Josy

See the Wikipedia article on polytopes.

Yes it is, but I think you were probably asking about the integral of [math]\frac{1}{x^2}[/math] which also goes to infinity. If you write it with a negative index it becomes much clearer. It converges if you let the upper limit go to infinity though. Have you read the question correctly? I've been caught out by that sort of mistake more often than I care to recall.

Since the proportion of the population that supports x is a continuous quantity, it's not simply a question of true or false; presumably you'd model the proportion of the parent population as a normal distribution with a mean of 45% (or 50% on Day 2) and a s.d. of something or other, then move forward from there. It's a while since I did any stats as well, so I can't really be much use on this.

See http://en.wikipedia.org/wiki/Binomial_theorem#Proof for a proof (by induction) of the binomial theorem. This appears only to prove the case where n is a positive integer. To be honest, I wouldn't mind seeing a proof of the general (real or complex) case myself.

It doesn't hold. For example, let n = 10 and r = 3.

[math]\frac{0}{0}[/math] is an indeterminate form. If we have two functions f and g such that [math]\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x)=0[/math], we have no idea whether [math]\lim_{x\to 0}\frac{f(x)}{g(x)}[/math] tends to a limit, to infinity or not at all. For example, [math]\frac{\sin x}{x}[/math] tends to 1, [math]\frac{x}{x^2}[/math] tends to positive infinity and [math]\frac{\sin{\frac{1}{x}}}{\tan{\frac{1}{x}}}[/math] doesn't do either.

The nice thing about RPN is that it doesn't need brackets or priority of operations at all; all you need (from a computing point of view) is a stack (a LIFO or last-in-first-out data structure). Essentially, a number means 'push this number onto the stack', + means 'pop the top two numbers off the stack, add them, and push the result', and *, - and / mean something analogous, making sure in the latter two cases to get the operands in the right order. RPN is terrible as far as human readability is concerned, though.

This is really more of a mathematical problem than a physical one. The confusion arises because of this ingrained tendency for people to think of infinity as being a number which is somehow greater than all other numbers. This can be a helpful way of thinking about it, but not always. In mathematical analysis, the formal definition of a function f(x) tending to infinity as x does the same is that for any K > 0, there exists a value of x such that f(x) > K. No actual number called 'infinity' is required. We use the notation [math]\lim_{x\to\infty}f(x)=\infty[/math] purely as a notational convenience to indicate a property that pertains purely to finite quantities. I suggest that limits are the best way to think about your pieces of string.

Defacing coins of the realm is technically treason in Britain. In the States I'm not sure but I believe it's illegal in some way. I doubt you'd be prosecuted in either country though. It simply isn't worth anyone's while.

Finally I think I see where you're coming from, and can formulate some kind of useful reply. For one thing, as far as I'm aware, the Beal conjecture is a conjecture, and thus not proven; indeed, Andrew Beal has offered a prize of US$100 000 for a proof or counterexample. Even assuming it does hold, it deals with equations of the form [math]a^x+b^y=c^z[/math] with all of x, y and z greater than two. Considering your initial equation, [math]Ta^x+Tb^y=Tc^z[/math], which is equivalent to the above, certainly all three terms have a common factor of T; however, there is nothing indicating that they do not have additional common factors; that is, [math]a^x[/math], [math]b^y[/math] and [math]c^z[/math] are not necessarily coprime. If they are coprime, then the Beal conjecture is false, and the discoverer of the counterexample might be well advised to send it to Andrew Beal and claim hir prize money.

I never made any such claim; as I said before, cancellation is a deep property of the real numbers; if this equation contradicts it, which it does not, then the error must lie in the equation, not in the law of cancellation. The real numbers, over which I will assume we are working, though we can allow the variables to take complex values with equal ease, form a field. Among the properties of a field is that it forms a group (an Abelian group in fact, but that is not relevant here) under multiplication. The definition of a group is thus: a set [math]G[/math] under an operation which I shall denote by simple juxtaposition forms a group if and only if all of the following conditions are met: [math]\forall a,b\in G,ab\in G[/math] (closure; this is fairly trivial, and is often taken as implicit in the definition of the operation itself) [math]\forall a,b,c\in G,(ab)c=a(bc)[/math] (associativity) [math]\exists e\in G:\forall a\in G,ae=ea=a[/math] (identity) [math]\forall a\in G,\exists b\in G:ab=ba=e[/math] (inverses, and we usually write [math]a^{-1}[/math] for [math]b[/math]) Note that these four conditions are taken as axiomatic; they are accepted as true by definition. Now, suppose that [math]a,b,c\in G[/math], and that: 1. [math]ab=ac[/math] Now, proceeding from first principles, i.e. the above four axioms: 2. [math]a^{-1}(ab)=a^{-1}(ac)[/math] (inverses) 3. [math](a^{- 1}a)b=(a^{-1}a)c[/math] (associativity) 4. [math]eb=ec[/math] (inverses again) 5. [math]b=c[/math] (identity) Thus the law of cancellation, one of the basic results of group theory, is derived. Your example actually corresponds to stage 2 above; stage 1 would feature the denominator of your [math]\frac{T}{T}[/math] appearing as part of the numerator of the RHS. Nevertheless, cancellation most certainly holds for it. I have tried to keep this to the mathematics, and not to make it personal, but I would like to comment that I find your tone rather more patronising than I'm entirely comfortable with. This may well be accidental; online conversations are rather notorious for misapprehensions of emotional content.

AFAICT the point is that if the equation in question held in general, it would prove that cancellation of common factors doesn't work. Since it doesn't hold, it pretty clearly proves nothing of the sort. This is unsurprising, since cancellation arises very fundamentally from the way the real numbers work, and is pretty obviously correct.

Is there a riddle outstanding?

I'm not sure if you are looking for a solution or just a few pointers. The bit about the violin is pretty clearly a red herring, though the third statement as a whole is not; think about what other information the master can gain from this statement. Also important is the fact that the two bits of information given first by the student are _not enough_ to determine the answer on their own, but with the information from the third statement, it is possible.

I'm Josy. I'm a third-year mathematics undergrad at the University of Glasgow in Scotland.