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Re: Rudin and nonsense
Posted:
Apr 30, 2012 2:54 AM
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On Sun, Apr 29, 2012 at 9:51 PM, Joe Niederberger
<niederberger@comcast.net> wrote:
> Clyde Greeno says;
>>when every professional mathematician would know that saying (a,0)=a leads to a logical contradiction...
>
> Joe N said previously...
>>I would say Paul well armed now to go prove absolutely anything in mathematics.
>
> Ha ha! Perhaps he will solve every Millenium Prize Problem!
>
> Cheers,
> Joe N
>
> p.s. all of course, under certain interpretations...
But you are the ones denying Rudin when he says that (a,0) = a.
Yes, he says it, since "saying" a theorem entails "saying" what all its substitution instances and their results say.
He actually wrote and published that the equality (a,b) = a + bi is true for all real a,b, as theorem 1.29.
But this theorem entails all of its substitution instances and their results, including replacing b with 0, which gives, for all real a,
(a,0) = a + 0i = a + 0 = a,
(a,0) = a.
So when you deny the equality (a,0) = a for all real a, you deny the general theorem of which this equality is an instance: You deny that Theorem 1.29 is true - you deny that (a,b) = a + bi is true for all real a,b.
Still laughing?
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