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Re: Rudin and nonsense
Posted:
Apr 30, 2012 12:54 AM
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Well, denying what Rudin actually wrote down on paper and published is clearly crank territory, yet we have people doing it here at math-teach.
Look at their denials all throughout the thread
"Just what is equality in mathematics, anyway?"
http://mathforum.org/kb/thread.jspa?threadID=2367800
These denials included you evidently, when you said, "Proving the unprovable not is likely to be really successful."
That is, you evidently claimed the equality (a,0) = a for all real a,b to be unprovable even though I proved it to be simply a partial instantiation (substitution instance) of Rudin's Theorem 1.29, the equality (a,b) = a + bi for all real a,b. That is, simply replace b with 0, and we have for all real a,
(a,0) = a + 0i = a + 0 = a,
(a,0) = a.
There is no reason why this partial instantiation in the form of this last equality may not be viewed as a corollary to Theorem 1.29, yet we have people denying it. Truly amazing.
On Sun, Apr 29, 2012 at 10:34 PM, Wayne Bishop <wbishop@calstatela.edu> wrote:
> I have been busy and not following this thread but, from what little I have
> read, Shakespeare summarized it decades ago, "Much Ado about Nothing".
>
> Wayne
>
> At 09:55 AM 4/29/2012, Paul Tanner wrote:
>
> On Sun, Apr 29, 2012 at 4:05 AM, Didacticus <ddidacticus@yahoo.com> wrote:
>> Nothing I wrote suggested Rudin's choice was *arbitrary* -- I simply
>> pointed out it was a *choice*.
>
> Well, then, the entire textbook that Rudin wrote is the result of
> choice. And so your talk of "choice" means nothing.
>
> And this "choice" gives us Theorem 1.29, the *equality* - and not mere
> equivalence - (a,b) = a + bi.
>
>>
>> The nonsense I alluded to is you going in circles ...
>>
>> ... and then accusing people of "denying" Rudin when they dare to point
>> out the silliness.
>
> Why do you think Rudin is "going in circles"?
>
> If I'm going in circles, then so is Rudin, since all I'm doing is
> laying out in detail what Rudin did, where the sequence of
> *equalities* - and not mere equivalences -
>
> (a,0) = a + 0i = a + 0 = a
>
> and so
>
> (a,0) = a
>
> is the end of what Rudin did with definitions and theorems 1.26 through
> 1.29.
>
> The equality (a,0) = a that follows from Theorem 1.29 is so trivially
> and clearly true as an application of Theorem 1.29, Rudin did not of
> course bother to actually write it out - he left that to us. And so
> the equality (a,0) = a may be viewed as simply an unwritten corollary
> to Theorem 1.29.
>
> You need to process what I in my prior post
>
> "Re: Rudin and nonsense"
> http://mathforum.org/kb/message.jspa?messageID=7807322
>
> actually wrote. Here is the most relevant part that proves that your
> thinking that Rudin is going in circles is pure silliness:
>
> Quote: "That is, taken as a whole, Rudin proved isomorphism in Theorem
> 1.26 by which (a,0) and a for all real a are at least equivalent (note
> that I did not say "equal"), but then by replacement and by Theorem
> 1.29, we find that not only are (a,0) and a equivalent, but that they
> are equal as well."
>
> Your thinking that Rudin is going in circles results from your mistake
> in thinking that equivalence is equality. It is not. Isomorphism is an
> equivalence relation. By definition, equivalence is necessarily a
> relation that is reflexive, symmetric, and transitive but it not
> necessarily antisymmetric, while by definition equality is necessarily
> all four.
Message was edited by: Paul A. Tanner III
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