3 ms·
ln(-1)/sqrt(-1)
by kdavis 14y ago
ln(-1)/sqrt(-1)
- scott_s 14y agoThe 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?
- kdavis 14y agoTwo 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.
- scott_s 14y agoRegarding 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.
- ColinWright 14y agoWhich ln are you using? Or for that matter, which sqrt(-1) are you using?
- mhartl 14y agoMaybe 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.
- kdavis 14y agoI'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.