On 09/28/2013 04:20 PM, Hetware wrote: > On 9/28/2013 4:24 PM, Richard Tobin wrote: >> In article <9cSdnYGhH7CMpNrPnZ2dnUVZ_sCdnZ2d@megapath.net>, >> Hetware <firstname.lastname@example.org> wrote:
>>> So the answer is consensus among mathematicians holds that F(t) = (t^2 - >>> 9)/(t - 3) is undefined at t=3? >> >> Yes. >> >>> Perhaps what I should have said at the >>> outset is something along the lines of: on any given day, if I'm setting >>> up an equation in physics, and produce an expression such as F(t) = (t^2 >>> - 9)/(t - 3), I treat it as t+3, and do not expect any adverse >>> consequence from doing so. >> >> Your simplification is not valid for t=3.
> I believe most mathematicians solving for x as a function of t given > > t^2 - 9 = x (t - 3) > > would not hesitate to factor the left hand side and divide both sides by > t - 3 without treating t = 3 as a special case.
This is high-school stuff. Literally.
I just pulled my text from Algebra II (tenth grade) off the shelf.
Turning to Section 3.6, "Rational Expressions: Reduction to Simplest Form", I see:
It should be observed that since division by zero is excluded as an operation, a fraction whose denominator is zero has no meaning. For example, the fraction 9/(y-7) has no meaning when y=7. Also, the fraction (x+5)/(x^2-9) has no meaning when x=3 or x=-3.
We sometimes encounter fractions such as (x-3)/(3-x) (x!=3). [...]
The authors then proceed to work through several examples of fractions with polynomials in the numerator and denominator. In every case, when the fraction is specified, they include:
(I'm using d(x) here to represent whatever the denominator polynomial happens to be.)
If your high school math classes neglected to point out that this exclusion is necessary, they were faulty. However, the errors and omissions of your past teachers do not change the fact that it is.
 _Algebra and Trigonometry: A Modern Approach_, Peters & Schaaf, van Nostrand, (c) 1965 -- Michael F. Stemper The name of the story is "A Sound of Thunder". It was written by Ray Bradbury. You're welcome.