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

*To*: mathgroup at smc.vnet.net*Subject*: [mg63278] Re: [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2-b^2]*From*: "Steven T. Hatton" <hattons at globalsymmetry.com>*Date*: Thu, 22 Dec 2005 00:04:40 -0500 (EST)*References*: <200512210435.XAA14733@smc.vnet.net> <CC6AC592-A4C5-4DA0-9C5E-E69AEF81E882@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

On Wednesday 21 December 2005 02:52, Andrzej Kozlowski wrote: > 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]. Yes, that is what I had intended. I was tired and not paying close attention. At least I have the solace that greatter mathematicians than I am make such errors. > Even so, I hope it will not be convinced *too easily* since: > > Sqrt[a - b]*Sqrt[a + b] /. {a -> -1, b -> 0} > > -1 ... ACK > Simplify[Sqrt[a + b]*Sqrt[a - b], {a >= b}] Hmmm...Seems I was trying too hard. I was trying thing with Assuming[]. Things such as a \[Element] Reals && b \[Element] Reals &&... Thanks for the suggestion. That works. Steven

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