the tree

@Athena: I don't know why you quoted my whole post there since you didn't address a single point mentioned in it. The problem with trying to measure cognitive development in something that more or less has to be taught is extreme experimental bias, every child is going to find algebra and geometry easier than number theory and graph theory because every child has been taught the former and not the latter from an early age - we can't decide that they are incapable of comprehending something we've never tried teaching them. @Gozonji: Everyone knows that Pythagoras was batshit crazy, but that doesn't mean that there aren't any potential applications of what he spouted, if you're smart enough to find them. @Schrödinger's hat: That is a fantastic idea, you should see what I mean, if something is fun you shouldn't have to "make it fun" and there really is no need to contrive a game out of something as inherently satisfying as building something. @khaled: Are you going to keep this method secret from us or what? And I'm going to throw a Citation Needed on that entire post, as it goes.

Though don't two algebras of two different arities essentially have the same properties assuming they both have an arity that is somewhere between two and infinity? So whilst they exist, the ones that have been studied are more 0,1,2,3,many,infinity than 0,1,2,3,4....?

For a ternary (yes, real terms do exist, people) operation, the typical notation would typically be something along the lines of f(x,y,z).

No matter what the question here is, the answer is Wolfram Alpha. Aaand when you just need to look something up, Wikipedia.

The Half Plane doesn't include the real axis or anything below it, in the same way that the Poincaré disk doesn't include the edge or anything outside of it.

Positive, negative and imaginary on a Poincaire half plane? Vector triple products? Scalar triple products?

I know you are but what am I?

One of the infinite possible answers: [math]a_x = {\frac {737}{1200}}\,{x}^{5}-{\frac {127}{15}}\,{x}^{4}+{\frac {9107}{ 240}}\,{x}^{3}-{\frac {2087}{60}}\,{x}^{2}-{\frac {16931}{100}}\,x+354 [/math]

I've never been too keen on patronising children of any age, for the most part they are far too smart to fall for it and "making it into a game" just gives the impression that they are supposed to enjoy it but not really expected to ("if you eat these sprouts then you'll get your desert" immediately makes the descision for the child that sprouts are disgusting and desert is the best thing ever, so you haven't done them a favour even if it happens to work once or twice). I think probably nearly everyone here has read Lockhart's Lament and I've got to say that I sympathise with his lack of patience when it comes to cutesiness - the exception in this case is that Schwartz is trying to teach an actual concept rather than how to numb your mind with routine calculations - the difference being that rather than trying to make maths interesting with contrived games and stories, Schwartz here is actually trying to show that it already is interesting. It's incredibly hard to show that something is interesting when the interesting parts are far from what you're actuallly teaching ("hey, look at the Mandlebrot Set, if you work hard for the next 15 years or so then you might have some idea of what the hell you're looking at") so the challenge is to find something that is both interesting and concieveable to teach to a young child. That's why I think elements of discrete maths should be introduced a lot earlier, there is nothing for instance particuarly difficult about the travelling salesman problem but it is a damn sight more interesting than factoring quadratics, even though the latter is still important it can be taught along with it's wider implications relating to classical mechanics or as a direct follow on from factoring integers to see how they are essentially two faces of the same coin. If you try to sell maths and arithmetic in one bundle then obviously maths is going to come across as incredibly dull, if anything arithmetic should be taught seperately as it is "useful" but it shouldn't be allowed to ruin maths.

Well everything that you said could be generalised to larger finite fields which can be described as [imath]\mathbb{Z}_p[/imath] for any prime [imath]p[/imath]. However the concept of parity does extend beyond just looking at the integers.

It's pretty much garunteed that you'll want to have a grounding in formal logic, but looking at your course reading list couldn't hurt either - if you know what programming languages are used on your course then you can definitely try playing with them at first and maybe even try out some Project Euler challenges.

