Feb 20, 2008 12:10:13 (281 days ago)
haha,
I just solved one of my abstract algebra questions because I was walking in a circle. I’d spent a little while trying to bang out the numbers but then I realized walking in a circle was like doing a permutation of a set(sort of...it’s really over-simplifying it). Basically it’s this complicated and big disjoint set. for example
(1 2 3 4 5 6 7)( 8 10 9)
This basically means 1 becomes 2, 2 becomes 3, etc. and 8 becomes 10, and 10 becomes 9, and 9 becomes 2.
ANYWAY, I was asked to find the lowest positive number where B^n=B^-5. Of course I started by finding out what B^-5 was, which....was a long process. and then I figured I might as well go forward to find it out, but it’d probably take a while. As I was walking in a circle. I realized that i was basically doing this. one step forward is a permutation of the set one step back is it’s inverse. This particular circle has 21 segments. So, If I take 5 steps back from the start. How many steps forward is the same thing? 16 steps of course. I find it funny something I thought was going to be so much work boiled down to walking in a circle.
I think i’m pretty stupid for not realizing this earlier, but whatever.
Lol. No one probably followed that. >__>