Chapter 345 (1/2)

Chapter 348

inspiration is always coming, and I can't prevent it!

Cheng Nuo's mouth slightly tick, turn the page back to the original page.

Since the proving process of Bertrand hypothesis given by Chebyshev is so complicated, I'd like to challenge myself to see if I can prove Bertrand hypothesis in simpler mathematical language.

By the way, to verify, this year's in-depth study, their ability to what extent.

A simple proof method of Bertrand hypothesis.

This paper title alone is enough to be called a district level paper. Of course, the premise is that Cheng Nuo can really explore that simple solution.

As Cheng Nuo had assumed before. The proving process of every conjecture or hypothesis in the mathematical field is a process from the starting point to the end point. Some routes are tortuous, others are straight.

Perhaps, Chebyshev found the more tortuous route, while Cheng Nuo needs to open up a more simple road on the basis of predecessors.

But it's simpler than proving Bertrand's hypothesis alone.

After all, it is standing on the shoulders of giants to look at the problem. With the proof scheme proposed by Chebyshev, the ”pioneer”, Cheng Nuo can more or less learn something from it and make a unique understanding.

Do what you want!

Cheng Nuo is not such a hesitant person. In any case, there is plenty of time for Cheng Nuo to find another paper direction after finding out that ”this road is blocked”.

If we want to put forward a more simple scheme, we should first understand the ideas of proof put forward by predecessors.

Instead of rushing into his own research, he bent down and read the dozen pages of Bertrand's hypothesis from beginning to end.

Two hours later, Cheng closes the book.

After a few seconds of contemplation, he took out a pile of blank draft paper from his schoolbag, picked up the black carbon pen on the desk, and concentrated on his deduction:

to prove Bertrand hypothesis, he must prove several auxiliary propositions.

Lemma 1: [lemma 1: let n be a natural number and p be a prime number, then the highest power of P that can divide n! Is: S = ∑ I ≥ 1floor (NPI) (where floor (x) is the largest integer not greater than x)]

here, we need to arrange all (n) natural numbers from 1 to N on a straight line, and stack a column of Si markers on each number. Obviously, the total number of marks is s.

The relation s = ∑ 1 ≤ I ≤ NSI means that the number of marks (i.e. SI) of each column is calculated first and then the sum is obtained. The relation thus obtained is lemma 1.

Lemma 2: [let n be a natural number and p be a prime number, then Π P ≤ np4n]

use mathematical induction. When n = 1 and N = 2, the lemma obviously holds. Suppose lemma holds for NN (n; 2), let's prove the case of n = n.

If n is even, then Π P ≤ NP = Πp ≤ n-1p, the lemma is obvious.

If n is odd, let n = 2m + 1 (m ≥ 1). It is noted that all prime numbers m + 1p ≤ 2m + 1 are factors of the combinatorial number (2m + 1)! M! (M + 1)! On the other hand, the combinatorial number (2m + 1)! M! (M + 1)! Appears twice in the binomial expansion (1 + 1) 2m + 1, so (2m + 1)! M! (M + 1)! ≤ (1 + 1) 2m + 12 = 4m.

in this way, we can

Cheng Nuo thought smoothly, almost without much effort, he used his own method to prove these two auxiliary propositions.

Of course, this is just the first step.

According to Chebyshev's idea, we need to introduce these two theorems into the proof step of Bertrand hypothesis.

Chebyshev's method is hard to gather, yes, it's hard to gather!

Through the continuous transformation between formulas, one or several necessary and sufficient conditions of Bertrand's hypothesis are transformed into lemma one or lemma two, and then the solution is simplified and integrated.

Of course, Cheng can't do that.