Re: Canceling square roots with Simplify

*To*: mathgroup at smc.vnet.net*Subject*: [mg18467] Re: [mg18211] Canceling square roots with Simplify*From*: Lars Hohmuth <larsh at wolfram.com>*Date*: Wed, 7 Jul 1999 23:08:35 -0400*Organization*: Wolfram Research, Inc.*References*: <7l3n9l$ggj@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"Ersek, Ted R" wrote: > Earlier I replied to a message where Everett Farr asked why > Simplify[ Sqrt[b^2]*Sqrt[1/b^2] ] > doesn't simplify to 1. > > As already noted by Adam Strzebonski the expression above isn't equal to 1 > for all values of (b). One counter example is when (b=I). > > For basically the same reason the rule I thought is missing from the > Simplify routine isn't true in general and shouldn't be included. > > The rule that isn't true in general is: > ---------------- > MyRules= { > ((zb_^(-1*z1_))^(z2_)):>((zb^z1)^(-z2)), > ((zb_^p_?Negative)^z2_):>(zb^(-p))^(-z2) }; > ---------------- > > As Adam Strzebonski noted one can use PowerExpand get the expected result. > Also using version 4 one can use Simplify/FullSimplify and indicate certain > variables are part of a specific domain to get the expected result. > > Regards, > Ted Ersek > As Ted pointed out already, Mathematica 4 also accepts assumptions in Simplify. For example: In[110]:= Simplify[Sqrt[b^2]*Sqrt[1/b^2]] Out[110]= Sqrt[1/b^2]*Sqrt[b^2] In[111]:= Simplify[Sqrt[b^2]*Sqrt[1/b^2], b > 0] Out[111]= 1 Lars Hohmuth Wolfram Research, Inc.