# Elementary insight into Bertrand's postulate

• Jun 18th 2009, 01:11 PM
Bruno J.
Elementary insight into Bertrand's postulate
Bertrand's postulate states that for any positive integer $n>1$ there is a prime with $n.
The following is not a rigorous proof but rather something I have thought of which makes the theorem intuitively evident, unlike the rigorous proofs which exist (all of which I have seen are very elegant, but tackle the problem in a roundabout fashion).

Suppose we are given an interval $I$ of integers. Call $s_p$ the proportion of integers in $I$ which are multiples of $p$. Then

$s_p\leq \frac{1}{p}$
is trivial. In perticular

$R(n) = \prod_{p\leq n}\Big(1-\frac{1}{p}\Big)$

can be thought of as the proportion of all integers which are not divisible by any of the primes less than $n$. If we have an interval containing $k$ numbers, we can expect approximately $kR(n)$ of those numbers to be divisible by none of the primes less than $n$.

Note that we have the ridiculously bad bound

$R(n) = \prod_{p\leq n}\Big(1-\frac{1}{p}\Big) \geq \prod_{j=2}^n\Big(1-\frac{1}{j}\Big) = \frac{(n-1)!}{n!}=\frac{1}{n}$

so that in perticular any interval containing $n$ numbers should certainly be expected to contain an integer not divisible by any prime less than $n$, since $nR(n)\geq 1$. In perticular, when this is applied to the interval $]n,2n]$, we see that it should very reasonably contain such an integer, which, in this case, would mean a prime.