Prime between n and 9 n divided by 8

Theorem
Let $n \in \Z$ be an integer such that $n > 48$.

Then there exists a prime number $p$ such that $n \le p \le \dfrac {9 n} 8$.

Proof
Let $P \left({n}\right)$ be the property:
 * there exists a prime number $p$ such that $n \le p \le \dfrac {9 n} 8$.

First note that $\dfrac {9 \times 48} 8 = 9 \times 6 = 54$.

We have that:

Thus $P \left({48}\right)$ holds.

We also have:

Thus $P \left({n}\right)$ holds for $33 \le n \le 37$.

Next:

Thus $P \left({n}\right)$ holds for $38 \le n \le 41$.

Then:

Thus $P \left({n}\right)$ holds for $42 \le n \le 47$.

But note that:

and so there is no prime between $32$ and $\dfrac {9 \times 32} 8$.

Hence it has been demonstrated that $P \left({n}\right)$ holds at least for all $n$ such that $33 \le n \le 48$, but not for all $n$ such that $n \le 32$.

It appears that the cutoff point is not $48$ but is instead $32$.