Forgive poor formatting, I don't know how to use mathjax very well. I am a 10th grade student with little experience writing proofs looking for constructive criticism.
I found this problem in the book Solving Mathematical Problems by Terrence Tao. In the book it says that Bertrand's Postulate (for every $n$, where $n$ is an element of $Z^+$, there exists a prime $p$, such that $n<p<2n$) is required to prove it, but--interested--I tried to see if it can be proved in a more elementary manner. I believe I have found one and wish to have it verified and critiqued. I shall proceed:
Theorem. For every $n, q$, where $n$ and $q$ are elements of $Z^+$, $2<=n$, it is true that $1/1+1/2+...+1/n$ does not equal $q$. (Could someone please tell me how to use summation notation in mathjax, thanks).
Proof. We will assume that the sum is equal to an integer, and reach a proof by contradiction with a counterexample. Multiplying the individual terms by each other to make them common in denominator then adding them together results in the fraction:
$2$ x $3$ x ... x $n+1$ x $3$ x ... x $n+ ...+1$ x $2$ x ... x $n-1 /n!$
Assuming this is equal to an inteqer, $q$, then:
$n!q = 2$ x $3$ x ... x $n+1$ x $3$ x ... x $n+ ...+1$ x $2$ x ... x $n-1$
$2(n!/2)q = 2$ x $3$ x ... x $n+1$ x $3$ x ... x $n+ ...+1$ x $2$ x ... x $n-1$ ($n$ is greater than $2$, therefore $n!$ is divisible by $2$).
Therefore, if $q$ is an integer, the numerator must be even for all positive integer $n$. However, taking the counterexample of $n=3$ the numerator equals 11, which is not an even integer. This is a contradiction. Therefore the negation of our assumption must be true, and there do not exist positive integers $q, n$ such that the $nth$ harmonic number equals $q$, as desired.