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In this post I proved that ln(3)/ln(2) is transcendental and an immediate corollary is that ln(x)/ln(y) is transcendental where x and y are natural numbers $ x,y \neq 0,1$ if x is odd and y is even or if x is even and y is odd, i.e. when a and b have opposite parities.

In an answer, SmileyCraft generalised this result further, stating

Looks correct to me, and I think it generalizes to ln(a)/ln(b) being transcendental for any positive integers a,b that admit no solution to $a^p=b^q$ for positive integers p,q .

Whether such a pair p,q exists can be determined from the prime factorization of a,b . The pair exists if and only if the ratio between the powers of each prime is the same.

Now SmileyCraft specifically states a and b are positive integers, but I am wondering if this result can be further generalised to:

ln(a)/ln(b) being transcendental for any positive rational numbers a,b that admit no solution to $a^p=b^q$ for positive integers p,q.

An initial attempt to explore extending prime factorisation to cover rational numbers would be to extend the prime factors to the left of one like so:

$.., \frac 1 {11}, \frac 1{7}, \frac 1{5}, \frac 1{3}, \frac 1{2}, 2, 3, 5, 7, 11,..$

so e.g. $7*3/49$ could be expressed as $7^1 * 3^1 * {(1/7)}^2$ but this a bit messy and does not have a unique representation. We can however note that $1/11, 1/7$ etc can be stated as ${11}^{(-1)}$ and $7^{(-1)}$ and so on. Our expression $7*3/49$ can now be expressed as $7^{(-2)} * 3^{(1)} * 7^{(1)} = 3^1 * 7^{-1}$ which I think is now a unique prime power factorisation of a rational number, although it is not specifically stated in the fundamental theorem of arithmetic.

Edit: I later found this statement in the Wikipedia article:

Allowing negative exponents provides a canonical form for positive rational numbers

Now since we are extending to rational values of a and b we can declare a = m/n and b = r/s where m, n, r, s are positive integers.

Is it safe to conclude that $ln(a)/ln(b) = ln(m/n)/ln(r/s)$ is transcendental if $a^p = b^q$ admits no solutions, simply by using the extended definition of prime power factorisation of rational numbers and SmileyCraft's method?

KDP
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  • I think you are trying to make the point that for example ln(r) where r is a rational can be represented in several different ways e.g. ln(a/b) or ln((ps)/(qr)) but I would argue that if r is represented as a "prime factorisation" then r has a unique representation even when r is not an integer (as long as we restrict ourselves to positive rational numbers) . For example is there any other way to represent 5/9 as a prime factorisation, other than $r = 3^{-2}*5^1$ ? Any counter examples? – KDP Oct 18 '23 at 01:35

1 Answers1

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Suppose $x \ln(b) = \ln(a)$, where $a$ and $b$ are positive and not equal to $1$. Thus $a = b^x$.
There are three possibilities:

  • Case 1: $x$ is rational. Then we can write $x = m/n$ for integers $m$ and $n$, and then $a^n = b^m$.
  • Case 2: $x$ is an algebraic irrational. Then at most one of $a$ and $b$ can be algebraic, because if $b$ is algebraic, the Gelfond-Schneider theorem says $a$ is transcendental.
  • Case 3: $x$ is transcendental.
Robert Israel
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