Re: Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]

*To*: mathgroup at smc.vnet.net*Subject*: [mg63285] Re: [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]*From*: Renan <renan.birck at gmail.com>*Date*: Thu, 22 Dec 2005 00:04:51 -0500 (EST)*References*: <200512210435.XAA14733@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 12/21/05, Steven T. Hatton <hattons at globalsymmetry.com> wrote: > Is there a way to convince Mathematica to multiply Sqrt[a+b]Sqrt[a-b] to > produce Sqrt[a^2+b^2]? Just my thoughts: Assuming[a > 0 $B"J(B b > 0, FullSimplify[Sqrt[a + b]Sqrt[a - b]]] gives me Sqrt[a^2-b^2], which seems to make sense to me, given that (a-b)(a+b) = a^2-b^2 Maybe this is what you want? What you said, Sqrt[a+b]Sqrt[a-b] doesn't seen to equal Sqrt[a^2+b^2]: Sqrt[a+b]Sqrt[a-b] == Sqrt[a^2+b^2] /. {a->11,b->9} gives me False. Obviously, I may have commited errors above. Hints anyone?

**References**:**Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]***From:*"Steven T. Hatton" <hattons@globalsymmetry.com>