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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 10,082 of 10,432   
   antispam@math.uni.wroc.pl to Nasser M. Abbasi   
   Re: solving sqrt(y)=x for   
   12 Jan 21 17:41:31   
   
   Nasser M. Abbasi  wrote:   
   >   
   > Should CAS solve   
   >   
   >      sqrt(y)=x   
   >   
   > for y by giving solution as y=x^2 without saying this is valid only for x>=0   
   ?   
   >   
   > Only Maxima would ask the user if x was negative,positive or zero when   
   > solving this equation.   
   >   
   > Maple:   
   > =======   
   > eq:=sqrt(y)=x;   
   > PDEtools:-Solve(eq,y);   
   >             y = x^2   
   >   
   > Mathematica:   
   > =============   
   > eq = Sqrt[y] == x;   
   > Solve[eq, y]   
   >                {{y -> x^2}}   
   >   
   > Maxima   
   > =======   
   > (%i1) solve(sqrt(y)=x,y)   
   > Is x positive, negative or zero?   
   >    positive   
   >                                          2   
   > (%o1)                              [y = x ]   
   >   
   > Fricas   
   > ========   
   > solve(sqrt(y)=x,y)   
   >               2   
   >    (4)  [y = x ]   
   >   
   >   
   > At school, if I wrote in the exam that the solution of   
   > sqrt(y)=x is y=x^2 without saying this is for x>=0, the   
   > teacher will take one or more point off.   
   >   
   > Why do then most CAS systems get away with it then? Is this just by   
   convention   
   > then?   
      
   When you solve for 'y' assumption (given) is that sqrt(y) = x.   
   By any reasonable definition of square root you have than   
   x^2 = y.  The sitation is different when you solve for x,   
   that is want to compute square root.  Than you actually   
   have equation to solve, namely equation x^2 = y.   
   Quadratic equation in algebraicaly complete field have   
   two solutions (counting double root (when y = 0) as two   
   solution).  You have than multiple conventions.  One   
   is that 'sqrt' represents one of solutions without   
   saying which one.  This is reasonable because by Galois   
   theory there is no _algebraic_ way to distinguish   
   solutions.  In analysis we deal with functions and   
   natural requirement is that we want continuous functions.   
   Then convention is that root which makes your function   
   continuous is correct one (this is also good convention   
   when solving physical and technical problems).  For   
   holomorphic function this still leaves open which   
   value to take at single point, but once you made   
   choice at single point per connected component values   
   in corresponding connected components are uniquely determined.   
   In numerical computations routines are expected to return   
   a number, so square root function must make a choice   
   and usual convention is to take "principal value".   
   However, this should be seen not as virtue, but   
   as unavoidable limitation: numerical square root   
   may be wrong for actual (say physical) problem   
   that you are solving, and users may be forced to   
   change root to correct one.   
      
   Another problem is that you may work in something   
   which is not algebraically closed, like real numbers.   
   Then eqation y = x^2 is not solvable for x when   
   y < 0.  High school typically avoids talking   
   about complex numbers and only uses real numbers.   
   Hence square root of negative numbers is left   
   undefined (treated as error).  In real numbers   
   there is algebraic distincion between root:   
   for nonzero y one x is negative, other is positive.   
   Positive numbers are defined in algebraic way,   
   because nozero real number y is positive <=>   
   equation y = x^2 is solvable for x.  So in   
   highs schools frequently use convention that   
   square root is nonnegative (forcing choice of   
   square root in this way).  However, even   
   if you start from real numbers you may get   
   complex square roots in formulas and normally   
   results obtained via complex numbers are   
   valid.  And complex way is easier than real   
   one.  Famous "casus irredutiblis" for equations   
   of degree 3 nicely illustrates difficulties   
   you have when you insist on staying within   
   real numbers (there are 3 real solutions,   
   but usual formulas produce complex numbers   
   in intermediate steps).   
      
   Anyway, in high school checking that x >= 0   
   when solving sqrt(y) = x for y is really   
   checking if your initial assumption is valid.   
   Of course, it makes sense to do extra checks   
   to verify that you made no mistake, but   
   checking assumptions is normally not considered   
   part of solution.  Outside high school   
   nonreal roots are useful, so actually there   
   is no extra assumprion to check.   
      
   --   
                                 Waldek Hebisch   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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