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Mar
17
comment Summation involving totient function: $\sum_{d\mid n} \varphi(d)=n$
@Sabyasachi: sorry, I had not finished with my comment. Also, on the contrary, I think your proof is extremely elegant. I'm making the case for inelegance here.
Mar
17
comment Summation involving totient function: $\sum_{d\mid n} \varphi(d)=n$
$\dots\;\;$ These means that $\nset$ can be partitioned according to the $d$ factor of these products, and the size of each partition is $\varphi(n/d)$. But don't get me wrong: such proofs are OK, as quick ways to convince oneself that something is true, but one needs to go beyond them to get at the essence of why something is true.
Mar
17
comment Summation involving totient function: $\sum_{d\mid n} \varphi(d)=n$
What sticks in my craw about this argument is that fractions are really inessential to the problem; they get brought in only because they come with a built-in, widely recognized "simplified to lowest terms" operation. IOW, fractions show up in this theorem only as, basically, a cute hack, and IMO they obscure what's really happening in the theorem, namely that (1) every $m \in \def\nset{\{1,\dots,n\}}\nset$ can be written uniquely as the product $du$ where $d\mid n$ and $\gcd(u, n/d) = 1$, and (2) every such product $du$ belongs to $\nset$. $\;\;\dots$
Mar
16
comment On last digit of 4 consecutive primes less than 10 apart
I see. Thanks for the clarifiication!
Mar
16
comment Artin 2nd Ed. Problem 12.5.3
As a follow-up to Ted Shifrin's comment, you may find this answer useful.
Mar
16
comment Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
@TonyK: thanks for your comment! Bill Dubuque's answer, however, convinced me that the traditional notation is more practical than I'd realized. (Now I see that my exposure to this notation is limited only to the simplest situations.)
Mar
16
comment Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
OK, I'm convinced. :)
Mar
16
comment Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
@rschwieb: AFAIC, "widely recognized" is the same as "conformance to convention." What's hyperbolic/melodramatic about that? Besides, every obsolete notation was "widely recognized" before it was replaced, so, as sole selling point, it strikes me as rather weak.
Mar
16
comment On last digit of 4 consecutive primes less than 10 apart
If I understand it correctly, your argument looks incomplete to me, because, as far as I can tell, it covers only the case where $x$ is congruent to $0$ modulo $3$, whereas $x$ could also be congruent to $1$ (e.g. $x=10$) or to $2$ (e.g. $x = 20$) modulo $3$.
Mar
16
comment On proving $n = \sum_{d\mid n}\varphi(d)$
@GregMartin: I suppose that the proof I wrote above closely parallels the one you mention, but somehow I find it more natural and useful to partition the set of integers $\{1,\dots,n\}$ than to partition the set of fractions $\{\frac{1}{n},\dots,\frac{n}{n}\}$, even though both procedures are basically equivalent.
Mar
14
comment How to convert the ln part of this equation to log10?
Yes, the equation should be as you have it, although I'd go ahead and multiply together the two constants you have there ($3101.420903$ and $2.302585093$). Also, the problem of getting $y$ when $x = n$, does not require converting to $\log_{10}$. All you need is to have an actual numeric (as opposed to symbolic) value for $x$, and then you just replace $x$ in the original equation with this actual value, take its natural logarithm, and off you go.
Mar
7
comment Natural uses for the co-product of sets?
@Rahul: and my contention is that the essential feature of the co-product is that it is a union of disjoint sets (tagged or not), and therefore the issue of tags is a red-herring. I understood Martin Brandenburg's $\mathrm{Humans} = \mathrm{Men} \sqcup \mathrm{Women}$ as saying that the sets $\mathrm{Men}$ and $\mathrm{Women}$ are indeed disjoint, and that their (standard set) union equals the set $\mathrm{Humans}$. Quibbling over how well these assertions represent the real world is like quibbling over whether cows are really spherical.
Mar
7
comment Natural uses for the co-product of sets?
@MJD: I think that tagging may be an inessential "implementation detail", an artifact of adopting a "synthetic", as opposed to an "analytic", point of view; I've written at length about this in an "Epilogue" to my original question.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: I see your point. Also Callus just pointed out an even bigger hole in my reasoning. I think that I can straighten out the mess in my head now. Thanks again.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@Callus: OK! I see! That's it. Thank you!
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: the book (whether correctly or not) writes "...use $b^8$ and $b^{12}$ as defining relators, and show that $b^4$ can be derived. Hence, by the preceding exercise [i.e. exercise 4], $b^8$ and $b^{12}$ are a set of defining relators." Are you saying the book is wrong, or are you saying that I am misinterpreting it? (I figure it's the latter, but if so I have to concede that I just don't get it, and that it's completely beyond my ken.) Anyway, thanks for your patience. (And likewise to Callus too.)
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@Callus: You may be right. I find the text very vague on the subject of picking relators. For one thing, it discusses the defining of relators in terms mappings from sets of symbols to group elements. If one uses the mapping $a \mapsto 2 \in \mathbb{Z}_4$, then $a^2 = 1$ (i.e. $2 + 2 = 0$ in $\mathbb{Z}_4$), so $a^2$ is a relator of the cyclic group of order $4$. And I can derive $a^4$ from it.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: the "then by exercise 4 blah blah" bit comes basically verbatim from the book. You can see the exact quotation in my post.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: the reasoning is exactly the same one that the authors give in the hint to exercise $5$(c). In a nutshell: 1) pick some relators $x, y, z...$; 2) show that $b^4$ can be derived from $x, y, z, ...$; 3) then, by Exercise 4, $x, y, z...$ are defining relators of $C_4$. Now, for my $x, y, z...$ I pick the single relator $b^2$, proceed to derive $b^4$ from it, and conclude that $\langle b; b^2 \rangle$ (i.e. $\langle a; a^2 \rangle$) is a presentation of $C_4$.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@Callus: you're right; I really wish I could delete the post, and maybe do it over...