Is there an alternative approach to the definition of prime numbers based on the definition of the natural logarithm? These are my thoughts about it, the questions are at the end:

Basically when a natural number $n \gt 1$ is composite at least there are two natural numbers $a,b$ such as: (1) $1 \lt a \le b \lt n$, and (2) $a \cdot b = n$.

If that is the case:

$a\cdot b=n$ then

$Ln(a\cdot b) = Ln(n)$ so

$Ln(a)+Ln(b)=Ln(n)$ and by the definition of the natural logarithm


Now (not sure about this step) this would be equivalent to:

(E) $\int_{1}^{a}{\frac{1}{x}dx} = \int_{b}^{n}{\frac{1}{x}dx}$

So by definition a prime number $p$ would not comply with the expression (E).

$\not \exists a,b, 1 \lt a \le b \lt n, \int_{1}^{a}{\frac{1}{x}dx} = \int_{b}^{p}{\frac{1}{x}dx} \implies p \in \Bbb P$

I would like to ask the following questions:

  1. Is the expression (E) correct or the last step is wrong?

  2. I tried to find this kind of alternative definitions unsuccessfully. Are there any papers or references about this kind of approach (e.g. not specifically the natural logarithm)? Thank you!

  • $\begingroup$ $\int_b^n \frac1x dx = \ln n - \ln b = \ln (n/b) = \ln a$ yes, and if you find nothing on that it is because it is not so helpful : why would it be ? $\endgroup$ – reuns Feb 9 '16 at 8:32
  • $\begingroup$ $ab=n \iff \int_{1}^{a}{\frac{1}{x}dx} = \int_{b}^{n}{\frac{1}{x}dx}$ may be true, but I do not see anything special about the right-hand side in terms of integers rather than real numbers $\endgroup$ – Henry Feb 9 '16 at 8:37
  • $\begingroup$ @user1952009 thanks for the confirmation! probably is not very helpful but sometimes a different point of view provides new insights. For instance the definition of primes turns into a problem of equivalence of areas in this domain. $\endgroup$ – iadvd Feb 9 '16 at 8:38
  • 2
    $\begingroup$ in some sense where it becomes helpful is with the Riemann zeta function, even it is only partly related to what you wrote $\endgroup$ – reuns Feb 9 '16 at 8:38

To answer the first part, this is a valid manipulation of integrals.

To answer the second, there aren't many interesting statements logically equivalent to "$p$ is prime". Yours suffers from a big problem: $a$ and $b$ are integers. Integrals, and calculus in general, really doesn't care about integers, so it doesn't seem like a good place to go looking for an alternative definition of prime, IMO


(E) is correct. But you should not forget about the case where a and b are non-natural divisors of n, so $\not \exists \{a,b\}\subset \mathbf{N}, \ldots$

  • $\begingroup$ thank you for the confirmation! please may I ask you to elaborate that point about the non natural divisors a little bit more? $\endgroup$ – iadvd Feb 9 '16 at 8:46
  • $\begingroup$ For example $n = 2 = \sqrt 2 \sqrt 2$. Then you have $\int_1^{\sqrt 2} 1/x \mathrm{d}x = \int_{\sqrt 2}^2 1/x \mathrm{d}x$, but this does not mean that 2 is not prime. $\endgroup$ – hkBst Feb 9 '16 at 8:53
  • $\begingroup$ understood, in my case I am strictly speaking about the natural numbers. $\endgroup$ – iadvd Feb 9 '16 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.