Bug in a solve function - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Bug in a solve function (/thread-13955.html) Bug in a solve function - rombva - 11-07-2019 10:04 AM Try to solve the inequality: (x^3-1)*(x^2-4*x+4)/(x^2-5*x+6)>=0 The Prime's CAS returns the wrong answer {x>=1 AND x<=2, x>3}. The right answer must be a {x>=1 AND x<2, x>3}. Xcas returns the same result. Point x=2 must not be in the solution list. The same issue appears with the equation x/x=1. Prime's CAS and Xcas return the wrong answer {x}. Point x=0 must not be in the solution list. NOT A Bug in a solve function - parisse - 11-07-2019 10:13 AM This is for the same reason as for your other posts, the expressions are considered like polynomials, therefore simplification happens. And this is in my opinion much better, because excluding these values just because they are roots of the denominator, without taking care of the numerator, is artificial. RE: Bug in a solve function - rombva - 11-07-2019 10:36 AM (11-07-2019 10:13 AM)parisse Wrote:  This is for the same reason as for your other posts, the expressions are considered like polynomials, therefore simplification happens. And this is in my opinion much better, because excluding these values just because they are roots of the denominator, without taking care of the numerator, is artificial. Let's substitute point x=0 to the equation x/x=1. We'll get undef=1, but if x=0 is the solution of this equation, then the left side must be equal to the right side. Simplifying (Reducing) fractions: Dividing the numerator and denominator of a fraction by the same non-zero number will also yield an equivalent fraction. In these cases (my examples) simplification cancels this rule. Why? RE: Bug in a solve function - parisse - 11-07-2019 07:45 PM When you type x/x and run a simplification, x is a polynomial. This is implicit when you run most of the commands of a CAS, like solve. This is an algebraic process, but is also true if you consider the limit as x->0 or look at the graph of 'x/x'. That's the reason why I say that excluding x=0 is artificial (0 is a removable singularity). RE: Bug in a solve function - rombva - 11-08-2019 03:57 AM Thanks.