Harmonic Number is not Integer/Proof 2

Proof

 * $(\text P): \quad \exists m \in \N: H_m \in \Z$
 * $(\text P): \quad \exists m \in \N: H_m \in \Z$

By the definition of the harmonic numbers:


 * $1 + \dfrac 1 2 + \dfrac 1 3 + \cdots +\dfrac 1 m = H_m$

$m$ is either prime or composite.

If $m$ is prime, we have that:

If $k \le n$, then $k$ divides $n!$, so all terms are integers.

Every term on the is divisible by $m$ except for one, namely $\paren {m - 1}!$.

Hence, the is not divisible by $m$.

But the is divisible by $m$, so this is a contradiction.

If $m$ is composite, let $p$ be the largest prime which is less than $m$.

Then we have:

For $k < p$, we have that $m!$ is divisible by $p$ and $k$ is not.

So from Euclid's Lemma, every term $\dfrac {m!} k$ is an integer divisible by $p$.

there exists $k \in \Z$, with $p< k< m$ such that $\dfrac {m!} k$ is not divisible by $p$.

Because $m!$ is divisible by $p$ and $\dfrac{m!} k$ is not, it follows from Euclid's Lemma that $k$ is a multiple of $p$.

Since $k$ is a multiple of $p$ which is greater than $p$:
 * $2 p \le k$

Thus:
 * $p < 2 p \le k < m$

From Bertrand's Postulate, there exists a prime $q$ such that $p < q < 2 p$.

Hence:
 * $p < q <m $

This contradicts the fact that $p$ is the largest prime less than $m$.

Hence, the assumption that there exists $k \in \Z$ with $p < k < m$ such that $\dfrac {m!} k$ is not divisible by $p$ is false.

Therefore, every term of the, except perhaps $\dfrac{m!} p$, is a multiple of $p$.

From above:
 * $p < m < 2 p$

so $p$ is the only positive integer less than $m$ which is divisible by $p$.

Since $m!$ is the product of positive integer less than or equal to $m$, it follows that $m!$ is divisible by $p$ only once.

Therefore, $\dfrac {m!} p$ is not divisible by $p$.

Because every term of the is divisible by $p$ except for one, it follows that the  is not divisible by $p$.

Because the is a multiple of $m!$, and hence a multiple of $p$, this is a contradiction.

So the assumption that there exists $m \in \N: H_m \in \Z$ is false.