Re: Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
- To: mathgroup at smc.vnet.net
- Subject: [mg63285] Re: [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
- From: Renan <renan.birck at gmail.com>
- Date: Thu, 22 Dec 2005 00:04:51 -0500 (EST)
- References: <200512210435.XAA14733@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 12/21/05, Steven T. Hatton <hattons at globalsymmetry.com> wrote:
> Is there a way to convince Mathematica to multiply Sqrt[a+b]Sqrt[a-b] to
> produce Sqrt[a^2+b^2]?
Just my thoughts:
Assuming[a > 0 $B"J(B b > 0, FullSimplify[Sqrt[a + b]Sqrt[a - b]]]
gives me Sqrt[a^2-b^2], which seems to make sense to me, given that
(a-b)(a+b) = a^2-b^2
Maybe this is what you want?
What you said, Sqrt[a+b]Sqrt[a-b] doesn't seen to equal Sqrt[a^2+b^2]:
Sqrt[a+b]Sqrt[a-b] == Sqrt[a^2+b^2] /. {a->11,b->9}
gives me False.
Obviously, I may have commited errors above. Hints anyone?
- References:
- Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
- From: "Steven T. Hatton" <hattons@globalsymmetry.com>
- Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]