On 25/08/2014 20:42, rkg wrote:
Will the following do as a proof? Just look!
Draw a rectangle with 2n rows and n columns, Area 2n^2.
In each column i (1 \leq i \leq n) draw two lines at levels a_i, b_i, so that there is just one line at every level from 1 to 2n. Then the answer is the area between the n pairs, i.e. the total area, 2n^2, minus the area above the higher lines, 0+1+2+...+(n-1) and the area below the lower lines, 1+2+3+...+n. 2n^2 - n^2 = n^2.
What seems to be missing is the proof that the higher lines are exactly {0,1,...,n-1} and the lower ones exactly {n,n+1,...,2n-1}. It turns out that this is true, and so far the neatest proofs all seem to go via this observation, but it's not instantly obvious to me and the proofs I've seen don't make it so. (It does have short clear proofs; so this may just be an observation about the limits of my own sense of the obvious.) -- g