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:(.

Yeah, unfortunately in my experience not that many are willing to invest that time.

Agree that more teachers should write their own exams, but I don't have much faith in the quality of paper that the average high school maths teacher could put together...

Re: HSC 2013 4U Marathon Computational complexity. This algorithm is horrendously inefficient. More efficient algorithms are exactly how they found the current largest known primes.

no worries, good luck...some parts of graph theory are pretty fun.

Re: HSC 2013 4U Marathon $As Sy123 proved, $f(m,d)$ is $1$ if $d$ divides $m$ and $0$ otherwise.\\ \\ By the definition of a prime as a positive integer greater than $1$ that is indivisible by all smaller positive integers that are greater that $1$, we get:\\ \\...

Here is a proof for an even number of vertices, try to prove the odd case yourself (it is almost exactly the same). Our claim is that the graph on 2n vertices can have at most n^2 edges without containing a triangle, we prove this inductively. 1. The claim is clear for n=1, we simply join...

Depends how rigorous you want to be. You could say something like: Suppose distinct vertices 1,2,3 form a triangle and without loss of generality assume 1 is in set A. Then by the way we defined our graph, 2 (being adjacent to 1) must be in set B. Similarly, 3 must be in set A. But now 1 and 3...

\left\lfloor\frac{x}{2}\right\rfloor^2 is a tight upper bound which you can prove by induction. Intuitively, the most efficient way of constructing such graphs is splitting your vertex set into two sets A and B of roughly equal size and then joining every vertex in A to every vertex in B.

Re: HSC 2013 4U Marathon No, that's just rewriting the same fallacy in a different form.

Re: HSC 2013 4U Marathon Which book? Lagrange multipliers slay it pretty easily, but don't have the time to play around with MX2 tools right now. If it is something that can be done with elementary methods (I suspect it can, but nothing immediately jumps out at me) and no-one answers it by...

Re: HSC 2013 4U Marathon Uhhh same issue, division will not preserve inequality in general. Eg. 2 > 1 & 5 > 2 does not imply that 2/5 > 1/2.

Re: HSC 2013 4U Marathon Is it? How does finding an upper bound for the numerator and an upper bound for the denominator help us find a lower bound for a fraction?

Re: Australian Maths Competition 2013 What was the question?

Medieval 2: Total War with the Stainless Steel mod...such a fun game to play!

She likes a sport you like? omg snatch her up now, you will never find another with that much in common.

lol

don't ask advice if you can't handle constructive criticism.

two words: friend zone. be a man and make your intentions clear from the outset.

because of honours? or other reasons?