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MathGroup Archive 2000

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Re: Simplification shortcomings?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24644] Re: [mg24621] Simplification shortcomings?
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Mon, 31 Jul 2000 09:23:17 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

on 7/28/00 11:23 PM, Bob Harris at nitlion at mindspring.com wrote:

> Howdy,
> 
> I'm a relative novice to Mathematica.  While working with it today, I had
> occasion to want to know if a result was equal to (1 + Sqrt[5])/2.  The
> result was shown as Sqrt[(3 + Sqrt[5])/2].  After some pancil and paper
> work, I figured out that these two are equal.  Or, I should say, that the
> former is one of the values that the latter can have.
> 
> I was frustrated in my attempts to get Mathematica to answer that question
> for me.  Simplify[(1 + Sqrt[5])/2 - Sqrt[(3 + Sqrt[5])/2]] didn't provide
> any improvement.  Calculating this value to many decimal digits showed it
> was near zero (probably close enough that I could have applied the
> techniques shown in Scheinerman's recent article in American Mathematical
> Monthly).  The only way I got Mathematica to show the equality was to square
> both numbers;  Simplify[((1 + Sqrt[5])/2)^2 - (3 + Sqrt[5])/2] is zero.
> 
> Is there any better way to do this?  I have some other, more complicated
> numbers that I need to compare.
> 
> Thanks,
> Bob Harris
> 
> 
> 
> 
> 
Yes.

In[3]:=
FullSimplify[(1 + Sqrt[5])/2 - Sqrt[(3 + Sqrt[5])/2]]

Out[3]=
0

or

In[5]:=
RootReduce[(1 + Sqrt[5])/2 - Sqrt[(3 + Sqrt[5])/2]]

Out[5]=
0

(Actually FullSimplify uses RootReduce so it is the latter that really does
the job here).

Andrzej
-- 
Andrzej Kozlowski
Toyama International University, JAPAN

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