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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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magidin@math.berkeley.edu

Posts: 11,155
Registered: 12/4/04
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 13, 2013 2:01 AM
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On Saturday, October 12, 2013 7:47:12 PM UTC-5, Hetware wrote:
> On 10/8/2013 12:43 AM, Arturo Magidin wrote:
>

> > On Monday, October 7, 2013 7:52:32 PM UTC-5, Hetware wrote:
>
>
>

> >> Too late, I already have. I now realize I was asserting my
>
> >> assumptions
>
> >>
>
> >> in the wrong order.
>
> >
>
> > It wasn't a problem of order. The problem was that you were asserting
>
> > assumptions without warrant.
>
> >
>
>
>
> I can assert a function to be continuous,


You can assert whatever you want, true.

Doesn't stop you from being wrong when you assert things which you have no warrant to assert, as is the case here.

All the rest is just you going through incredible mental acrobatics to try to convince yourself that you were not *wrong* for making the unwarranted assertion in the first case.

> and then give a rule for
>
> determining its mapping in some part of the domain of reals. There may
>
> be times when such an assertion leads to a contradiction, and thus the
>
> definition of the function must be abandoned. In the case under
>
> discussion, there is no contradiction if I define f(x)=(x^2-9)/(x-3)
>
> where x!=3, and then use the assumption of continuity to determine f(3).


And by asserting that the function is continuous EVERYWHERE (which is something different from just asserting that it is continuous, as I explained to you), you are making an UNWARRANTED assumption.

Yes, you are free to do so, just like you are free to stable your forefinger and thumb together. Doesn't stop you from being wrong for asserting it.

Yes, you can set up an entirely new and different system of assumptions from the one you are working in. But that would make it a new and different system.

You came here to ask (form us "mere mortals", I may add) why your conclusions were not matching what you were finding. We told you why: you were assuming without warrant something about the function. That sent you into the pompous and fatuous equivalent of a rage, providing all sort of silly, unwarranted, ignorant, and incorrect justification for your assumptions. It's nothing but a dance to try to justify yourself, rather than simply accept what your mistake was. You even create an incorrect reason for your mistake.

> According to Thomas a function f(x) which is continuous in some
>
> neighborhood of x = c
>
>
>
> 1) has a definite finite value at f(c), and
>
>
>
> 2) limit[f(x),x->c] = f(c).
>
>
>
> If I assert that f(x) is a continuous function over the domain of real
>
> numbers I am asserting that it has a definite finite value at f(c).


And if you assert this for a function which is NOT DEFINED AT c, then you are asserting something false, and making and unwarranted assumption.

Simply, no? Apparently not.


> If I
>
> then provide a rule for determining f(x) for all reals x != c, and
>
> limit[f(x),x->c] exists and is finite, I can then use that fact and the
>
> definition of continuity to determine the value of f(c). To wit
>
> limit[f(x),x->c] = f(c).


Then you are talking about a DIFFERENT function, not the function you were given which happens to not be defined at 3, despite all your mental acrobatics.



>
> Therefore f(x) provides a continuous mapping from the domain of all
>
> reals onto a range of reals.


No. A DIFFERENT FUNCTION which agrees with f other than at 3 is.


> The definition
>
>
>
> f(x) = (x^2-9)/(x-3) where x!=3 and f(c) = limit[f(x),x->3],



defines a DIFFERENT FUNCTION from the one you were given.

> is synonymous with the definition
>
>
>
> f(x)=(x^2-9)/(x-3) where x!=3 and f(x) is continuous at x = 3.


and is a DIFFERENT FUNCTION from the one you were given.

You can talk all you want about this DIFFERENT function; it's still a DIFFERENT FUNCTION FROM THE ONE YOU WERE GIVEN.

By asserting that the function you were given was continuous at 3, you were making an INCORRECT and UNWARRANTED assumption. Period.


> >> Axiomatic set theory, though it's been a long time since I went
>
> >> through
>
> >>
>
> >> the entire exercise. Some concepts must be taken /a priori/, such
>
> >> as the
>
> >>
>
> >> dictates of logic.
>
> >
>
> > And where do you think that axiomatic set theory came from? From some
>
> > immortal? You said you did not accept things handed down to you from
>
> > mortals... but apparently you do; so it wasn't so much that you were
>
> > against accepting things "from mortals", but rather something else.
>
> > Which was... what?
>
>
>
> I am Neoplatonic in this respect. That is to say, the foundations of
>
> mathematics are a-priori, and the essence of mathematics is not
>
> invented, it is discovered.


Now you are just pitifully and unsuccessfully attempting to cover for your prior pompous statement, shown for the fatuous emptiness that it was.

