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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: [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


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