Re: fullsimplify problem

• To: mathgroup at smc.vnet.net
• Subject: [mg52933] Re: [mg52908] fullsimplify problem
• From: DrBob <drbob at bigfoot.com>
• Date: Fri, 17 Dec 2004 05:18:57 -0500 (EST)
• References: <200412160841.DAA27350@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```>> Given ( - a / b^2) = =  ( + a / b^2), it can be written as (-2 a / b^2)...

WHAT can be written as (-2 a / b^2)? The equation? No, rewrite an equation and you still have an equation. Maybe you mean lhs == rhs implies 0 = lhs-rhs = (-2 a / b^2). Maybe. Hard to tell.

>> when I use FullSimplify on that equation, I get the incorrect result ( a /
>> b ) = = 0

IF a/b == 0, your lhs and rhs would both be 0/b or 0, so the equation would be true. On the other hand, the equation can't be true UNLESS a/b == 0, since if a/b isn't zero we can divide both sides by a/b, getting -1/b == 1/b. But that means 2/b == 0, and that can't happen (except for b = Infinity, -Infinity, or ComplexInfinity, but then a/b would be zero as well).

>> ( a / b ) = = 0.  How is this possible?

It's true if a == 0. In Mathematica (but not mathematics) it's true if b is Infinity, ComplexInfinity, or -Infinity (unless a is also one of those).

Bobby

On Thu, 16 Dec 2004 03:41:11 -0500 (EST), symbio <symbio at has.com> wrote:

>
>
> Given ( - a / b^2) = =  ( + a / b^2), it can be written as (-2 a / b^2), but
> when I use FullSimplify on that equation, I get the incorrect result ( a /
> b ) = = 0.  How is this possible?
>
> In[2]:=
> -a/b^2 == a/b^2
> % // FullSimplify
> Out[2]=
> \!\(\(-\(a\/b\^2\)\) == a\/b\^2\)
> Out[3]=
> \!\(a\/b == 0\)
>
> In[5]:=
> -2a/b^2 == 0 // FullSimplify
> Out[5]=
> \!\(a\/b == 0\)
>
>
>
>

--
DrBob at bigfoot.com
www.eclecticdreams.net

```

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