The things given to you by "mere mortals" that you are willing to accept (because they don't make you be wrong) are 'reality'. The things given to you by "mere mortals" that show you were WRONG are dismissed as "appeals to authority", words of "mere mortals", etc. Fatuous and pompous.

> Extra terrestrial mathematics is isomorphic
>
> to terrestrial mathematics.


More fatuous statements.

> >>> Again: get off the high horse; because you are being either
>
> >>> ignorant,
>
> >>
>
> >>> or hypocritical.
>
> >>
>
> >>
>
> >>
>
> >> In this context /argumentum ad hominem/ is clearly a fallacy.
>
> >
>
> > Sigh...
>
> >
>
> > First: saying that you are being ignorant or hypocritical is not, in
>
> > and of itself, an "argumentum ad hominem". Contrary to common
>
> > misconception, "argumentum ad hominem" is not merely the saying of
>
> > bad things about others.
>
>
>
> There's something about a "high horse" in that as well.


One has to climb to reach the heights of your ignorant arrogance, otherwise you won't hear.

In any case, you did not know what an argumentum ad hominem was.

By the way: shall I cry like you did that you deleted part of what I wrote?



> It is quite legitimate for a mathematician or anybody, for that matter,
>
> to appeal to native first principles rather than the authority of others
>
> as the first and final arbiter of Truth. Any other position would be
>
> unacceptable (mental slavery).


Fatuous statements, justifying willing ignorance.

Go stroke your ego somewhere else; it's becoming obscene.


>
> > You wrote: "I don't answer to the authority of mortals."
>
> >
>
> > Yet you admit that you take as authority axiomatic set theory, which
>
> > comes from, surprise, mortals.
>
>
>
> See above.


Indeed. See above for evidence of your hypocrisy.


> > (i) It is defined at a; (ii) The limit of f(t) as t approaches a
>
> > exists; and (iii) the value of the function at a equals the value of
>
> > the limit of f(t) as t approaches a.
>
> >
>
> > A real valued function of real variable f(t) is "continuous at all
>
> > real numbers" or "continuous everywhere" if and only if it is
>
> > continuous at x=a for all real numbers a.
>
> >
>
> > A real valued function of real variable f(t) is "continuous" if and
>
> > only if it is continuous at all points in its domain.
>
>
>
> Is limit[(t^2-9)/(t-3), t->c] continuous for all real numbers c?



Now you are being dishonest. The quote above was about the difference between saying a function is "continuous" and saying the function is "continuous at all real numbers." You used the two terms interchangeably, they are not interchangeable.

Fatuous. Ignorant. And dishonest. The triple crown of the internet troll...

> Saying that f(t) is continuous everywhere is the same as saying that
>
> f(t) has a finite limit for all real numbers t, and that f(t) =
>
> limit[f(x), x->t].


Still missing the point.

"f is continuous everywhere"

"f is continuous at all real numbers"

"f is continuous"

Two of them mean the same thing always. The third does not.

AS I EXPLAINED.


> Is f(u) = limit[(t^2-9)/(t-3), t->u] a continuous function of u over the
>
> reals?


Is f(x) = sin(x) continuous at all real numbers?

I mean, if you are going to change the question in order to try to justify your answer, why stick so close to the original? Just go crazy, man. If you're going to lie, do so freely.

> \> You quoted yourself and made claims about my responses, but cited none
>
> of those responses.


Those responses were contained in the post I was replying to. Do you have as much of a problem with short term memory as you have admitting your obvious errors and your ignorance?

Are you going to just spout more fatuous empty sentences about Truth(tm) and "mere mortals"?

Just go stroke yourself somewhere else, instead of bothering with what us "mere mortals" have to say. You've "decided" you know where your error was. You've "decided" you know how to fix it.

In the opinion of apparently many "mere mortals" you were wrong to begin with, and you are even wrong about what you were wrong about. But it's just us "mere mortals", so why would you care?

--
Arturo Magidin





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10/10/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Virgil
10/18/13
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Hetware
10/19/13
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Peter Percival
10/19/13
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fom
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Peter Percival
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Hetware
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Peter Percival
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Hetware
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
fom
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
magidin@math.berkeley.edu
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Hetware
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
magidin@math.berkeley.edu
10/20/13
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Hetware
10/20/13
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quasi
10/20/13
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quasi
10/20/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Hetware
10/20/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Peter Percival
10/20/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
magidin@math.berkeley.edu
10/20/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Hetware
10/20/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Arturo Magidin
10/20/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
Hetware
10/20/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
magidin@math.berkeley.edu
10/19/13
Read Re: Is (t^2-9)/(t-3) defined at t=3?
fom

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