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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 11   Last Post: Oct 20, 2013 11:15 AM

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Ben Bacarisse

Posts: 1,150
Registered: 7/4/07
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 14, 2013 5:36 PM
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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^2-9)/(t-3) 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^2-9)/(t-3) = 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.



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