About Bertrand's postulate

4

ABOUT BERTRAND’S POSTULATE

M. I. Bashmakov

Among the 69 works of Pafnuty Lvovich Chebyshev, two works played a particularly important role in the development of number theory. These works: “On determining the number of prime numbers not exceeding a given value” and “On prime numbers,” written by him in his youth, are related to the most difficult question in number theory: the distribution of prime numbers.

In the second of these works, inequalities are indicated that make it possible to fairly accurately estimate the number of prime numbers in a given segment of the natural series and, in particular, the following fact is proven: between $x$ and $2x$ (where $x>1$) there is always at least one prime number. This statement was made by the famous French mathematician Bertrand, who tested it within the existing tables of prime numbers, but was unable to prove it and accepted it as a construct.

We will trace Chebyshev’s main ideas that led him to these estimates, and then propose a series of problems whose solutions allow readers (familiar enough with the properties of logarithms) to reconstruct Ukazyv’s proof of Bertrand’s construction and obtain inequalities for the number of prime numbers not exceeding a given value.

The reader may wonder what does logarithm have to do with prime numbers? This is the merit of Chebyshev, that he pointed out how to use the general functions of analysis - roots, logarithms - in the study of problems about integers. If we want to find out whether there is at least one prime number on the interval from $n$ to $2n$, then we can proceed as follows:

1. come up with some function $y=f(x)$ of the real argument $x$ that would not change its value while $x$ runs through the segment between neighboring prime numbers, and would take on a larger value when passing through a prime number;

2. prove inequalities for the relative function $f$, which would help to estimate the growth rate quite accurately.

For a function chosen in this way, the difference $f(2n) - f(n)$ will be a positive number if and only if the function $f$ returns when moving from $n$ to $2n$, that is, if there is at least one prime number between $n$ and $2n$.

The first condition is easy to satisfy. We can consider the following function: let us denote by $\pi(x)$ the number of prime numbers not exceeding $x$.

For example, $\pi(1) = 0$ (one is not considered a prime number), $\pi(3) = 2$, $\pi(9) = 4$, $\pi(9,5) = 4$, $\pi(10,9) = 4$, $\pi(11) = 5$, etc.

Of course, if we could prove that

$$\pi(2x) - \pi(x) > 0,$$

we would have proved Bertrand to act, since

$$\pi(2x) - \pi(x)$$

just denotes the number of prime numbers contained between $x$ and $2x$.