Simplify Question
- To: mathgroup at smc.vnet.net
- Subject: [mg41633] Simplify Question
- From: "Dana DeLouis" <delouis at bellsouth.net>
- Date: Thu, 29 May 2003 08:14:00 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hello. Could someone offer an explanation or a solution on the following? I am trying to simplify an equation that has a Sqrt in both the numerator and denominator. I would like to have just one Sqrt function. Sqrt[2]/Sqrt[1 + 1/x] Returns... Sqrt[2]/Sqrt[1 + 1/x] I am aware that one has to be careful when simplifying Sqrt functions. However, I know that x will always be greater than zero, so I thought this would work. FullSimplify[Sqrt[2]/Sqrt[1 + 1/x], x > 0] Returns the same equation... Sqrt[2]/Sqrt[1 + 1/x] I even mention that x is Real... FullSimplify[Sqrt[2]/Sqrt[1 + 1/x], x > 0 && Element[x,Reals]] Sqrt[2]/Sqrt[1 + 1/x] I still get a Sqrt over a Sqrt However, if I change the number 2 in the numerator to a variable say t, to represent two, then it works... FullSimplify[Sqrt[t]/Sqrt[1 + 1/x], x > 0] Sqrt[(t*x)/(1 + x)] The above has only 1 Sqrt function. :>) I can't seem to simplify the above by keeping 2 as 2, and not as a variable (t). Help on FullSimplify has an example of a Sqrt in both the numerator and denominator. The help example did not make any assumptions on x. FullSimplify[Sqrt[(2 - x)/(3 + x)]/Sqrt[2 - x]] Sqrt[1/(3 + x)] Thank you in advance. This simple example has caused me a lot of grief. -- Dana DeLouis Windows XP Mathematica $VersionNumber -> 4.2 ng_only at hotmail.com = = = = = = = = = = = = = = = = =