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visits member for 3 years, 5 months
seen Nov 24 at 14:24

Apr
5
asked What are integrating factors, really?
Mar
19
awarded  Notable Question
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
accepted On proving $n = \sum_{d\mid n}\varphi(d)$
Mar
16
comment On last digit of 4 consecutive primes less than 10 apart
I see. Thanks for the clarifiication!
Mar
16
accepted On last digit of 4 consecutive primes less than 10 apart
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
answered Find the point on the y-axis which is equidistant from the points $(6, 2)$ and $ (2, 10)$.
Mar
16
revised Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
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Mar
16
revised Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
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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
accepted Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
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
revised Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
added 40 characters in body; edited title
Mar
16
asked Is there any advantage to the $a \equiv b\;\;(\mathrm{mod}\;c)$ notation?
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
revised On last digit of 4 consecutive primes less than 10 apart
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