Infinity of primes — proof using calculus

Let us look at a proof that the number of primes is infinite — a proof that uses only elementary calculus.

Let \pi (x) = \# \{ p \leq x: p \in P\} be the number of primes that are less than or equal to the real number x. (Note: here P is the set of primes). We number the primes P = \{ p_{1}, p_{2}, p_{3, \ldots}\} in increasing order. Consider the natural logarithm \log {x} defined as \log {x}=\int_{1}^{x}\frac{1}{t}dt.

Now we compare the area below the graph of f(t)=\frac{1}{t} with an upper step function. Thus, for n \leq x < n+1 we have

\log {x} \leq 1+\frac{1}{2}+\frac{1}{3}+ \ldots + \frac{1}{n-1}+\frac{1}{n} \leq \Sigma \frac{1}{m} where the sum extends over all m \in N which have only prime divisors p \leq x.

Since every such m can be written in a unique way as a product of the form \Pi_{p \leq x}p^{k_{p}}, we see that the last sum is equal to

\Pi_{{p \in P}_{p \leq x}}(\Sigma_{k \geq 0}\frac{1}{p^{k}})

The inner sum is a geometric series with ratio \frac{1}{p}, hence,

\log {x} \leq \Pi_{{p \in P}_{p \leq x}}\frac{1}{1-\frac{1}{p}}=\Pi_{{p \in P}_{p \leq x}}\frac{p}{p-1}=\Pi_{k=1}^{\pi(x)}\frac{p_{k}}{p_{k}-1}

Now, clearly p_{k} \geq k+1 and thus

\frac{p_{k}}{p_{k}-1}=1+ \frac{1}{p_{k}-1} \leq 1 + \frac{1}{k}=\frac{k+1}{k}

and therefore,

\log {x}\leq \Pi_{k=1}^{\pi(x)}\frac{k+1}{k}=\pi(x)+1.

Everybody knows that \log {x} is not bounded, so we conclude that \pi(x) is unbounded as well, and so there are infinitely many primes. QED.

Ref: Proofs from THE BOOK (Martin Aigner and Gunter M. Ziegler) (Third Edition).

More about primes later,

Nalin Pithwa

 

 

 

 

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