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

[Page, Bill]

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.