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).

Anyone aware of a calculator out there on the internet...

No matter what the question here is, the answer is Wolfram Alpha.

Alternatively, is there a...

Aaand when you just need to look something up, Wikipedia.

Positive, negative and imaginary on a Poincaire half plane?

Wouldn't this also have negative imaginary making it a quadruplet, or would there be a good reason to exclude that possibility?

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?

...dammit.

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.

So, it's Modular Arithmetic in [imath]\mathbb{Z}_2[/imath]

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.

assume E is any even number, and O is any odd number

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.

 

[a given odd number] can only be factored by odd numbers. what do you think ..?

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.

 

• odd*odd + odd + odd + 1 = odd+1 = even
  
• odd*odd + odd = odd+odd = even
  
• even*even + even + even + 1 = odd
  

• 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]).

however, is it true that not all formulas are transformable to linear system ..?

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.

Where did the k go?

It disappears around the point that you take it out as a common factor.

[imath]e^{u} = k\left( 1-\frac{1}{1-e^v}\right)[/imath]...Take logs....[imath]u = \ln(k) + \ln(1 - \tfrac{1}{1-e^v} )[/imath]

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]

Tony McC's conjecture:-

 

(1) In every correct sum a^n +b^n = c^n, if n is a whole number >2 and a, b and c are positive numbers >0 then one or more of the terms a, b or c is irrational.

 

A consequence of this is that if a triangle is constructed using a, b and c at least one side will be irrational.

Because of the irrationality no similar triangle can be constructed with all three sides integers.

Therefore there is no integer solution to the equation x(a^n) + x(b^n) = x(c^n) where x is any number.

This proves Fermat's Last Theorem.

 

Please comment.

(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.

 

How does that relate to The Tree saying - "There are definitely no rational solutions to aⁿ+1=cⁿ, it's a fairly trivial special case." Can The Tree elaborate on that.

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.

 

 

Nor am I trying to. My interest is in why such a simple looking problem is so difficult and to do that I am trying to understand where the actual problem lies. In other words, what is the point at which anybody attempting to resolve this realises they have hit an obstacle that cannot be overcome?

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.

 

I should declare that it is a professional interest for a business solutions methodology I a working on for which I am using FLT as an unambiguous reference case.

o___________O

I agree that a^n +1 = c^n cannot have an integer solution.

Thanks but do you think that it can also be true for a and c being positive rational numbers?

There are definitely no rational solutions to aⁿ+1=cⁿ, it's a fairly trivial special case.

 

However, it seems to me that Fermat may have proved (or thought he had proved) that ALWAYS at least one of the three terms a, b, or c will be irrational if n is a whole number >2.

As you can read above, "D H" seems to think that is the case and "the tree" seems to think that it may not be the case. In other words the idea is [contentious].

How exactly did you infer, from me pointing out that I'd said it first, that I was disagreeing?

Well, as I see it, if you tell me that c is irrational and a and b are rational in a^n + b^n = c^n (n>2) then it follows that c is a number followed by an infinite number of digits after the decimal point.

Yes. Not exactly the most rigorous, formal or unambiguous phrasing in the world - but yes.

 

The elf mutters to himself "This doesn't seem to make sense!"

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.

 

I would very much like to see an example of a^n + b^n = (c/d)^n where a, b, c and d are whole numbers.

Well you wont, as we've established - such a thing doesn't exist.