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

*To*: mathgroup at smc.vnet.net*Subject*: [mg63268] Re: [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 22 Dec 2005 00:04:30 -0500 (EST)*References*: <200512210435.XAA14733@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 21 Dec 2005, at 13:35, Steven T. Hatton wrote: > Is there a way to convince Mathematica to multiply Sqrt[a+b]Sqrt[a- > b] to > produce Sqrt[a^2+b^2]? I hope that it will never become convinced of that! Perhaps you meant to convince it that Sqrt[a+b]Sqrt[a-b] is Sqrt[a^2 - b^2]. Even so, I hope it will not be convinced *too easily* since: Sqrt[a - b]*Sqrt[a + b] /. {a -> -1, b -> 0} -1 Sqrt[(a - b)*(a + b)] /. {a -> -1, b -> 0} 1 However, if you give Mathematica reasonable assumptions then it will (usually) also be quite reasonable, e.g. Simplify[Sqrt[a + b]*Sqrt[a - b], {a >= b}] Sqrt[a^2 - b^2] Andrzej Kozlowski

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