# Elementary insight into Bertrand's postulate

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• Jun 18th 2009, 01:11 PM
Bruno J.
Elementary insight into Bertrand's postulate
Bertrand's postulate states that for any positive integer $\displaystyle n>1$ there is a prime with $\displaystyle n<p<2n$.
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 $\displaystyle I$ of integers. Call $\displaystyle s_p$ the proportion of integers in $\displaystyle I$ which are multiples of $\displaystyle p$. Then

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

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

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

Note that we have the ridiculously bad bound

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

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