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>
- Reply-to: drbob at bigfoot.com
- 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
- References:
- fullsimplify problem
- From: "symbio" <symbio@has.com>
- fullsimplify problem