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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 10,087 of 10,432    |
|    Richard Fateman to All    |
|    Re: solving sqrt(y)=x for    |
|    16 Jan 21 13:04:43    |
   
   From: fateman@gmail.com   
      
   >   
   The question is, what should a CAS provide as an answer.   
   y=x^2 is, by common usage in CAS, incomplete or outright wrong.   
   Consider sqrt(y)=-5 (that is, x=-5)   
   y=25 would seem to be the specific "answer".   
      
   Yet sqrt(25) is commonly held to be 5, not -5, and so the given   
   answer does not satisfy the equation.   
      
   If you view sqrt as multivalued, and sqrt(25) is therefore the   
   set {-5,5}, then it is also false that the set {-5,5} is equal to   
   5 (or the set {5} ).   
      
   Maxima seems to be headed in the right direction, though perhaps the answer   
   could   
   be stated as some kind of conditional, if (x is real and x>=0 ) then y=x^2   
   else....   
      
   The rule of thumb I was taught was that you can square (etc) both side of an   
   equation at   
   the risk of introducing extraneous solutions, and so they must be checked.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   
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