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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: [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




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