Okay let's just stop you there. The easiest way to denote every even number ever and every odd number ever is with the modulo classes [imath]\{ \bar{0},\bar{1} \}=\mathbb{Z}_2[/imath], and all those equations that you just listed are really quite obvious. Well straight from the schoolish definition of odd number, if it's odd then it cannot be divided by 2, so neither can any of it's factors so it can only be factored by odd numbers. The argument from modular arithmetic is more subtle: since you are working in Z2 and 2 is prime it is a known result that Z2 is a field with no zero divisors. So there is no [imath]a \in \mathbb{Z}_2[/imath] such that [imath]a\cdot\bar{0}=\bar{1}[/imath] mod 2. As it happens, 2 is also a really small number so you can also get that from inspection. 1x1+1+1+1=4=0 mod 2, so yes. 1x1 + 1 = 2 = 0 mod 2, so yes 0*0 + 0 + 0 + 1 = 1 mod 2, so yes Modular arithmetic: it really works.

It wouldn't lead to a proof, but if you're simply looking for an answer then plotting [imath]y(x)=||x^2-4x+3|-2|-m[/imath] for various positive values of [imath]m[/imath] will give it away. (you're looking for an answer of the form [imath]m>c[/imath] for some [imath]c[/imath]).

mmmyeah, kind of, I guess. For pretty much any system you can linearise it at individual points and use linear techniques at those individualise points but such techniques aren't useful at all points and finding the points where they are useful isn't necessarily possible.

Assuming that amongst 195 countries, they are on average covered by roads over less than 0.5% of their respective surface areas (although I'd imagine much less than that), that works out okay. In short, there just aren't that many roads.

Honestly it's much less of an issue now that calculators aren't limited to those tiny LCD screens, Shaq O'Neal barely fitted onto my little Casio pocket calculator.

Apparently it is not only possible, it already exists and is avaliable on the consumer market. http://www.thinkgeek.com/stuff/41/wec.shtml although EM radition and an electrical current aren't exactly the same thing so you could argue that the wireless stage isn't strictly electricity.

Yes and yes, although I did do that a year ago so I think it's too late to be overly bothered about it.

I'm going to guess that x@y = x*a/y, so 1@1=a, but I wouldn't be able to say if that's a unique solution or not.

It disappears around the point that you take it out as a common factor. Then [imath]\frac{\mbox{d}u}{\mbox{d}v}= \frac{\mbox{d}}{\mbox{d}v} \ln(k) + \frac{\mbox{d}}{\mbox{d}v} \ln(1 - \tfrac{1}{1-e^v} )[/imath] And obviously: [imath] \tfrac{\mbox{d}}{\mbox{d}v} \ln(k)=0[/imath]

(1) is a slightly awkward rephrasing of FLT - so you are assuming the thing you're trying to prove. Not a good start at all. DH was right, you are, both you and Slick - approaching this from the wrong direction. You have a much more satisfying experience if you work with some basic algebra before trying to tackle one of the hardest problems in mathematical history.

I did say that it was a special case, so yes - proving it would be equivalent to FLT, that's kind of the point - and the fact that you put that into one line is what makes it trivial.

Elaborate on that one statement? No, not really. 1). FLT states that there are no integers [a,b,c] (all not equal to zero) that satisfy an+bn=cn for a positive integer n>2. 2). A corollary of (1) is that there are no rationals [a,b,c] (all not equal to zero) that satisfy the same equation. 3). 1 is both an integer and a rational number 4). (2) and (3) mean that there are no rational pairs that satisfy any of these equations 1+an=cn, an+1=cn or an+bn=1. Well it's fairly simple nowadays - with automatic proof programs, you can brute force your way through various axioms and known results and it will indeed turn out that no conventional method is going to get you there. And at the same time you can set another computer to look for a counterexample and fail just as miserably. If you're doing it by hand then I guess the typical mathematician knows that they are in trouble once the coffee runs out. o___________O

There are definitely no rational solutions to an+1=cn, it's a fairly trivial special case. How exactly did you infer, from me pointing out that I'd said it first, that I was disagreeing?

Yes. Not exactly the most rigorous, formal or unambiguous phrasing in the world - but yes. Well it just does. Limits and irrational numbers and all of that are well established, well defined and possibly more exposed to "excessive rigour" than any other object in mathematics. The fact that they don't stretch neatly into the physical realms with elves and such like is sort of secondary or beside the point. Well you wont, as we've established - such a thing doesn't exist.