Joe ST

Hey there Feeki, great to have you aboard. I hope we are as friendly as you hope we are Oh, and did you keep your ears?

Hey there Kerry, great to have you aboard. As Lyssie says, watch out for the admins, and dont accept the waffles They are delicious! Have some!

Hey there, those aren't the waffles you're looking for

Hey there JH great to have you aboard, that is some scary stuff there.

If there were space for widgets and such-like, and we had a table of facts stored somewhere, it would be fairly easy to implement either or both of those. The best place for the countdown clocks might actually be in the news item related to the release, and/or maybe a thread on the forums (or a page on the site) dedicated to all the release-dates. I hear that there might be a 'library' section coming to the site possibly in the future, so it might be a good idea to put the release dates on those pages too, prior to the arrival of the book. The facts thing could be a bookmarklet or another site which offers a javascript widget. I like the idea of a daily fact(oid)

Indeed, and I'm sorry for seeming like I was targeting you Kurk, it was more for Link's benefit

how would they, a 'new' and 'undiscovered' creators, get ahold of a Wii U devkit. If they're developing a 'serious' game, why would they even think about Nintendo? and how would a serious Mistborn game even work on a Wii-style console? If they've developed it to use motion control then the controls are either gimiky, or risk making multiplatform unbalanced due to the differences in interfacing, or they've spent 90% of their time trying to make a balanced motion control system rather than a good game (which I doubt, since Brandon seems to like them). And I concur, the LEGO franchise seem to be pretty good

indistinctive might not be a word *shrugs* indistinct or nondescript "The officer looked surprised. But, apperantly uncertain what else to do, he saluted Jason. Jason nodded, then turned to leave." should be apparently sunglassed might not be a word switchover might not be a word vacationing might not be a word "She's amnisiatic," should be amnesiatic His ponderings led him nowhere. can be pondering without loss of meaning Tis speech but: "Someone tried to get ahold of a Varavax host—but something went wrong." ahold can just be hold He reveled in his returned Sense for a wonderful, silent moment. Should probably be revelled Lanna stood dissatisfiedly a short distance away. Dissatisfied would do the job disclaimer: this is just using chrome's en_UK dictionary and google. that's all I found going through a copy-paste job I have. It's such a great little story too

Hey there Shardbinder, have you read any of his other works? They're fairly equal in terms of awesomeness (says the fanboy )

I like the quote

not just because of the distance of the sky?

Hey there Unclogged great to have you aboard, since you seem to have read all the series' there shouldn't be any spoilers if you visiting *any* of the boards below to discuss how awesome Brandon's worlds are

Indeed, a very merry day to miss Berry and to miss Hyde They're evidently the same persons. And of course, there are no girls on the internet

SHHH, dont reveal yourself...they'll only spike you harder

Rake >> Moiraine

It should also map a all elements of G into unique elements of H*K (and |G| = |H*K|)

Well you could, instead, define a homomorphism between G and H*K that uses the predefined functions to prove that it's a bijection and thus G and H*K are isomorphic. That's possibly the best (or fastest) way to get to the answer.

Yes that's basically what abstract algebra is, it's defining how different actions work on sets of arbitrary elements. It get's pretty abstract pretty darn fast. And I was talking to Eric about the zero stuff. He only knows Fields and VectorSpaces, he's a n00b Oh, and an Algebra (in particular, a Lie Algebra) is something else entirely. Wikipedia is pretty good at explaining Abstract Algebra, once you know the terminology. H*K is defined as {(h,k) for all h in H, k in K} then you can define 'similar' functions, f1' f2' g1' and g2': - f1': H*K -> H as f1'(h,k) = h - f2': H*K -> K as f2'(h,k) = k - g1': H -> H*K as g1'(h) = (h,e) - g2': K -> H*K as g2'(k) = (1,k) where 1 and e are Identity elements in H and K respectively Then you can show that these 'similar' functions act the same as the first ones: - f1' o g1' = Ih OR g1'(f1'(h,e)) = g1'(h) = (h,e) - f2' o g2' = Ik OR g2'(f2'(1,k)) = g2'(k) = (1,k) - f1' o g2' = 0' OR g2'(f1'(h,k)) = g2'(h) = (1,e) - f2' o g1' = 0' OR g1'(f2'(h,k)) = g1'(k) = (1,e) and - (f1' o g1') + (f2' o g2') = i' OR g1'(f1'(h,k)) + g2'(f2'(h,k)) = (h,e) + (1,k) = (h,k) Where 0' is the zero map and i' is the identity map over H*K. I can't remember the correct words to use though, since it's been a year or so since I did this stuff. Basically these new 'similar' functions are enough to let you prove that G and H*K are similar, aka isomorphic. I'd suggest to not take anything I've said as rigorous proof, but I'm fairly sure that this is the way one should approach such a problem.

Technical niggle, you misuse the term zero element where you mean identity element. The zero element is the one that sends every element to itsself. It gets confusing because the identity in the (R,+) group of addition over R, is the zero in (R,*) group of multiplication over R. So you're right in saying it's the zero element of R, but not of the addition group. I'll come back and try your groups problem at lunch Voidbringer, seen as it sounds like I'm the most qualified in abstract algebra in this place

Hey there Sara great to have you aboard, though it seems you've already been got by the spikers :\ Three of them in fact

I second that motion!

You're fricking useless, Eric... First you dont update the forums, and then you dont update the wiki. And NOW ... well now you're just making a mockery of the place I say he should have his spikes revoked!

oh, no, I just quack randomly sometimes

Sorry but that's also not possible.