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Number Theory Nerd
Feb. 22nd, 2011 11:10 amSo, last night as I was walking home from Porter, I was in a little bit of a state. See, I shut down when it gets cold enough and I have to walk, just burrowing into myself and the internal monologue becomes something akin to "I hate this, I hate this, I hate this." One of the things that helps a little bit with that pure-misery mood is to count steps, or other things, as it gives my brain something to focus on and makes the trip much shorter.
The last couple times I've been miserable and returning home from Porter Exchange or Square, I've been counting intersections, and streets I have to cross. The assumption of course is that the blocks are consistent length, so by knowing how many there are (eight), I can get a rough idea of how much farther I have to walk --I am an eighth of the way home, two eighths, three eighths...
While I was walking last night, I started wondering if there was a pattern in the numerators of that sequence. See, if you reduce all fractions to their smallest version, you get:
With sixteenths, you have:
It's the same sequence. It works for 1/32 as well --they match at least to "1/4" (or 8/32)-- and it almost works on 1/4 if you continue the sequence into greater-than-one fractions:
So, I haven't figured out why this happens --I need to sit down with a pen and paper in order to try and make a proof or something-- but I find it kindof interesting, and I like that the pattern seemingly repeats across negative powers of two. I'd also like to figure out if there's a similar pattern across, say, negative powers of three, or of five or some other number.
As for things I did figure out, near the end of this, I somehow got stuck wondering if a power of two would ever be divisible by three, and more importantly, if there was a way to prove it. It took me a couple minutes, and I felt rather embarrassed that I hadn't hit it, but there is indeed --reduce the number to its prime factorization. It's impossible for a power of two to be divisible by three, because there is no three in the prime factorization --it's twos all the way down.
And by the time I arrived back to my dorm, I was in rather a good mood. Walk well spent!
~Sor
MOOP!
The last couple times I've been miserable and returning home from Porter Exchange or Square, I've been counting intersections, and streets I have to cross. The assumption of course is that the blocks are consistent length, so by knowing how many there are (eight), I can get a rough idea of how much farther I have to walk --I am an eighth of the way home, two eighths, three eighths...
While I was walking last night, I started wondering if there was a pattern in the numerators of that sequence. See, if you reduce all fractions to their smallest version, you get:
[1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, 1]
or, just the numerators:
[1, 1, 3, 1, 5, 3, 7, 1]
With sixteenths, you have:
[1/16, 1/8, 3/16, 1/4, 5/16, 3/8, 7/16, 1/2...]
or
[1, 1, 3, 1, 5, 3, 7, 1]
It's the same sequence. It works for 1/32 as well --they match at least to "1/4" (or 8/32)-- and it almost works on 1/4 if you continue the sequence into greater-than-one fractions:
[1/32, 1/16, 3/32, 1/8, 5/32, 3/16, 7/32, 1/4]
[1, 1, 3, 1, 5, 3, 7, 1]
[1/4, 1/2, 3/4, 1, 5/4, 3/2, 7/4, 2]
[1, 1, 3, 1, 5, 3, 7, 2]
So, I haven't figured out why this happens --I need to sit down with a pen and paper in order to try and make a proof or something-- but I find it kindof interesting, and I like that the pattern seemingly repeats across negative powers of two. I'd also like to figure out if there's a similar pattern across, say, negative powers of three, or of five or some other number.
As for things I did figure out, near the end of this, I somehow got stuck wondering if a power of two would ever be divisible by three, and more importantly, if there was a way to prove it. It took me a couple minutes, and I felt rather embarrassed that I hadn't hit it, but there is indeed --reduce the number to its prime factorization. It's impossible for a power of two to be divisible by three, because there is no three in the prime factorization --it's twos all the way down.
And by the time I arrived back to my dorm, I was in rather a good mood. Walk well spent!
~Sor
MOOP!