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Rob Arthan

 Discussion on question by Arbuja: Is …

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Rob Arthan
Jan 20 10:26
@Malady: good observation, but those limits need not exist. I suppose you could take the measure of the discontinuity at a point to be $\infty$ in that case. However, I don't see any way to extend this to give a measure that somehow embraces all the points in the domain and is finite unless $f$ is hyper-discontinuous.
Rob Arthan
Jan 20 10:26
I don't know of any standard definition of such a measure. You can easily create one: just take $M(f) = 0$ if $f : X \to Y$ is continuous and $\infty$ otherwise. However, you are probably looking for something more subtle than that. You need to say more about how you want the measure to behave on functions that are neither continuous nor hyper-discontinuous.
 

 Discussion on question by Smiley1000:…

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Rob Arthan
Jan 16 11:20
I think math.stackexchange.com/questions/93409/… is highly relevant here. The ticked answer and comments thereunder show how to construct groups that could not serve as $X_0^\times$.
 

 Discussion on question by Nico Zaczko…

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Rob Arthan
Dec 31, 2024 19:02
@anankElpis: this post is about arithmetic. $0^0 = 1$ is the standard reading in arithmetic.
 

 Discussion on question by Davide Mara…

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Rob Arthan
Dec 4, 2024 20:55
As @julio_es_sui_glace points out, this is very close to (in fact of a special case of) the notion of semi-algebraic variety studied in real algebraic geometry. The existence of the cylindric algebraic decomposition show that your set will have finitely many components (not necessarily convex as otheer have pointed out). The proof will likely give more detailed information on the number of components.
 

 Discussion on question by urhie: defi…

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Rob Arthan
Oct 11, 2024 17:03
The text of your definition is impenetrable (i.e., I don't have a clue what you are talking about). Your inequality is certainly not the right way to go: try it out with the limit of $f(x) = x^2$ as $x$ tends to $0$. I think you should persevere and try to understand how the standard definition works and is used.
 

 Discussion on question by Elvis: Is $…

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Rob Arthan
May 24, 2024 00:33
@BrianTung: your string of beads suggests that the rational numbers form a discretely ordered set, but they don't: between any two rationals there is yet another rational. But they're gappy: there is no rational that separates the sets $\{q \mid q^2 < 2\}$ and $\{q \mid q^2 > 2\}$.
Rob Arthan
May 24, 2024 00:33
@BrianTung: if it helps with your intuitions about the Dedekind cut construction, I would say that a bounded above set of rationals is a structure that packs a lot of information.
Rob Arthan
May 24, 2024 00:33
In Landau's delightful book The Foundations of Analysis, he works through the constructions of $\Bbb{N}$, $\Bbb{Z}$, $\Bbb{Q}$, etc. and says that at each stage we "throw out" the earlier number system and replace it with the new one. This means that $\Bbb{N}$ really is a subset of $\Bbb{Z}$. you could could also replace the new things that represent the old things. Either way you end up with the inclusions that you want. So this is essentially an implementation detail and not worth worrying about.
 

 Discussion on question by 4-4: Which …

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Rob Arthan
Feb 24, 2024 14:28
This looks like an interesting question. But you will get downvotes and likely have the question closed if you don't provide some background: where does the problem come from and what have you tried?
 

 Discussion on question by AbysmalMath…

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Rob Arthan
Feb 11, 2024 21:46
This seems to me to be (a) opinion-based and (b) wrong: the most fundamental algebraic structure is one with a single monadic function and this is very widely studied and discussed (usually under the name of a set with a successor function).
 

 Discussion on question by Rob Arthan:…

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Rob Arthan
Dec 23, 2023 21:41
@XanderHenderson The instructions in that link didn't work for me using Safari on Mac OS.
Rob Arthan
Dec 22, 2023 23:47
Ah no it didn't. I'll have to work on that.
Rob Arthan
Dec 22, 2023 23:46
I see. I am not sure if I have figured about how to use MathJax. Does this render right: $a = b$.
Rob Arthan
Dec 22, 2023 23:34
@XanderHenderson: my recollection from attempts to engage with chat many years ago, is that it didn't seem to be as smooth as that. If pointers from questions to any associated chat chains appear reliably now, then that answers most of my concerns.
Rob Arthan
Dec 22, 2023 23:34
@XanderHenderson: thanks to you and everyone else for your comments. Chat feels less shared and public to me because I don't get to see chat chains unless I actively look for them, whereas I see all new questions and can filter for the ones on my favourite topics. Maybe there is a better way of knowing what is going on in the chat system that I am unaware of. The "step outside" thing is to do with the fact that viewers of the original question and comments may not be aware that there is an interesting chat chain relating to the question. Again, that may be just my ignorance.
Rob Arthan
Dec 22, 2023 23:34
Chat may be public, but I never go near it. @XanderHenderson your comment is helpful but doesn't actually address my question.
 

