# Mills' Theorem/Lemma 1

## 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.

$p_{n + 1} - p_n < K {p_n}^{5 / 8}$

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:

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

$\blacksquare$