number such as \(\sqrt{2}\) is an irrational number which is not a rational number. Come to think of it, I am even convinced that no irrational number is at all a rational number, purely by definition of those concepts. What prompts you to ask such a question? Also you are interfering with an existing thread, seemingly trying to support some kind of private agenda. If you insist on not answering the questions that are asked to you, the moderators may ban you.

Nice catch +1. In the context of the thread it is not relevant though, because we are dealing with just positive distances.