# Goldbach implies Bertrand

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## Theorem

Bertrand's Postulate is a logical consequence of the truth of the Goldbach Conjecture:

Bertrand's Postulate: $\forall n \in \N: n > 1:$ between $n$ and $2n$ lies a prime number $p$:
$n < p < 2n$
(Strong) Goldbach Conjecture: Every even number greater than $2$ can be written as a sum of two prime numbers.

## Proof

Assume the truth of the Goldbach Conjecture.

Since $n > 1$, we have that $2 n$ is an even integer greater than $2$.

From the Goldbach Conjecture, $2n = p_1 + p_2$ for some prime numbers $p_1$ and $p_2$.

We have that $p_1, p_2 < n \implies p_1 + p_2 < 2n$.

Therefore, at least one of $p_1, p_2 \ge n$.

That is, $n \le p_i < 2n$.

If $n$ is not a prime, the strict inequality holds.

On the other hand, suppose $n$ is prime.

$(1): \quad$ Consider the case $n = 2$.

For $2 < 3 < 2 \times 2$ (and $3 \in \Bbb P$) Bertrand's Postulate holds with $p=3$.

$(2): \quad$ Otherwise $n > 2$, and so $n$ is an odd prime.

Then $n + 1$ is composite, because of being even.

So $2 \left({n + 1}\right) = p'_1 + p'_2$, where $p'_1$ or $p'_2$ must be greater than $n + 1$.

That is, either $p'_1 > n$ or $p'_2 > n$.

Without loss of generality, suppose $p'_2 > n$.

Then, $n < p'_2 < 2n + 2$.

But $p'_2 \ne 2n + 1$, or else $p'_1 = 1$.

Also $p'_2 \ne 2n$ since $2n$ is composite.

So $n < p'_2 < 2n$.

So, if the Goldbach Conjecture were to be true, it would logically imply the truth of Bertrand's Postulate.

$\blacksquare$