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Re: Just what is equality in mathematics, anyway?
Posted:
Apr 16, 2012 4:54 PM
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On Mon, Apr 16, 2012 at 4:24 PM, Joe Niederberger
<niederberger@comcast.net> wrote:
>>And before anyone wishes to blow off what Walter Rudin proves in a textbook some call the bible of real and complex analysis,
>
> Paul, I'm blowing off your confused interpretation of what he said. When he said "we can identify (a,0) with a" he does so without any formal proof, but rather based on the observation that the system of numbers of form (x,0) have the same "arithmetic properties" as the Reals. His "identification" here amounts to an proof he wants you to accept because of his arithmetic properties argument, so fine, its proved already! I already SAID in the SAME post that you pick & choose from, that I accept it!
>
> (When really persnickety mathematician make these sorts of identifications, they throw some stock boilerplate around it, like "up to isomorphism"...)
>
> And yet you think you can pick up on what he is saying and come up with your "proof" that (x,0) = x for all real x! And furthermore, you think it somehow contradicts anything I hadn't already said?
>
> It's called "assuming that which is to be proved", "begging the question", "petitio principii" and I knew you were doing this even before your attempted come back. The point is you want to start from here: "(x,0) = x for all real x" and go on to "prove" that "(x,0) = x for all real x".
>
>>And before anyone wishes to blow off what Walter Rudin proves...
>
> I'll note this is another trick of yours to misunderstand something and then pretend that authorities share your misunderstanding.
>
> Joe N
It is not a trick. Again:
"Re: Just what is equality in mathematics, anyway?"
http://mathforum.org/kb/message.jspa?messageID=7769795
see Rudin's Theorem 1.29 and his proof of Theorem 1.29:
"1.29. Theorem. If a and b are real, then
(a,b) = a + bi.
Proof.
a + bi = (a,0) + (b,0)(0,1) = (a,0) + (0,b) = (a,b)."
In the first equality of the proof, using his stated equivalence of complex (x,0) and real x for all real x (see my link above to see what he says in his textbook), he uses the replacement property of logic and mathematics, which says that we can replace any well-formed formula A (that is in another well-formed formula P) with an equivalent well-formula B (and the resulting well-formed formula Q is equivalent to P). (This can work for equal well-formed formulas, since He replaces real a with complex (a,0), he replaces real b with complex (b,0), and he replaces i with (0,1).
You are denying Rudin; you are denying Theorem 1.29.
Again, we now have the following:
By Rudin's Theorem 1.29, by the ring theorem 0r = r for all elements r in the ring, by the replacement property of logic and mathematics, and by the additive identity property, we have for all real a, for complex i, and for additive identity 0,
(a,0) = a + 0i = a + 0 = a
and so by the transitive property of equality,
(a,0) = a.
The only way to deny this is to deny one of the following: Rudin's Theorem 1.29, the ring theorem 0r = r for allelements r in the ring, the replacement property of logic and mathematics, the additive identity property, or the transitive property of equality. Which do you deny, assuming you wish to continue to deny this?
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