 Discussion on answer by Ethan Bolker:…

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Rob Arthan
Dec 1, 2023 16:32
It's your reaction that seems to me to be very odd. I wrote "how do you prove it" using "you" in its informal sense of "one". Why are you informing me of something that I now know (and never thought was the case) and which Ethan has confirmed?
Rob Arthan
Dec 1, 2023 16:32
@BradyGilg: my comment begins "you seem to assuming ..." and goes on to ask a question. As Ethan had no objection to my comment, I don't see why you do.
Rob Arthan
Dec 1, 2023 16:32
@BradyGilg: nothing in my comment implied what you are objecting to: I just asked a question. Why are you objecting to what was intended to be a constructive comment that has led to interesting input from Ethan and others.
Rob Arthan
Dec 1, 2023 16:32
You seem to be assuming that the digits in the decimal expansions of $2^x$ will be uniformly distributed. That sounds very plausible, but how do you prove it?
 

 Discussion on question by Agent Zebra…

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Rob Arthan
Jul 31, 2022 12:36
@Cornman: no, that is not what I was trying to say. Please read what I wrote.
Rob Arthan
Jul 31, 2022 12:36
@Cornman: Gödel showed that there are true statements in the theory of arithmetic that cannot be proved from any given set of axioms for arithmetic (satisfying certain technical conditions). Your ill-informed simplification of the incompleteness theorem doesn't bear on Lubin's excellent comment and doesn't help the OP or his child.
 

 Discussion on question by Kevin Njoko…

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Rob Arthan
Jul 29, 2022 02:58
To understand groups, you will do much better to study the examples of groups that appear in the textbooks. If you just pluck some binary operation on a set out of the air, it is highly unlikely to be the operation of a group. "Wanting to believe nothing will go wrong" is the antithesis of mathematical thinking. You should look at known examples first.
Rob Arthan
Jul 29, 2022 02:58
Please edit your question if you have made a mistake.
 

 Discussion on question by Taha Khabou…

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Rob Arthan
Jun 13, 2022 20:11
That isn't right. Your $g$ might not be invertible. And saying $f(x, y) = x + g(y)$ is not the same as saying $z = x + g(y)$: the first assertion is a statement about $x$ and $y$, while the second introduces a new variable $z$.
Rob Arthan
Jun 13, 2022 20:11
What do you mean by "are in function with each others already"? I $x$, $y$ and $z$ are just parameters to your problem, why should there be any relation between them?
Rob Arthan
Jun 13, 2022 20:11
But what information do you have that suggests any of the three variables is a function of the others?
Rob Arthan
Jun 13, 2022 20:11
After reading your reply to my comment, I now suspect that what you want to do is to decompose $F(x, y, z)$ as a composite of binary operations. I don't think there is a useful way of doing it. In any case, you really should tell us more about the actual context and what you are actually trying to achieve.
Rob Arthan
Jun 13, 2022 20:11
Why do you want to use a Taylor expansion? How can the introduction of an infinite sum reduce the number of computations? I think it would be helpful if you explained what you mean when you say you want to "factorize" the function.
 

 Discussion on question by Mike Arrh: …

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Rob Arthan
Jun 1, 2022 07:07
I'm sorry Mike, but statements like "According to this, heuristic arguments are the enemy of existence for a proof – most (?) true statements have one or the other, not both" are just froth.
 

