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Do you know if you can do it without the axiom of choice? The only counterexample I've seen in detail is Banach-Tarski at a short talk a few months ago, and the guy giving the talk kinda skipped over the axiom of choice part. If you weren't to accept the axiom of choice, would there still exists...

I haven't done any university mathematics but I'm pretty sure a volume measure doesn't exist on R^3, even though an area measure exists on R^2. See http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox.

Oh capacity doesn't exist either. What I mean to say is, there's actually no volume-like quantity (doesn't matter what you name it, volume, capacity, etc) in three dimensional space. It just doesn't exist.

A strange feeling? Who'd have thought that? I'm pretty sure almost all the scholarships are awarded starting in year 7, and I'm not sure what it's like in lower years but there are quite a few full scholarship holders in my year but I'm not sure of the exact number. I think there's around 20...

Interestingly I guess I got something pretty different :P I'm not entirely sure its correct though. I got (after a bit of manipulation): F(n+1) = \frac{(n^2 + 2n)!(2n+1)! + n!(n+1)!(n^2 + 2n)!}{n!(n+1)!(n^2+3n+1)!} I'm pretty sure our answers are out by an order of magnitude. I only...

Nah I meant that legitimately, volume is not actually a property of 3 dimensional space.

By using similar triangles with the triangles on the top and AB, you get \frac{AB}{20} = \frac{100-x}{100} Hence AB = 200 - \frac{x}{5} Then taking a slice near x, you have: \delta V = \frac{h(AB + CD)}{2} \delta x = \frac{20(20 - \frac{x}{10} + 20 - \frac{x}{5})}{2} = (400 - 3x)\delta x So...

Nice, but could you help me with this one first mate? "To take a bite of chicken while suited up, that's something that rings true! (3,4,5)"

"Mathematics, rightly viewed, possesses not only truth, but supreme beauty — a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only...

Yep thats correct ^^ It's a kinda nice result, there are lots of nice problems out there with similar methods. For a muuuuch harder and more abstract thing a bit like it, if you feel like trying it, try (but I really meant it with much harderr): For some fixed n, let a1, a2, a3, ..., an be real...

Thats how my teacher did it. I dont really like it that much coz its a bit random, like theres no clear motivation to go from line 1 to line 2 other than guesswork. If you do auxiliary angles, you can sorta metagame the question the whole way through. And artosis, I haven't done the HSC yet...

Yeah that's right. Try either the polynomial question or that neat probability thing on the page before, theyre both fantastic questions in my opinion

Uh, thank you? Maths is my favourite subject though by a way. And yes I am. I can see why you'd want to draw a little diagram explaining the relationship between deltaV, deltax and x, but I really don't get why you need a limiting sum. It's not really helped by the fact that the statement with...

I'm going to do it anyway in case I get a silly marker

I can't remember if theres a part (i), but the answers to (ii) and (iii) are wrong. Just because a polynomial Q[x] has roots a,b,c,... doesn't mean Q[x] = (x-a)(x-b)(x-c)... .

No I think I understand the concept perfectly well. My argument is that "limit as delta x approaches zero of the limiting sum" and definite integration are EXACTLY THE SAME PROCESS. When I write integral from a to b of f[x] dx, what that statement means is the limiting sum when delta x...

I think that mightve even been part [i] in that question in our trial, but I can't remember at all whether it had 2 or 3 parts

Yeah alright Ill probably have to do that in the external exam, but I maintain its unnecessary and stupid mathematically. The fact the Board of Studies would even consider requiring you to do that is indicative that their course isnt very good at actual mathematical understanding; it's as if the...

It's not hard to understand at all. It's just that we were never taught it in class, I've always completely got away with not doing it, and as far as I'm concerned its a completely unnecessary step logically. The actual integration itself means exactly the same thing as the sum with the limits...

its a cool question to think about. the answers really random anyway, its always 1/2 lol um try this, from q8 in my trial but its considerably easier than it looks I reckon Factorise z^{2n} + z^{2n-1} + ... + z + 1 into quadratic factors with real coefficients, and hence prove that: 2^n...