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2h
answered Basic doubt on Stirling numbers of Second Type
2h
comment Is there a connected linearly ordered space of size $\leq 2^\omega$ that looks nowhere like the reals?
@ForeverMozart: A Suslin line has cellularity $\omega$, and it’s a general result that if $X$ is a LOTS, then $|X|\le 2^{c(X)}$.
3h
comment Understanding of the topology of pointwise convergence
@Jack: I figured out what happened to the previous version and corrected it.
3h
revised Understanding of the topology of pointwise convergence
Corrected really silly error.
3h
comment What is the remainder of $(14^{2010}+1) \div 6$?
@user103816: Not really, no.
17h
comment Are Homogenous countable complete metric spaces always discrete?
@Josh: You're very welcome.
1d
revised What is the limit of this sequence as n->infinity?
added 4 characters in body
1d
comment Is $\mathbb R^N$ an $C$-distinguished topological space?
Such spaces are usually called completely Hausdorff or functionally Hausdorff.
1d
answered Why is $^nC_r$ not equal to $ ^{n-k}C_{r-k}\times ^nC_k$?
1d
answered Are Homogenous countable complete metric spaces always discrete?
1d
comment What is the remainder of $(14^{2010}+1) \div 6$?
@user103816: I’m simply using the fact that $2\cdot2^{2k}\equiv2\pmod6$ and the very basic fact that if $a\equiv b\pmod{m}$ and $c\equiv d\pmod{m}$, then $a+c\equiv b+d\pmod{m}$.
2d
comment Understanding of the topology of pointwise convergence
@Jack: Sorry: I apparently missed the ping and only just noticed this. You’re right that there’s clearly something wrong; I’ll take a look at it later tonight.
2d
comment What is the remainder of $(14^{2010}+1) \div 6$?
@user103816: The induction hypothesis is that $2\cdot2^{2k}\equiv2\pmod6$. Therefore $\left(2\cdot2^{2k}\right)\cdot2^2\equiv2\cdot2^2\pmod6$. And $2\cdot2^2=8\equiv2\pmod6$. That induction proves the general result that $2\cdot2^{2k}\equiv2\pmod6$. I don’t know where you got the idea that $2^{2009}\equiv2\pmod6$: it’s not in anything that I’ve said. However, $2^{2009}=2\cdot2^{2008}=2\cdot2^{2\cdot1004}\equiv2\pmod6$ by that general result proved by induction on $k$.
Jul
30
comment What is the remainder of $(14^{2010}+1) \div 6$?
@user103816: You’ll have to be more explicit, I’m afraid. Every step there is explained either in the original answer or in my previous comment, so I can’t tell exactly what part(s) you’re not understanding.
Jul
29
comment proving a implication with regular expressions
@ohyaboon: You’re welcome.
Jul
29
revised proving a implication with regular expressions
added 46 characters in body
Jul
29
answered proving a implication with regular expressions
Jul
29
answered Generalization of Cantor Pairing function to triples and n-tuples
Jul
29
comment What is the remainder of $(14^{2010}+1) \div 6$?
@user103816: The first follows by induction on $k$: it’s true for $k=1$, and if it’s true for $k$, then $2\cdot(2^2)^{k+1}=\left(2\cdot(2^2)^k\right)\cdot2^2\equiv 2\cdot 2^2\equiv 2\pmod6$, so it’s true for $k+1$. It doesn’t say that $2^{2009}+1\equiv 2\cdot2+1\pmod6$; it says that $2\cdot2^{2009}+1\equiv 2\cdot2+1\pmod6$, which is true because $2^{2009}=2^{2\cdot 1004+1}\equiv 2\pmod6$.
Jul
29
comment Preference relations and the existence of extensions of functions representing them
@Kolmin: You’re welcome!