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Topic:
Is (t^29)/(t3) defined at t=3?
Replies:
11
Last Post:
Oct 20, 2013 11:15 AM




Re: Is (t^29)/(t3) defined at t=3?
Posted:
Oct 14, 2013 5:36 PM


Hetware <hattons@speakyeasy.net> writes:
> On 10/14/2013 4:01 AM, Peter Percival wrote: >> Hetware wrote: >> >>> >>> The statement "Let f(t) be a continuous function for all real numbers t" >>> has a concise meaning. >> >> There's nothing wrong with that (one can easily prove that there are >> such functions). What is verboten is defining f with some expression in >> t and _then_ assuming it is continuous. >> > > The more I think about this, the more ambiguous it seems. > > One way to prove something in mathematics is to propose the opposite, > and show that the proposition leads to a contradiction. So let's try > that with the definition f(t) = (t^29)/(t3) without giving > continuity as part of the definition. > > Now, for the purpose of finding a contradiction, let us propose that > f(t) is globally continuous. What does that entail? f(t) has a > definite finite value for every real number t, and that value is f(t) > = limit[f(p), p>t]. > > Now, I concede that (t^29)/(t3) = 0/0 when t = 3, and 0/0 is > meaningless. It is not, however, a contradiction.
That seems to me as much of an explicit contradiction as you could want. It may seem odd because the contradiction comes immediately from the definition of f rather than after some few lines of reasoning, but it's a contradiction none the less: assume f continuous everywhere, oops f not even defined at 3. How is that not a contradiction?
<snip>  Ben.



