e is not rational.

I'm confused - he's addressing the point made in the article by Ed Pegg that things are interseting if you allow log and sqrt. Surely that means he's allowed to use log and sqrt, and in particular, to use them in rational expressions.

So I don't really understand what your point is.

The point of the exercise is to find a short approximation of pi, no? If you allow the use of e, then you can define pi. What he wrote is not an approximation, it is pi. Surely at that point you've defeated the point of the exercise.

This is more-or-less my point.

Once you allow sqrt and ln (or sqrt, log, and e) the problem is silly. He explicitly allows, see the bottom of the article, sqrt, log, and irrational numbers.

I think it's reasonable to assume he did not introduce imaginary and transcendental numbers.

Not really.

Once he introduces the square root he introduces imaginary numbers, sqrt(-1), and transcendental numbers, for example the Gelfond–Schneider constant 2^sqrt(2).

Now we're really down the rabbit hole of someone else's intent, but personally, I assumed he was still trying to maintain some restriction. So, no imaginary numbers, no trascendental numbers. It's easy to restrict what we take the root of, and what we do with the potential irrational result of such roots, to ensure that. As you pointed out, to not do so defeats the purpose of the exercise, and I think my assumption is both reasonable and charitable.

ln(-1)/sqrt(-1)

Maybe you're just trolling, but rational numbers are numbers that can be expressed as p/q, where p and q are integers. Neither i e nor i is an integer, so your proposed quotient has no bearing on the (ir)rationality of e.

I'll assume you are not trolling. My point is that

log(-1)/(sqrt(-1)*log(e)) = ln(-1)/sqrt(-1)

I am just writing the same equation using a log with another base so I don't have e in the equation explicitly.

The natural log - ln - is not rational, as it is the logarithm with base e. That is, ln(x) answers the question, to what power would we have to raise e in order for it to equal x?

Two points:

1. As soon as you allow square roots, you allow irrational numbers. (This sqrt(2).)

2. "The natural log - ln - is not rational" is a different statement than "e is irrational". A rational function is one that can be written as the ratio of two polynomials.

Regarding point 1, you had many things wrong. I picked what was the most obvious to me at the moment - in order for it not to apply, there only needs to be one thing wrong with it. And my point with the natural logarithm is that once you introduce it, you have introduced an irrational number. Overall, I'm not sure what your point has been.

Which ln are you using? Or for that matter, which sqrt(-1) are you using?