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. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $48$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $48$