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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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