ProofOfInfinitudeOfPrimes 2007-05-19 16:23:42 2007-05-19 20:17:56 Proof prime 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here We begin by noting a fact about factorizations. Suppose that $n > 0$ is an integer which has a prime factorization \[ n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m} . \] Then, because $2$ is the smallest prime number, we must have \[ p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m} \ge 2^{k_1} 2^{k_2} \cdots 2^{k_m}, \] so $n \ge 2^{k_1 + k_2 + \cdots + k_m}$. Assume that there were only a finite number of prime numbers $p_1, p_2, \ldots p_m$. By the above-noted fact, given an integer $j > 0$, every integer $n > 0$ could be expressed as \[ n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m} \] with \[ k_1 + k_2 + \cdots + k_m \le j . \] This, however, leads to a contradiction because it would imply that there exist more integers than possible factorizations despite the fact that every integer is supposed to have a prime factorization. To see this, let us over-count the number of factorizations. A factorization being specified by an $m$-tuplet of integers $k_1, k_2, \ldots, k_n$ such that $k_1 + k_2 + \cdots + k_m \le j$, the number of factorizations is equal to the number of such tuplets. Now, for all $i$ we must have $0 \le k_i \le j$, so there are not more than $(j+1)^m$ such tuplets available. However, for all $m$, one can choose $j$ such that $2^j > (j+1)^m$. For such a choice of $j$ we could not make ends meet --- there are not enough possible factorizations available to handle all integers, so we conclude that there must be more than $m$ primes for any integer $m$, i.e. that the number of primes is infinite. To make this exposition self-contained, we conclude with a proof that, for every $m$, there exists a $j$ such that $2^j > (j+1)^m$. We begin with the case $m=1$ and showing that, for every integer $a \ge 2$, we have $2^a > a+1$. This is an easy induction. When $a = 2$, we have $2^2 = 4 > 3 = 2 + 1$. If $2^a > a+1$ for some $a > 2$, then $2^a + 1 > a + 2$. By our hypothesis, $2^a \ge 1$, so $2^a + 1 \ge 2^a + 2^a = 2^{a+1}$, hence $2^{a+1} > (a+1)+1$. From this starting point, we obtain the desired inequality by algebraic manipulation. Setting $a = m - 1$, we have $2^{m-1} > m$, or $2^m > 2 m$. Raising both sides to the $2m$-th power, $2^{2 m^2} > 2^{2m} m^{2m} = 2^m (2 m^2)^m \ge 2 (2 m^2)^m$. Setting $j = (2 m^2 - 1)$, this becomes $2^j > (j+1)^2$.