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 August 8th, 2010, 05:59 PM #1 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 not an integer let n be integers in Z with n>1. Show that ?(^n)(_i=1) (1/i) is not an integer. So i am trying to prove that 1/i is not an integer. I put some number in to see if this is correct and it is. however i don't know how to generalize it if n >1 for all n. thanks.
 August 8th, 2010, 06:26 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: not an integer You're trying to show that H_1 = 1 is the only integer harmonic number. The key step is looking at the denominator and seeing how many primes divide it.
 August 8th, 2010, 06:40 PM #3 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: not an integer Hi again tinynerdi. It's pretty clear that for all n > 1, 1/n is not an integer, right? Because no positive integer greater than 1 divides 1. Now, for the sum, you probably want to use induction on n. Clearly, if n = 2, then $\sum_{i=1}^{2}\frac{1}{i} = \frac{3}{2}$ is not an integer. Now, assume that for some n > 2, the statement is true, i.e. suppose that $S_n= \sum_{i=1}^{n}\frac{1}{i}$ is not an integer. Clearly, it is a rational number, though, so let's say $S_n= p/q$ in lowest terms. We want to show that $S_{n+1}$ is not an integer. But of course $S_{n+1}= S_n + \frac{1}{n+1} = \frac{p}{q} + \frac{1}{n+1} = \frac{pn+p+q}{qn+q}$ Now, see if you can prove that the above fraction cannot be an integer. [Hint: Assume the contrary and derive a contradiction.] Best of luck. Addendum. Sorry, I noticed just after I posted that CRGreathouse had already replied. I did not mean to try to "one-up" his advice, which is just as good.
August 8th, 2010, 06:53 PM   #4
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Re: not an integer

Quote:
 Originally Posted by Ormkärr Addendum. Sorry, I noticed just after I posted that CRGreathouse had already replied. I did not mean to try to "one-up" his advice, which is just as good.
Yours is much more useful; I was in too much of a hurry to give a real answer (but figured I'd at least leave a hint),

 August 8th, 2010, 07:28 PM #5 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 Re: not an integer thanks two both

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