Re: [Axiom-developer] Zero divisors in Expression Integer

[William Sit]

On Thu, 4 Jan 2007 05:47:29 +0100 (CET)
 Waldek Hebisch <address@hidden> wrote:
> I have already written that due to incomplte 
> simplification we
> may get zero divisors in Expression Integer.  Below an 
> easy
> example that multiplication in Expression Integer is 
> nonassociative
> (or, if you prefer, a proof that 1 equals 0):
>
> (135) -> c1 := sqrt(2)*sqrt(3*x)+sqrt(6*x)
>
>            +--+    +-+ +--+
>    (135)  \|6x  + \|2 \|3x
>                                                      Type: 
> Expression Integer
> (136) -> c2 := sqrt(2)*sqrt(3*x)-sqrt(6*x)
>
>              +--+    +-+ +--+
>    (136)  - \|6x  + \|2 \|3x
>                                                      Type: 
> Expression Integer
> (137) -> (1/c1)*c1*c2*(1/c2)
>
>    (137)  1
>                                                      Type: 
> Expression Integer
> (138) -> (1/c1)*(c1*c2)*(1/c2)
>
>    (138)  0
>                                                      Type: 
> Expression Integer

But this is not just an Axiom problem. Mathematica does 
the same thing, with a slight variation on input:
a1 = Sqrt[2]*Sqrt[3 Sqrt[5x + 7] + 6] - Sqrt[6Sqrt[5x + 7] 
+ 12]
a2 = Sqrt[2]*Sqrt[3 Sqrt[5x + 7] + 6] + Sqrt[6Sqrt[5x + 7] 
+ 12]
(1/a1)*a1*a2*(1/a2)  (* answer 1 *)
(1/a1)*(a1*a2 // Simplify)*(1/a2)  (*answer 0, Simplify is 
needed to get this *)

The problem seems to be the lack of a canonical form for 
radical expressions and an algorithm to reduce expressions 
to canonical form. A related problem is lack of algorithm 
to test zero. Another is denesting of a nested radical 
expression. These problems have been studied by Zippel, 
Landau, Tulone et al,  Carette and others.

William