 Discussion on question by 147pm: Any …

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Rob Arthan
Dec 3, 2021 20:21
Well Haskell's $n$-fold products are not iterated binary product as this transaction with Haskell shows :λ (1, 2, 3) == ((1, 2), 3) Couldn't match expected type ‘(Integer, Integer, Integer)’ with actual type ‘((Integer, Integer), Integer)’ In the second argument of ‘(==)’, namely ‘((1, 2), 3)’ In the expression: (1, 2, 3) == ((1, 2), 3)
Rob Arthan
Dec 3, 2021 20:21
What programming language are you using? If it's a typed functional programming language like ML or Haskell, then the $n$-ary products definitely aren't iterated binary products.
Rob Arthan
Dec 3, 2021 20:21
Thanks to @ArturoMagidin for his useful expansion on my previous comment. It is a (seemingly necessary) oddity in the usual set-theoretic foundations of mathematics that we first define the notion of ordered pairs $(a, b)$ and then use it to define the notion of functions and then use functions to supersede the original definition of ordered pairs to define $n$-tuples.
Rob Arthan
Dec 3, 2021 20:21
I don't think it is standard to consider an $n$-fold Cartesian product as an iteration of binary products (as doing so would introduce an unnecessary asymmetry).
 

 Discussion on question by Ten O'Four:…

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Rob Arthan
Sep 25, 2021 15:13
You have not stated what you are trying to prove. Your proof is not made clearer by introducing all those definitions - in mathematics we are often required to exceed Hrair's limi in understanding an argument, but we should try to stay within reasonable limits. Also, what is your "corollary" a corollary of. Frank Adams used to offer the advice that the best way learn how to write mathematics was to read a lot of (well-written) mathematics. I suggest you take that advice.
 

 Discussion on answer by Mark Saving: …

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Rob Arthan
Sep 1, 2021 08:40
That wasn't quite my point. Let me put it more directly: what do you mean by "we can still prove that $P$ cannot be neither true nor false"? What statement of IL denotes that? The double negation of LEM?
Rob Arthan
Sep 1, 2021 08:40
I think that part of your answer would be much clearer if you gave that formal statement. Note that by applying one negation, you have mapped into the world where intuitionistic and classical logic agree.
Rob Arthan
Sep 1, 2021 08:40
+1: nice answer, but I don't see what you mean by the penultimate paragraph. In the usual algebraic semantics for IL using Heyting algebras, there are typically lots of "middle values".
 

 Discussion on question by Graviton: H…

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Rob Arthan
Aug 21, 2021 18:50
What is the formal definition of your set $Z$. Your axioms don't justify defining a set by giving an infinite list of its elements.
Rob Arthan
Aug 21, 2021 18:50
@seayellow: there is a rich literature on the foundations of mathematics. Russell and Whitehead were pioneers in this subject and their Principia Mathematica is a seminal work, but other systems, most notably Zermelo-Fraenkel set theory, have been more widely used and studied
Rob Arthan
Aug 21, 2021 18:50
You need to give us much more information on the axioms of set theory that you are using. What is the definition of finiteness in this set theory? How do these axioms force an order on the way you define and prove things?
 

 Discussion on question by Istiak: Why…

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Rob Arthan
Jul 4, 2021 11:56
You haven't showed the definition of $f$ from your book (probably $f(x) = |1 - x|$).
 

 Discussion on question by PiKindOfGuy…

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Rob Arthan
Jun 23, 2021 10:56
This is not a sociology forum.
Rob Arthan
Jun 23, 2021 10:56
@JackGallagher: this has nothing to do with the law of the excluded middle. The OP is using heated words like "gibberish", not me. Your observations about the word "or" in English are wrong.
Rob Arthan
Jun 23, 2021 10:56
Mathematics has to deal with all kind of situations, including the ones that you put aside as "gibberish". Please don't ask wind-up questions on MSE.
 

 Discussion on question by Tyma Gaidas…

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Rob Arthan
May 8, 2021 14:57
Your edits do not address any of my concerns: your notation in your proposed definition of $P(x)$ still makes no sense and you are still talking about the minimal polynomials of real numbers that may be transcendental.
Rob Arthan
May 8, 2021 14:57
The notation in your defining equation for $P(x)$ makes no sense. Also, note that most computer algebra systems work on the (very bad) garbage-in/garbage-out principal, so if you ask for the minimal polynomial of a transcendental number you won't get a useful result.
 

 Discussion on question by Nickotine: …

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Rob Arthan
Mar 18, 2021 22:28
You have two questions here. The second one (about MathJax) does not belong on MSE, try math.meta.stackexchange.com or tex.stackexchange.com. (And don't be afraid of maths: it won't eat and it can often help you eat $\ddot{\smile}$.)
 

 Discussion on question by Jim: Why ar…

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Rob Arthan
Mar 8, 2021 23:07
@user4894: I disagree with your algebra professor. If you become really famous, then your name becomes absorbed into the vocabulary.