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proof of infinitude of primes∗
rspuzio†
2013-03-21 22:54:57
We begin by noting a fact about factorizations. Suppose that n > 0 is an
integer which has a prime factorization
n = pk11 pk22 · · · pkmm .
Then, because 2 is the smallest prime number, we must have
pk11 pk22 · · · pkmm ≥ 2k1 2k2 · · · 2km ,
so n ≥ 2k1 +k2 +···+km .
Assume that there were only a finite number of prime numbers p1 , p2 , . . . pm .
By the above-noted fact, given an integer j > 0, every integer n > 0 could be
expressed as
n = pk11 pk22 · · · pkmm
with
k1 + k2 + · · · + km ≤ 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 k1 , k2 , . . . , kn such that k1 + k2 + · · · + km ≤ j, the number of factorizations
is equal to the number of such tuplets. Now, for all i we must have 0 ≤ ki ≤ j,
so there are not more than (j + 1)m such tuplets available. However, for all
m, one can choose j such that 2j > (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 2j > (j + 1)m . We begin with the case m = 1
and showing that, for every integer a ≥ 2, we have 2a > a + 1. This is an easy
∗ hProofOfInfinitudeOfPrimesi created: h2013-03-21i by: hrspuzioi version: h39404i
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induction. When a = 2, we have 22 = 4 > 3 = 2 + 1. If 2a > a + 1 for some
a > 2, then 2a +1 > a+2. By our hypothesis, 2a ≥ 1, so 2a +1 ≥ 2a +2a = 2a+1 ,
hence 2a+1 > (a + 1) + 1.
From this starting point, we obtain the desired inequality by algebraic manipulation. Setting a = m − 1, we have 2m−1 > m, or 2m > 2m. Raising both
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sides to the 2m-th power, 22m > 22m m2m = 2m (2m2 )m ≥ 2(2m2 )m . Setting
j = (2m2 − 1), this becomes 2j > (j + 1)2 .
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