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Nov
1
comment In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$?
@Steven Stadnicki why not? Exponentiation is defined there
Oct
30
comment A definable hyperreal system
@Ian Mateus yes this is exactly what I saw before making this question.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
You are right but this also gives it as an axiom: faculty.uml.edu/klevasseur/courses/92.421/PS1_Fall2011.pdf
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Well the definition of recursive hyper operator just postulates it to be 1 at n=3, b=0 as a special case. It is just another formulation of the same axiom.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Well according to en.wikipedia.org/wiki/Exponential_field, the E(0)=1 axiom is required from a set to be considered exponential field. This hints at that it may be an independent axiom over real numbers.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
By the way, in abstract algebra $x^0=1$ is usually cansidered an axiom en.wikipedia.org/wiki/Exponentiation#In_abstract_algebra
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
"It can be derived from the definitions/axioms for the natural number case." Can you please give a derivation from the standard axioms?
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Re limits: usually, but not necessary. Where do you see it is necessary to define it so?
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Re the link: as I see the definitions on that page are in fact axioms or equivalent to axioms.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
"is not a well formed statement because the limit..." How on earth existence or the value of the limit makes the statement about value of a function in a point well formed or not?
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
"true statement because it is an empty product" that empty product is equal 1 an axiom? If not, how to prove it?
Oct
28
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@Graphth Gnome calculator on Linux. Interestingly, the KDE calculater for a while returned 0 while Gnome calculator returns 1. The KDE calculator has been fixed since following a bugreport.
Oct
28
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@glebovg I thought that the limit exists on the complex plane either. Anyway, existence of the limit IS NOT a proof of the value of the expression.
Oct
28
comment Sex distribution
yes. Can you elaborate on this, that is take into account the typical deviation?
Oct
27
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@Brian M. Scott the limit does not exist. For a limit in a point exist, there should be both the left and right limits.
Oct
27
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@Graphth Windows calculator, for example.
Oct
27
comment Sex distribution
Yes but it only evident from your answer that you're more likely to find yourself in a group with slightly more persons of your sex (that is by 1). But I wonder whether the difference is more sufficient if the distribution is far from uniform.
Oct
27
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@Brian M. Scott -1
Oct
27
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@Brian M. Scott is is very much strange that you do not want to test your claim with actual values.
Oct
27
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@Brian M. Scott "0^x=0 for all non-zero x" - no.