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Submitted by : (unknown) at: 2007-11-17T22:15:40-08:00 (14 years ago)
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I just ran across the following astonishing bug:

s :=-x^3+1/6*(-2*sqrt(6)+2*sqrt(3)+3*sqrt(2))*x^2+1/6*((sqrt(3)+sqrt(2))*sqrt(6)-2*sqrt(2)*sqrt(3))*x-sqrt(2)*sqrt(3)*sqrt(6)/6

-{{x}^{3}}+{{{-{2 \ {\sqrt{6}}}+{2 \ {\sqrt{3}}}+{3 \ {\sqrt{2}}}}\over 6}\ {{x}^{2}}}+ 
{{{{{\left({\sqrt{3}}+{\sqrt{2}}\right)}\ {\sqrt{6}}}-{2 \ {\sqrt{2}}\ {\sqrt{3}}}}\over 6}\  x}-{{{\sqrt{2}}\ {\sqrt{3}}\ {\sqrt{6}}}\over 6}
Type: Polynomial(AlgebraicNumber?)
factor s

\label{eq2}- 1(2)
Type: Factored(Polynomial(AlgebraicNumber?))

This is the same problem as 191ExquoAndThereforeGcdCannotHandleUPXEXPRINT. Namely, the roots sqrt(2), sqrt(3) and sqrt(6) are dependent, which cause problems because sqrt(6)^2 = 6 = 23 = sqrt(2)^2sqrt(3)^2 but we do not know if sqrt(6) = sqrt(2)sqrt(3) or sqrt(6) = -sqrt(2)sqrt(3). This effectively creates ring with zero divisors, while factoring/GCD routines assume a field.

There are several things to notice, in fact:

  • The factorisation is nonsense
  • I think that AlgebraicNumber? should be able to simplify \sqrt{2}\sqrt{3}\sqrt{6} to \sqrt{36}
  • shouldn't sqrt{36} be simplified to 6? Usually, sqrt denotes the positive square root.


In fact, the problem shows already with

s :=x^2-sqrt(2)*sqrt(3)*sqrt(6)

\label{eq3}{{x}^{2}}-{{\sqrt{2}}\ {\sqrt{3}}\ {\sqrt{6}}}(3)
Type: Polynomial(AlgebraicNumber?)

and it seems to occur in 'InnerAlgFactor?':

  )tr InnerAlgFactor  )ma

Status: open => duplicate

  Subject:   Be Bold !!
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