Prime between n and 9 n divided by 8

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

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

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 $P \left({n}\right)$ does not hold for the following $n < 48$:
 * $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 14, 15, 19, 21, 23, 24, 25, 31, 32, 47$

Taking $47$ as an example:

We have that:
 * $\dfrac {9 \times 47}8 = 52 \cdotp 875$

and so it is seen that $P \left({47}\right)$ does not hold.