       Re: Simplification shortcomings?

• To: mathgroup at smc.vnet.net
• Subject: [mg24649] Re: Simplification shortcomings?
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Mon, 31 Jul 2000 09:23:20 -0400 (EDT)
• References: <8lsutf\$27c@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Bob,

FullSimplify[(1 + Sqrt)/2 - Sqrt[(3 + Sqrt)/2]]

0

--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Bob Harris" <nitlion at mindspring.com> wrote in message
news:8lsutf\$27c at smc.vnet.net...
> 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)/2.  The
> result was shown as Sqrt[(3 + Sqrt)/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)/2 - Sqrt[(3 + Sqrt)/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)/2)^2 - (3 + Sqrt)/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
>
>
>
>

```

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