# Number of Primes is Infinite/Proof 3

## Theorem

The number of primes is infinite.

## Proof

Aiming for a contradiction, suppose that there are only $N$ prime numbers.

Let the set of all primes be:

$\Bbb P = \set {p_1, p_2, \ldots, p_N}$

By the Fundamental Theorem of Arithmetic, every integer $k > 1$ can be expressed in the form:

$k = {p_1}^{a_1} {p_2}^{a_2} \dotsm {p_N}^{a_N}$

Let $n > 1$ be fixed.

Let $a$ be the largest exponent occurring in the prime decomposition of all positive integers $k \le n$.

Then:

$\displaystyle \sum_{k \mathop = 1}^n \frac 1 k \le \prod_{j \mathop = 1}^N \paren {\sum_{k \mathop = 0}^a \frac 1 { {p_j}^k} }$

is expressible in the form:

 $(1):\quad$ $\displaystyle 1 + \frac 1 2 + \frac 1 3 + \dotsb + \frac 1 n$ $\le$ $\displaystyle \paren {1 + \frac 1 {p_1} + \frac 1 { {p_1}^2} + \dotsb + \frac 1 { {p_1}^a} }$ $\displaystyle$  $\, \displaystyle \cdot \,$ $\displaystyle \paren {1 + \frac 1 {p_2} + \frac 1 { {p_2}^2} + \dotsb + \frac 1 { {p_2}^a} }$ $\displaystyle$  $\, \displaystyle \cdot \,$ $\displaystyle \ \dotsb$ $\displaystyle$  $\, \displaystyle \cdot \,$ $\displaystyle \paren {1 + \frac 1 {p_N} + \frac 1 { {p_N}^2} + \dotsb + \frac 1 { {p_N}^a} }$

which can be seen by multiplying out the factors on the right hand side.

$1 + x + x^2 + \dotsb = \dfrac 1 {1 - x}$

for all $x$ such that $\left\vert{x}\right\vert < 1$.

Thus the factors in $(1)$ are less than the numbers:

$\dfrac 1 {1 - 1 / p_1}, \dfrac 1 {1 - 1 / p_2}, \dotsb, \dfrac 1 {1 - 1 / p_N}$

and so:

$1 + \dfrac 1 2 + \dfrac 1 3 + \dotsb + \dfrac 1 n < \dfrac {p_1} {p_1 - 1} \dfrac {p_2} {p_2 - 1} \dotsm \dfrac {p_N} {p_N - 1}$

We have chosen $n > 1$ arbitrarily, so this holds for every $n > 1$.

This contradicts the result Harmonic Series is Divergent.

Hence the result, by Proof by Contradiction.

$\blacksquare$

## Historical Note

This proof was devised by Leonhard Paul Euler.