# Why The Number of Primes Could Not Be Finite?

This is a little shortened version of the paper by Filip Saidak, __A note on Euclid’s Theorem concerning the infinitude of the primes__ (*Acta Universitatis Matthiae Belii*, series Mathematics, Volume 24 (2016), 59–60). As the paper shows, no finite number of primes could generate the whole set of natural numbers. So, the short answer to the question in the title is "There would not be enough of them!"

We investigate the factorization geometrically and consider the canonical representation as an operation (on exponents) in two dimensions, with single prime powers representing what we will call the "vertical" and their products the "horizontal" dimensions.

### Vertical Dimension

For a fixed prime number $p,\,$ and $0\le i\le m,\,$ there are $m + 1\,$ positive integers that can be written in the form $p^i,\,$ the largest of which is $p^m.\,$ Since, clearly, $m + 1\le (1 + 1)^m = 2^m \le p^m,\,$ many integers are not of this form; so for the proportion $\nabla(p^m)\,$ of these powers (up to $p^m)\,$ we not only have $\nabla(p^m)\lt 1,\,$ for all $m \gt 1\,$ (as well as $\nabla(p^m)\rightarrow 0,\,$ as $m\rightarrow\infty),\,$ but also $\nabla(p^m)\gt \nabla(p^{m+1}),\,$ because

(1)

$\displaystyle \frac{m + 1}{p^m}\gt\frac{m + 2}{p^{m+1}} \Longleftrightarrow 1 -\frac{1}{m + 2}\gt\frac{1}{p}.$

Thus, considered vertically, the proportions are monotonically decreasing.

### Horizontal Dimension

Recall that a function $f:\,\mathbb{N}\rightarrow\mathbb{C}\,$ is called multiplicative, if $f(1)=1\,$ and $f(ab)=f(a)f(b),\,$ for all coprime $a,b\in\mathbb{N}.\,$ A critically important property of the proportions $\nabla\,$ is their multiplicativity. For all $k\ge 2,\,$ let us define

$\displaystyle \nabla(p_1^{m_1}\cdots p_k^{m_k}) := \frac{\#\{n=p_1^{a_1}\cdots p_k^{a_k}:\,0\le a_j\le m_j,\text{for}\,1\le j\le k\}}{p_1^{m_1}\cdots p_k^{m_k}},$

then, for all permutations of exponents $m_j,\,$ we have

(2)

In other words, the multiplicativity of $\nabla\,$ implies the horizontal monotonicity.

Combining these two monotonic orthogonal trends is enough to prove the infinitude of the prime numbers. This is because the vertical dimension is (trivially) infinite, and (1) implies an ever increasing sparseness of integers represented by a given prime power; while from the monotonicity property of (2) it follows that the same will remain true upon any finite composition of such powers, and therefore only an infinite horizontal dimension could possibly compensate for the growing deficit and create a complete cover of $\mathbb{N},\,$ guaranteed by the unique factorization theorem.

- Infinitude of Primes
- Infinitude of Primes - A Topological Proof
- Infinitude of Primes - A Topological Proof without Topology
- Infinitude of Primes Via *-Sets
- Infinitude of Primes Via Coprime Pairs
- Infinitude of Primes Via Fermat Numbers
- Infinitude of Primes Via Harmonic Series
- Infinitude of Primes Via Lower Bounds
- Infinitude of Primes - via Fibonacci Numbers
- New Proof of Euclid's Theorem
- Infinitude Of Primes From Legendre's Formula
- Infinitude of Primes - via Bertrand's Postulate
- Infinitude of Primes - An Impossible Injection
- Infinitude of Primes - from Infinitude of Mutual Primes
- Infinitude of Primes Via Euler's Product Formula
- Infinitude of Primes Via Euler's Product Formula for Pi
- Infinitude of Primes Via Powers of 2
- Infinitude of Primes As a Quickie
- A One-Line Proof of the Infinitude of Primes
- Infinitude of Primes - A. Thue's Proof
- Why The Number of Primes Could Not Be Finite?

|Contact| |Front page| |Contents| |Algebra| |Up|

Copyright © 1996-2017 Alexander Bogomolny

62621329 |