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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

$\int_{1}^{a}{\frac{1}{x}dx}+\int_{1}^{b}{\frac{1}{x}dx}=\int_{1}^{n}{\frac{1}{x}dx}$

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!

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  • $\begingroup$ $\int_b^n \frac1x dx = \ln n - \ln b = \ln (n/b) = \ln a$ 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
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    $\begingroup$ 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
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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

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

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  • $\begingroup$ 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

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