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MathGroup Archive 2005

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63270] Re: [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Thu, 22 Dec 2005 00:04:32 -0500 (EST)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

expr=Sqrt[a+b]*Sqrt[a-b];

expr/.Sqrt[x_]*Sqrt[y_]:>Sqrt[x*y]//Simplify

Sqrt[a^2 - b^2]


Bob Hanlon

> 
> From: "Steven T. Hatton" <hattons at globalsymmetry.com>
To: mathgroup at smc.vnet.net
> Date: 2005/12/20 Tue PM 11:35:42 EST
> Subject: [mg63270] [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt [a^2+b^2]
> 
> Is there a way to convince Mathematica to multiply Sqrt[a+b]Sqrt[a-b] to
> produce Sqrt[a^2+b^2]?  
> -- 
> The Mathematica Wiki: http://www.mathematica-users.org/
> Math for Comp Sci http://www.ifi.unizh.ch/math/bmwcs/master.html
> Math for the WWW: http://www.w3.org/Math/
> 
> 


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