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MathGroup Archive 2002

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AW: nth roots

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36610] AW: [mg36581] nth roots
  • From: Matthias.Bode at oppenheim.de
  • Date: Fri, 13 Sep 2002 23:33:08 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello watcher2,

try:

(((((3^(1/2))^(1/2))^(1/2))^(1/2))^(1/2))/(6^(1/2))

Out:
1/(Sqrt[2]*3^(15/32))

You can see what happened:

1st: 6^(1/2) = 2^(1/2)*3^(1/2);
2nd: subtraction of the 3's exponents (1/2)^5 - 1/2 = 1/32 - 16/32 = -
15/32.

MATHEMATICA does these straightforward simplifications without you having to
invoke Simplify[].

Best regards,

Matthias Bode
Sal. Oppenheim jr. & Cie. KGaA
Koenigsberger Strasse 29
D-60487 Frankfurt am Main
GERMANY
Tel.: +49(0)69 71 34 53 80
Mobile: +49(0)172 6 74 95 77
Fax: +49(0)69 71 34 95 380
E-mail: matthias.bode at oppenheim.de
Internet: http://www.oppenheim.de





-----Ursprüngliche Nachricht-----
Von: watcher2 at sympatico.ca [mailto:watcher2 at sympatico.ca]
Gesendet: Freitag, 13. September 2002 07:14
An: mathgroup at smc.vnet.net
Betreff: [mg36581] nth roots


Simplify:

5nth sqrt (3)/sqrt (6)

5nth Sqrt(3)
------------
     Sqrt(6)

1st.. is this the correct notation for use on a computer

2nd.. how is the solution solved, step by step please.



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