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its a pretty cool question coz its like, calculus but not lol. and if my name is on the list, it could easily be inferred from my surname anyway I came across this awesome geometry problem the other day, its neat but pretty hard. hint is that its probably doable by year 8 or 9 students :P try...

Pshhhh people shouldn't be complaining about this at all, module B was in my opinion a fantastic question. It made you stop and think, and look at how something shapes and relates to the rest of your material, and it really throws out the people who did what you're not supposed to be doing in...

This is a fantastic question :P the first one obviously doesn't (choose example f[x] = x), but for the second you basically integrate what should be its derivative (well you dont actaully integrate it because you dont know its differentiable but thats what it amounts to): For some x,y real and...

(i) Each pair of lines intersects once (as no pair are parallel) and each point of intersection has exactly two lines through it (as there are no concurrent triples of lines). Hence the number of points of intersection is exactly 5C2 = 10. (ii) Note that each line has exactly 4 points of...

The second part goes something like this: a = n!e - (some integer). Just expand n! (1+1/1! + 1/2! + ... + 1/n!) and its obviously an integer, right? lets call this integer F[n]. Now suppose e were rational; e = p/q in lowest form (i.e. the gcd of p and q is 1). Then you have: a = \frac{n!p -...

13/27? you just count the possibilities?

Surely the way to do it would be, in the question you linked (and the way I did it in that exam): "By inspection, z = 3+5i and -3-5i satisfy the equation z^2 = -16 + 30i. Since this equation, a quadratic, has at most two roots, 3+5i and -3-5i are the complete set of square roots of -16+30i"

This sounds like an excellent addition to the syllabus if you ask me.

seanieg89 is completely right, although you have unique Pk and Qk assumed, you need to prove that there isnt a different P(k+1) and Q(k+1) that DOESNT satisfy the recurrence relation (obviously if it satisfied that same recurrence that'd be fine, but you need to check that it doesnt). seanieg89...

No. The problem explicitly asks to to prove there are unique Pk and Qk. What that then means is you have to show there aren't two different possible solutions for them. You don't need induction for this, and any use of induction is probably a bit silly; so when n=1, you just say "let (p,q) and...

That doesn't prove uniqueness. You have to add another second part of the proof to prove uniqueness. Suppose for a contradiction that for some n>=1, there exist distinct ordered pairs of integers (p,q) and (r,s) such that: (1+\sqrt{2})^n = p+q\sqrt{2} = r+s\sqrt{2} Now clearly q=/=s; if q were...

(a) Each triangle is formed by choosing 2 of those lines and the line x = 1. So if n < 4, n is one of 0,1,2,3 so there are 4C2 = 6 triangles (b) Same as (a) except n can be one of 10 things. So 10C2 = 45 triangles can be formed. (c)(i) We choose 2 of those triangles above, so 45C2 = 990. (ii)...

I think that question is actually reasonably hard, I can't think of a way to do it without using the "inclusion-exclusion" principle, something that isn't really in the course (although we did it in class in case it might come up in a problem like that). Essentially think of a venn diagram like...

It could possibly be an easy olympiad question. The difference between easy olympiad questions and hard "harder 3U" questions is almost nil though. The entire circle geometry topic, entire probability topic, most induction questions (probably not those involving trig functions), definitely all...

It's not really cheating lol, but yeah it feels like it. Its interesting to know that you absolutely need choice to find nonmeasurable sets. And this isn't really set theory is it? I was under impression Banach-Tarski and other such things were more about group theory, in particular rotation...

Nice, I did pretty much the same thing ^^ I think you can motivate it by (1) knowing you want to somehow telescope, and (2) after computing a few values you guess that it goes to 3. Which would make sense as one on a third, so you try telescoping with 1/un. Telescoping means that technique to...

[i][a] Yeah just do it on your calculator [b] Let 4223a = K + F(4223a) where K is an integer. 1.989 < 10^{F(4223a)} < 1.990 1.989 \times 10^K < 10^{K + F(4223a)} < 1.990 \times 10^K 1.989 \times 10^K < 10^{4223 \times 2136 log_{10} 2} < 1.990 \times 10^K 1.989 \times 10^K < 2^{4223 \times 2136}...

I did a bit of wikipedia-ing and found this; http://en.wikipedia.org/wiki/Solovay%27s_model. Apparently you need an 'inaccessible cardinal' though (and I'm still not sure what this means and any meaningful explanation is probably far beyond what understanding of set theory I might have), and...

The first question I posted is probably strictly more an easy olympiad question, until it turned up in my trial last month. The question expressed in terms of simultaneous equations is a proper olympiad question (from an Asia-Pacific mathematics olympiad a few years ago, I think) and the one I...

Oh I realised I got my method wrong, I defined Q incorrectly :P The minus one at the end should be multiplied by (x^2+1)...(x^2 + n); i.e. you need Q(x) = (x^2+1)(x^2+2)...(x^2+n)\left[x^2F(x) - 1\right] You then get the same thing, as in Q has the same 2n roots, except its degree is now at...