The Prime Number Theorem

So one of the few math books I’m still fairly interested in poking at sooner rather than later, since graduating and moving on to a computer programming job, is “The Prime Number Theorem”,  by Jameson. It claims to be accessible at the undergraduate level, so I’m sort of optimistic I can follow it. When I’m feeling ambitious, I figure I’ll read the book and do the exercises, and even write posts about what I learn as I go. And cook more. And eat healthier. And…

Right, so I started reading, and got to the first set of exercises. I know I’ve done the second and third before. The second (show there are arbitrarily large gaps between primes) I actually remember doing from my freshman year as an undergraduate, so didn’t put much into it. The third (show p_n\leq 2^{2^{n-1}} and that this implies \pi(x)\geq (\log\log x)/(\log 2)) I remembered doing while I was reading Hardy & Wright more recently, and decided to see if I could get the inequalities right again. It took a few minutes, but I think I got there.

The first exercise, though, took me an embarrassing amount of time. Good start, huh? The problem reads:

Let ≥ 1 and let = {30n+r : 0 ≤ r ≤ 29}. For which values of r is 30n+r not a multiple of 2, 3, or 5? By considering the possible positions of multiples of 7, show that E contains at most seven primes (seven cases, no short cuts!)

Ok, so the first part of that question I got without much difficulty (actually, I got wrong without much difficulty). It’s just all the other primes bigger than 5 but less than 30: 7, 11, 13, 17, 19, 23, 29 (see my error yet?). The second part got me though (since I got the first part wrong). There’s only 7 numbers in that list I just picked out, so what more is there to show? Any 30n+r that’s prime must have an r from that list, because everybody else is divisible by 2, 3, or 5, right?

Nope. Stupid 1. The case r=1 isn’t accounted for. So that means that primes could actually show up in any of 8 spots. So that’s where the hint about multiples of 7 comes in. The first multiple of 7 in the 30 consecutive integers comprising E corresponds to r = 0, 1, …, 6, and then for any of those choices, you can easily write down what the other multiples of 7 are. For example, if the first multiple of 7 in E is with r = 3, then the other multiples are at 10, 17, and 24. Notice that, in this case, the prime 17 would have to get taken out of the 8 possible spots for primes (since it is now known to be a multiple of 7), and we’d be down to 7 primes. The same thing happens in the other 6 cases about the multiples of 7, and we obtain the result.

So that slightly rocky start, those 5 pages of reading and 3 exercises, took just under a week, start to finish (not that I was actively working on it that every day). And, in all honestly, I only sorted out that first one when I sat down to start writing, here, that I couldn’t figure it out. But whatever. There’s my first post since graduation. Perhaps I’ll keep learning some math after all (in addition to the math/stats sorts of things I’m learning for work). Here’s to fewer errors in section 2!

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