matt grime

There is no debate, at least by people who know what they are talking about. Locked thread, please.

Or more algebraically it satisfies the equation x^2=x+1, which can be used to demonstrate simply all of the properties that people think are mystical but really are very straight foward. I see no one ever gets as worked up by that fact that 1/1 =1 , which is very odd: surely that is more amazing....

that is false there is no constraint that says the 4 digit numbers are prime.

that is not what is written in the first post. B is a multiple of A, although I am assuming that A is the top row 4 digit number and B the bottom row

Is that a question because you don't know the answer, in which case I'm not going to reply to it because you should try to work it out (the hint I've given is more than enough to work it out: how many two digit primes end in 2,4,6, or 8? Can you think of any other digit that a 2-digit prime cannot end in?) or because you know the answer and want to see if anyone else can do it? And your grid still has too many yellow shaded squares in it, which makes me think you don't know the answer, so why not explain what you've managed to work out and why.

Use a little intution, A and B are going to be (in all likelihood) the two 4-digit numbers, and probably A will be at the top and B at the bottom.

I think you've definitely shaded in one too many yellow squares. It is reasonalby easy to work out what digits must be in each row by the primality constraint, and then to use the multiple constraint. You might want to put these in the puzzles forum, since that is what it is more than a mathematics topic.

learn latex: [math]\frac{1}{\sqrt{2\pi\sigma^2}} e ^{-\frac{ (x-\mu)^2}{2\sigma^2}}[/math]

Q1 is 121 composite? What are its fenders? Have more confidence in your answers. Q3 just make it. If want a fender with 2 and 3 as factors I'd write down 6, the smallest number with 2 and 3 as factors. it also has factors 1 and 6, so i need only figure out how to make it have factors 4,5,7,8,9. More sophisticatedly, you might notice that you only need to worry about prime powers: if it is divisible by 9 and 4 then it is divisible by 2,3,6, 18 and 12.

Your original question is slightly ambiguous, but that isn't a criticism. It is common in combinatorics. Do you mean 1. how many ways are there to select two things from n. 2. how many ways are there to partition n into subsets of size p. this is subtly different. for example, if n=4, the first case gives 12,13,14,23,24,34 ie 6 pairs. however the second count classifies things like (12)(34), or (13)(24) or (14)(23) which are the only 3 ways to split 4 up into groups of 2 objects (assuming order unimportant). actually i think there is yet another way of thinking about this,and it is the correct way: order must be important to you. doing A then B will need a different animation to doing B then A, so yes, now I think about it the number you want is just n^2.

No. The of ways of picking 2 from n is n choose two, or n(n+1)/2, and that is the number of transition moves you need (one for each possible choice of pairs of moves).

on what? how will it help?

Ok, I have a drop of water, I add another drop of water, and oh look they coalesce and I have one drop of water.... perhaps 1+1 is not 2 after all..... never mind the other far more complex philsophical issues at stake here either. (Try reading Russell.) And how do you prove that even if you define 'thing' properly that addition *always* behaves properly? Remember 'well, it is cos it is' is not a *proof* in mathematics. You either need to accept something as an axiom, or try to prove it if at all possible in your model, or disprove it, or whatever you wish. Whilst no one would argue that 'of course 1+1=2' that is not a proof. And some people feel the need to prove it from simpler axioms. Look at Peano Arithmetic for a firm axiomatic description of the natural number system.

And what is a 'thing'? how do you knowthat if you put two things together you always get two?

Total is not a sqaure factor of n, unless I misunderstand what 'square factor' means, ie a factor that is a square. total is not a square. total will be a square root of a square factor. in anycase, how are you using this? because it seems to me that given N you put N in, and get N' out but then you need to put N' in again, and so on until it outputs what you input. This seems unwieldy. Firstly, your program should run something we'll label 'IsSquare( )', then it should at the very least do something like: if p divides N work out largest power of p that divides N, if even exponent do something if odd do sometihng however, this essentially factors N, so you might as well just ask for the factorization and read off the answer. what size input are you considering and what length of time do you need to do this in?

I just didn't wish you to think I was being crabby and rude; replies often appear that way because you write what you'd say in person, and the intent is less clear.

The important part of language and communication is that you manage to express to someone else what it is that you wish to express. I have no idea from what you asked what is that you wish to know, or communicate.

Why do so many people attempt to make inaccurate and unhelpful 'kind of like...' statements? It is what it is, nothing to do with rounding up or anything. "With Calculus you can take any given section or point (instanteous?) and find out the rate of change to a more accurate/percise calculation?" doesn't make sense as a sentence.

