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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?
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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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Mar
16
asked On last digit of 4 consecutive primes less than 10 apart
Mar
16
revised On proving $n = \sum_{d\mid n}\varphi(d)$
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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
16
revised On proving $n = \sum_{d\mid n}\varphi(d)$
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Mar
16
revised On proving $n = \sum_{d\mid n}\varphi(d)$
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Mar
16
revised On proving $n = \sum_{d\mid n}\varphi(d)$
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Mar
16
revised On proving $n = \sum_{d\mid n}\varphi(d)$
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Mar
16
revised On proving $n = \sum_{d\mid n}\varphi(d)$
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Mar
16
revised On proving $n = \sum_{d\mid n}\varphi(d)$
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Mar
16
asked On proving $n = \sum_{d\mid n}\varphi(d)$
Mar
14
revised Expectation expression
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Mar
14
revised Expectation expression
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Mar
14
answered Expectation expression
Mar
14
revised How to convert the ln part of this equation to log10?
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