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## RE: [Axiom-developer] [Q] Explanation of 1+z==z+1 impact ?

**From**: |
Page, Bill |

**Subject**: |
RE: [Axiom-developer] [Q] Explanation of 1+z==z+1 impact ? |

**Date**: |
Fri, 18 Feb 2005 06:06:51 -0500 |

On Friday, February 18, 2005 3:53 AM Vladimir Bondarenko wrote:
>* ...*
>* -> 1+z==z+1*
>* Compiled code for + has been cleared.*
>* 1 old definition(s) deleted for function or rule +*
>* *
>* Type: Void*
>
Wow! Did you really want to re-define the function + ??
What do you expect '1+z==z+1' to mean? As a function
definition it seems to have an infinite recursion.
Try this instead
-> 1+z==2*z
Then you can compute
-> 1+3
6
but
-> 2+3
is still undefined.
>* > integrate(1+z, z)*
>* Compiling function + with type (PositiveInteger,PositiveInteger)*
>* -> Float*
>* Compiling function + with type (Variable z,PositiveInteger)*
>* -> Float*
>* *
>* Compiling function + with type (PositiveInteger,Variable z)*
>* -> Float*
>* *
>* *
>* The function + is not defined for the given argument(s).*
>* *
With + defined as
-> 1+z==2*z
then
>* integrate(1+z, z)*
z^2
Does that make sense?
Regards,
Bill Page.