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## [Axiom-developer] [Q] How to classify radicalSolve(z^(1/1)=1) -> "There

**From**: |
Vladimir Bondarenko |

**Subject**: |
[Axiom-developer] [Q] How to classify radicalSolve(z^(1/1)=1) -> "There are..." ? |

**Date**: |
Sat, 12 Feb 2005 21:46:05 +0200 |

Good Day,
Should the behaviour (C) be interpreted as a bug or as a feature?
.......................................................................
(A) -> radicalSolve(z^1=1)
[z= 1]
.......................................................................
(B) -> solve(z^1=1)
[z= 1]
-> solve(z^(1/1)=1)
[z= 1]
.......................................................................
but
.......................................................................
(C) -> radicalSolve(z^(1/1)=1)
There are 4 exposed and 0 unexposed library operations named
radicalSolve having 1 argument(s) but none was determined to be
applicable. Use HyperDoc Browse, or issue
)display op radicalSolve
to learn more about the available operations. Perhaps
package-calling the operation or using coercions on the arguments
will allow you to apply the operation.
Cannot find a definition or applicable library operation named
radicalSolve with argument type(s)
Equation Expression Integer
Perhaps you should use "@" to indicate the required return type,
or "$" to specify which version of the function you need.
.......................................................................
The same with
radicalSolve(z^(2/2)=1)
radicalSolve(z^(3/3)=1)
radicalSolve(z^(4/4)=1)
radicalSolve(z^(5/5)=1) etc
Much thanks in advance for the help.
Best wishes,
Vladimir Bondarenko

**[Axiom-developer] [Q] How to classify radicalSolve(z^(1/1)=1) -> "There are..." ?**,
*Vladimir Bondarenko* **<=**