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Re: Mathematica 6.01 does not know one can not divide by

  • To: mathgroup at smc.vnet.net
  • Subject: [mg107256] Re: Mathematica 6.01 does not know one can not divide by
  • From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
  • Date: Sun, 7 Feb 2010 06:11:34 -0500 (EST)
  • References: <201002041124.GAA29654@smc.vnet.net> <hkgl81$6vv$1@smc.vnet.net>

Nothing unexpected here, I'd say. For all means and purposes a value
infinitesimally close to 5 is indeed a solution to the equation.

The error is due to both sides of the equation evaluating before they
are equated. If you do it like this:

Simplify[3*x + (1/(x - 5)) == 15 + (1/(x - 5))] /. {x -> 5.`}

You get True.

Cheers -- Sjoerd

On Feb 5, 10:35 am, Tomas Garza <tgarz... at msn.com> wrote:
> Does it, really?
>
> In[98]:= NSolve[3*x+(1/(x-5))==15+(1/(x-5)),x]
>
> Out[98]= {{x->5.}}
>
> In[99]:= 3*x+(1/(x-5))==15+(1/(x-5))/.{x->5.`}
>
> During evaluation of In[99]:= Power::infy: Infinite expression 1/0. encountered. >>
> During evaluation of In[99]:= Power::infy: Infinite expression 1/0. encountered. >>
>
> Out[99]= ComplexInfinity==ComplexInfinity
>
> Tomas
>
> > Date: Thu, 4 Feb 2010 06:24:12 -0500
> > From: zeno... at mindspring.com
> > Subject:  Mathematica 6.01 does not know one can not divide by 0??
> > To: mathgr... at smc.vnet.net
>
> > I had Mathematica 6.01 try and solve this " equation", and it gives an
> > answer back of 5, which if substituted back into the equation, one
> > would be dividing by 0. Code and result here...
>
> > NSolve[3*x + (1/(x - 5)) == 15 + (1/(x - 5)), x]
> > {{x -> 5.}}



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