Prime between n and 9 n divided by 8

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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 $\map P n$ be the property:

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


First note that $\map P n$ 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:

\(\ds 48\) \(=\) \(\ds 2^4 \times 3\) is not prime
\(\ds 49\) \(=\) \(\ds 7^2\) is not prime
\(\ds 50\) \(=\) \(\ds 2 \times 5^2\) is not prime
\(\ds 51\) \(=\) \(\ds 3 \times 17\) is not prime
\(\ds 52\) \(=\) \(\ds 2^2 \times 13\) is not prime

We have that:

$\dfrac {9 \times 47} 8 = 52 \cdotp 875$

and so it is seen that $\map P {47}$ does not hold.


This theorem requires a proof.
In particular: It remains to be shown that $\map P n$ holds for all $n \ge 48$ as well.
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Sources

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