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:

$N^3 < p_{n + 1} < \paren {N + 1}^3 - 1$

$\blacksquare$