# Question about Number Theory Text

September 12, 2007 9:12 PM Subscribe

Question about Hardy's "An Introduction to the Theory of Numbers".

On page 14, in Theorem 16, the series of equalities in which the numerator of the third term includes x^2^k -- is this unusual notation, or is this normal exponentiation? If normal exponentiation, then this is implying that 2^2^n^2^k == 2^2^(n+k), isn't it? I'm having no problems with the fact that no two fermat numbers are coprime, but I'd like to understand the steps behind this argument, too.

On page 14, in Theorem 16, the series of equalities in which the numerator of the third term includes x^2^k -- is this unusual notation, or is this normal exponentiation? If normal exponentiation, then this is implying that 2^2^n^2^k == 2^2^(n+k), isn't it? I'm having no problems with the fact that no two fermat numbers are coprime, but I'd like to understand the steps behind this argument, too.

Also that book has gone through a good 4 or 5 editions by now, some with different page numbers, I'd imagine.

posted by number9dream at 9:45 PM on September 12, 2007

posted by number9dream at 9:45 PM on September 12, 2007

I don't have the book either, but that looks like normal exponentiation to me. It should mean x^(2^k) and not (x^2)^k, since the second one could be written more simply as just x^2k.

posted by Sand Reckoner at 10:23 PM on September 12, 2007

posted by Sand Reckoner at 10:23 PM on September 12, 2007

Response by poster: a pdf of the book is available here. it's page 30 of the pdf (14 in the in book numbering.).

posted by tehgeekmeister at 10:51 PM on September 12, 2007

posted by tehgeekmeister at 10:51 PM on September 12, 2007

Best answer: Okay, I think I got it. Actually, the implication is that 2^2^(n+k) = 2^((2^n)(2^k)) = (2^2^n)^(2^k), which I believe is true and leads directly into substituting x in the next equality. I have definitely made mistakes with these kinds of hairy exponentiations before (I always

posted by crinklebat at 11:12 PM on September 12, 2007

*want*things to be true so badly!) so my word should not be taken as gospel, but I think that's where you got turned around. Hope this helps.posted by crinklebat at 11:12 PM on September 12, 2007

Just to clarify crinklebat's correct point above (and as an excuse to use Texify, which I spotted here)

2^2^(n+k) = 2^((2^n)(2^k)) = (2^2^n)^(2^k) made prettier

posted by sappidus at 4:42 AM on September 13, 2007 [1 favorite]

2^2^(n+k) = 2^((2^n)(2^k)) = (2^2^n)^(2^k) made prettier

posted by sappidus at 4:42 AM on September 13, 2007 [1 favorite]

Thanks for alerting me to the existence of Texify, sappidus! A friend and I were totally looking for exactly that tool about four months ago...

posted by crinklebat at 7:03 AM on September 13, 2007

posted by crinklebat at 7:03 AM on September 13, 2007

My house number theorist agrees that crinklebat/sappidus have it.

posted by LobsterMitten at 3:04 PM on September 13, 2007

posted by LobsterMitten at 3:04 PM on September 13, 2007

This thread is closed to new comments.

Perhaps you could post a little clarification for those of us who don't have access to Hardy? Seems like there must be more than a few people here who have taken at least a little number theory and ought to be able to puzzle out something from page 14...

posted by crinklebat at 9:39 PM on September 12, 2007