You get in a car and it accelerates away. What was your velocity at any given time? How could you work it out? Well, you could average the distance travelled over the time taken, but that would only be a rough estimate of your speed at any given instant, the so-called instantaneous thing you're thinking about. So, to get a better idea of your speed at some given time you could measure the distance travelled in a small time period about that instant dividing by the length of the time period. The key idea is that as we let that time period get smaller and smaller, the better and better that estimate will be of the actual velocity you were travelling at that instant. In less physical terms, suppose you want to estimate the tangent slope to a curve, then you should try to draw a little chord on the graph to estimate it, the smaller the chord the better the approximation. The limits as these little time periods tend to zero, or little chord lengths tend to zero, is the instanteous 'rate of change'. In the first case it is distance over time, ie velocity (ok, displacement), the second it is just the abstract y with respect to x.

Can any of the functions ever take negative values? ("Have an infinity range" doesn't make any sense as a sentence in the English language; all sentences in mathematics should make sense as grammatical constructs.)

for this question there is no difference, surely? so I believe you did just do his homework for him; the question is such that there was nothing else you could have done (apart from ask him what the method is he needs to use without specifying it).

I think you just did.

I think I must disagree with that statement. We have a very simple way to define the n'th prime in a simple function (or to encode them in an equation) it's just that there is no nice way to evalute it in any reasonable time for any reasonably large n. There is a difference between a function, or equation, and actually evaluating that equation/function at a point. It is a subtle difference but one that is important. Let's think on an example: Define a_n by a_1=2, and a_n is the smallest integer greater than a_1,..,a_{n-1} not divisible by any previous term. Define b_n by b_n=sin(n). Find a_{10^30} and b_{10^30}. The question here is what do we mean by "find"? a_(10^30) is the 10^30'th prime number, but I can't write that out as a decimal expansion. But then I can't write out b_(10^30) either as a decimal expansion (it is an irrational number in all likelihood) but you will probably happily accept sin(10^30) as "the answer", so why is one symbol better than the other? Sure I can find a huge number of the digits of sin(10^30), and very quickly, but I can never write them all out, I might even be able to give some kind of formula for them, whereas, although it takes a huge amount of time to compute the a_n for large n I could in theory, if the universe didn't end work out all the decimal digits for any one of them. There are plenty of functions whose positive values are exactly the set of prime, or whose roots occur at exactly the set of primes, for instance. And if anyone wants to say 'yeah, well, if there is a magic equation then what's the 3 trillionth prime number?' then you've missed the point of this post. I can no more work out large primes than I can large terms of te Fibonnacci sequence; it is just computationally infeasible, the difference is that one becomes infeasibly more quickly than the other; just because there is a formula for the n'th fibonacci number doesn't really mean that we have a magic equation that pops out any term your care to name, and this is all about the mathematical interpretation of inherently non-mathematical labels, like 'magic', so this is a statement of opinion not an assertion of fact. I would have been happy with you saying there was not 'a magic algorithm' to spew out the primes in milliseconds. Here is a function all of whose zeroes are exactly the set of primes, by the way: [math] f(x):= \prod_{p}(1-\frac{x}{p^2}) [/math] Of course if we could find a nicer expression for that (that didn't involve p explicitly) and a quick way to find roots we'd be laughing.

What's to explain? Let L be a lattice in R^n, that is just the set of coordinates with integer entries, at t=0 you usually start at the origin, and at each time t=n (n in N) you move in any one of the 2n possible directions with equal probability (1/2n). Ie you can't stand still. That's all it is. Perhaps you mean more complicated things like why it is 'recurrent' for n<3 and not otherwise. But that's just probability on n-nomial distributions which sounds far more complicated than it actually is, and is so well known it must be easily googled. Find an explanation on the web (googlefo random walk) and ask about what you don't understand.

Why on earth is latex a pain in the back to use? Produce a latex file, get a command line and latex filename.tex and woo-hoo, it's done. If you're usign emacs or vi you don't even need a command line (C-c C-c b, I believe, and I can't remember the vi command, something like :ll is all you need). It is absolutely straightforward, and not at all a pain in anywhere. even windows has a command line you can use to do things if you don't have emacs or vi(m) lying around, and mac OS X has TeXshop that makes life a breeze. There is no reason for latex to be a problem at all on any system that you might reasonably use. I admit I've not used OS/2 or amiga OS but I doubt that these are a problem in any sense.