Mills' Theorem/Lemma 1
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Lemma for Mills' Theorem
Let:
- $\N$ denotes the set of all natural numbers
- $\Bbb P$ denotes the set of all prime numbers.
where $K$ is an unknown but fixed positive integer.
Let $p_n$ be the $n$th prime number.
From Difference between Consecutive Primes:
- $p_{n + 1} - p_n < K {p_n}^{5 / 8}$
where $K$ is an unknown but fixed positive integer.
Then we have that:
- $\forall N > K^8 \in \Z: \exists p \in \Bbb P: N^3 < p < \paren {N + 1}^3 - 1$
Proof
Let $p_n$ be the greatest prime less than $N^3$.
\(\ds N^3\) | \(<\) | \(\ds p_{n + 1}\) | because $p_n$ is the greatest prime less than $N^3$ | |||||||||||
\(\ds \) | \(<\) | \(\ds p_n + K {p_n}^{\frac 5 8}\) | Difference between Consecutive Primes | |||||||||||
\(\ds \) | \(<\) | \(\ds N^3 + K N^{\frac {15} 8}\) | because $p_n < N^3$ | |||||||||||
\(\ds \) | \(<\) | \(\ds N^3 + N^2\) | because $N > K^8$ | |||||||||||
\(\ds \) | \(<\) | \(\ds N^3 + 3N^2 + 3N\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {N + 1}^3 - 1\) |
Therefore:
- $N^3 < p_{n + 1} < \paren {N + 1}^3 - 1$
$\blacksquare$