REPOST: Re: Pomerance/Crandall "Prime Numbers" book: inequality question
From: Robert Israel (israel_at_math.ubc.ca)
Date: 11/02/05
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Date: 2 Nov 2005 01:40:51 GMT
In article <11mg1esoqc3nb09@corp.supernews.com>,
yer mammy <yer_mammy@yahoo.com> wrote:
>The second edition of Pomerance and Crandall's "Prime Numbers" book is
>out, and before I buy it I figured I should read the first edition,
>which I already own (I bought it a few years ago and only read a few
>sections that were of interest to me at the time).
>
>Sadly, I can't figure out the following inequality, which occurs -- even
>more sadly -- on page 6:
>
>x <= PROD( floor( 1 + (ln x) / (ln p) ) ), where x is a positive integer
>and the product is taken over all primes p <= x.
>
>Why is this inequality true?
Call the right side P(x) = product_p f(x,p).
f(x,p) = floor(1 + ln(x)/ln(p) = n iff p^(n-1) <= x < p^n. Thus f(x,p)
is the number of powers of p that are <= x. Consider selecting
one of those powers of p for each prime p <= x and multiplying them
together. This gives us P(x) distinct positive integers, and each
of the x positive integers <= x can be obtained in this way, so
x <= P(x).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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Subject: Re: Pomerance/Crandall "Prime Numbers" book: inequality question
From: israel@math.ubc.ca (Robert Israel)
Date: Thu, 2 Nov 2005 22:02:57 GMT
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- In reply to: yer mammy: "Pomerance/Crandall "Prime Numbers" book: inequality question"
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