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

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   Message 8,517 of 10,432   
   Richard Fateman to Axel Vogt   
   Re: when is sqrt(a/b) not the same as sq   
   01 Mar 14 16:54:56   
   
   From: fateman@cs.berkeley.edu   
      
   There are simpler questions.   
      
   Like is  sqrt(x) - sqrt(x)  equal to 0  or 2*sqrt(x)  or -2*sqrt(x).   
      
   Recall that there are two square roots.  Which one did you mean   
   for the first sqrt?   Which for the second?  which for the third or fourth?   
      
   If you specify, as some systems seem to advise, that x is positive,   
   it still is ambiguous.  Isn't it possible that sqrt(4) has 2 values,   
   sqrt(-4) has 2 values as well?   
      
   Now if you or your favorite CAS change the meaning of sqrt, then who knows?   
      
   As Lewis Caroll wrote,   
      
   "'When I use a word,' Humpty Dumpty said, in rather a scornful tone, 'it   
   means just what I choose it to mean — neither more nor less.'   
      
   'The question is,' said Alice, 'whether you can make words mean so many   
   different things.'   
      
   'The question is,' said Humpty Dumpty, 'which is to be master — that's   
   all.' "   
      
   RJF   
      
      
      
   On 3/1/2014 1:35 PM, Axel Vogt wrote:   
   > On 01.03.2014 22:19, Nasser M. Abbasi wrote:   
   > ...   
   >   
   > As Edgar already said:   
   >   
   > [sqrt(a/b) , sqrt(a)/sqrt(b)]:   
   > subs(a=1, %);   
   >                                   1/2   1   
   >                             [(1/b)   , ----]   
   >                                         1/2   
   >                                        b   
   > subs(b=-1, %);   
   >                                 [I, -I]   
   >   
   > You need that for x * 1/x = 1   
   >   
   > The reason is the brunch cut: sqrt is 'only defined'   
   > except the negative real axis and continued to that   
   > by a special 'rule' in a non-continous way   
   >   
   > In Maple that is named "winding number", K:   
   >   
   >   
   > (z*w)^a = z^a*w^a * exp(-2*I*Pi*a*K(ln(z)+ln(w)))   
   >   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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