Albert Rich schrieb:   
   >   
   > On Wednesday, October 4, 2017 at 6:21:59 AM UTC-10, clicl...@freenet.de   
   wrote:   
   >   
   > > I believe that:   
   > >   
   > >   SQRT(x^2/(1 + SQRT(x^2 - 1))^2)*(1 + SQRT(x^2 - 1))/x = x/SQRT(x^2)   
   > >   
   > > FriCAS will probably squeal, but can Mathematica or Maple validate   
   > > or invalidate this simplification?   
   > >   
   >   
   > Dividing both sides of the proposed identity by (1+sqrt(x^2-1))/x   
   > gives   
   >   
   >   sqrt(x^2/(1+sqrt(x^2-1))^2) = sqrt(x^2)/(1+SQRT(x^2-1))   
   >   
   > which is an equation of the form sqrt(z^2)=z with   
   > z=sqrt(x^2)/(1+SQRT(x^2-1)).   
   >   
   > sqrt(z^2)=z is valid iff re(z)>=0 for all complex z. Thus to prove   
   > the proposed identity need to show re(sqrt(x^2)/(1+SQRT(x^2-1)))>=0   
   > for all complex x. Any ideas?   
   >   
      
   I think your equivalence must be amended as follows:   
      
     sqrt(z^2) = z iff re(z) > 0 or (re(z) = 0 and im(z) >= 0)   
      
   And for simplicity one may consider:   
      
     z = re(sqrt(w)/(1+sqrt(w-1)))   
      
   for all complex numbers w = x^2.   
      
   Martin.   
      
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