Show that the nth harmonic number is not an integer

http://quicklatex.com/cache3/ql_2a36...f837054_l3.png

Let $\displaystyle H_n$ be the $\displaystyle n$th harmonic number. I started with induction and easily arrived at $\displaystyle H_n$ is not an integer so assume that $\displaystyle H_n=\frac{p}{q}$ where $\displaystyle \gcd(p,q)=1$. Then $\displaystyle H_{n+1}=\frac{p}{q}+\frac{1}{n+1}$.

But I realize now that this is a dead end. How do I proceed? I know I am trying to show that the denominator of the sum is even while the numerator is odd.

Re: Show that the nth harmonic number is not an integer

While the internet is useful, the site you have linked seems to be a collection of rough sketches of proofs; which I had a bit of trouble fleshing out. When I tried to follow similar arguments to construct a proof, I found that things got very messy.

I did several internet searches before I made my post, but all the arguments I found seemed either: ad-hoc, or dependent on p-adic arguments. And I don't know what p-adic is so...

I have constructed the following proof. I didn't see anything like this in the online proofs that I found, so I'm worried that my proof is incomplete.

http://quicklatex.com/cache3/ql_02b8...d33783a_l3.png

Re: Show that the nth harmonic number is not an integer

If D={1,2}, then m- ½=1=an integer. It’s a trivial case.

m = r/s + 1/p

=>(pr+s)/ps=m

=>s = mps-pr

=>s=p(ms-r)

i.e. p|s which is a contradiction due to the defn of s. This might suffice the proof. Your approach was excellent.