       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?

```

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