Does inductive reasoning lead to knowledge?
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From: Marshall on 19 Jan 2010 21:11 On Jan 19, 9:31 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jan 19, 10:37 am, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > [...] but if we actually > > > express each value that way you will have engaged a > > > runaway system: > > > b i = 0 + bi = 0 + 0 i + 0 + b i = ... > > > So? What's wrong with the equation(s) above? Nothing > > that I can see. > > > Also note that: > > > 5 = 0 + 5 = 0 + 0 + 5 = ... > > > Don't you agree? Is addition of natural numbers a > > "runaway system?" I don't see how an infinite number > > of equations makes for any kind of problem. > > Why did you propose this resolution in the first place? You are now > defending it and I must resolve the issue by returning to the source > of your argument. I admit that the zeros are harmless, but then are > they doing anything at all? You built this, not me. Why did you build > it? The zeroes in the equations I wrote are harmless, just as the zeroes in the equation you wrote are harmless. This discharges any claim that there is any kind of problem to be had with a "runaway system." > It does not resolve my initial complaint and so we must return to > the initial point which caused this response. That is my own claim > that the product > b i > of the complex representation > a + b i > where a and b are real and i is not real is not a ring product. It's not a product in the usual real ring, sure. So what? It's a product in the complex ring. > > > > Doesn't your argument work just as well with the rational > > > > numbers? Thus: > > > > > Again, applying the closure principles to the "/" operator in > > > > the rational value > > > > a / b > > > > we see no agreement with the rational ring definition. It is > > > > this simple. a and b are integers. Therefore this quotient > > > > a/b > > > > is incompatible with the ring definition's product. > > > Marshall > > The operator is not associative. > ( 4 / 2 ) / 2 = 4 / ( 2 / 2 ) > fails the test with '/' as either operator of the ring definition > according to > http://en.wikipedia.org/wiki/Ring_mathematics#Formal_definition Huh? I thought your argument was about closure. Forget associativity for the moment; I probably phrased it badly, but the point was, we don't run into any problem with the rationals having two integer components. Neither do we run into any problems with the complex numbers having two real components. You seem to be unhappy with the idea of numbers having components in a construction; if so, you should be unhappy about the rationals, as well as the complex numbers. > By even > attempting to resolve it you have admitted the existence of this > conflict, and yet your mind will continue its denial, thus closing the > conversation. I admit to the existence of a misconception on your part about the complex numbers, and I deny that your claim of a problem indicates an actual problem. Thus I agree with your assessment that I have made an admission and am in denial about something. Marshall
From: Marshall on 20 Jan 2010 20:00 On Jan 20, 3:25 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > > Regardless of how many zeros you'd like to enforce you'll still have a > term > b i > which is a product and which is inconsistent with the closure > requirement of the ring product because b is real and i is not real. You are hallucinating: b is complex. You also are apparently unable to specify which ring, and hence which ring product, you are talking about. > The zeros are quite > meaningless. We can throw zeros in anywhere anytime we like so long as > they are superposed (summed). To rely upon these zeros for any > argument is a weak stance. I did not offend your requirement that b > carry a zero. I built a b' and it need not carry that requirement. The > requirement is pure silliness based on the logic of zeros which starts > this paragraph. I'll stand by my own falsification of your statement > on uniqueness, since you've allowed the additional components by > stipulating additional components for your construction. That's hilarious! You misread the original statement as calling for unique reals; you also claim that you can construct some other number b' that doesn't meet the requirements given; yet you still "stand by" your position. > The closure principle makes such a simple discrepancy of such > constructions yet noone bothers to criticize except me as far as I can > tell. Indeed you are the only one who sees a problem. The obvious conclusion is that everyone else is blind and only you can see the truth. http://en.wikipedia.org/wiki/Crank_%28person%29 > Shall I just declare a hoax and reboot the matrix? Are we all > just apes anyway? Seems like proof to me. I await falsification of my > statements. They were false when you first wrote them. The distinctions necessary for *you* to see how false they are are apparently not distinctions you are capable of making: specifically, the distinction between the product of the real ring and of the complex ring. Marshall
From: Tim Golden BandTech.com on 20 Jan 2010 22:04 On Jan 20, 8:00 pm, Marshall <marshall.spi...(a)gmail.com> wrote: > On Jan 20, 3:25 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: > > > > > Regardless of how many zeros you'd like to enforce you'll still have a > > term > > b i > > which is a product and which is inconsistent with the closure > > requirement of the ring product because b is real and i is not real. > > You are hallucinating: b is complex. You also are apparently Here I take the opportunity to falsify your statement above. b is not complex. b is real. Please falsify my own statements in such a direct manner and you will have something. - Tim > unable to specify which ring, and hence which ring product, > you are talking about. > > > The zeros are quite > > meaningless. We can throw zeros in anywhere anytime we like so long as > > they are superposed (summed). To rely upon these zeros for any > > argument is a weak stance. I did not offend your requirement that b > > carry a zero. I built a b' and it need not carry that requirement. The > > requirement is pure silliness based on the logic of zeros which starts > > this paragraph. I'll stand by my own falsification of your statement > > on uniqueness, since you've allowed the additional components by > > stipulating additional components for your construction. > > That's hilarious! You misread the original statement as calling > for unique reals; you also claim that you can construct some > other number b' that doesn't meet the requirements given; > yet you still "stand by" your position. > > > The closure principle makes such a simple discrepancy of such > > constructions yet noone bothers to criticize except me as far as I can > > tell. > > Indeed you are the only one who sees a problem. The obvious > conclusion is that everyone else is blind and only you can > see the truth. > > http://en.wikipedia.org/wiki/Crank_%28person%29 > > > Shall I just declare a hoax and reboot the matrix? Are we all > > just apes anyway? Seems like proof to me. I await falsification of my > > statements. > > They were false when you first wrote them. The distinctions > necessary for *you* to see how false they are are apparently > not distinctions you are capable of making: specifically, the > distinction between the product of the real ring and of the > complex ring. > > Marshall
From: Marshall on 20 Jan 2010 22:24 On Jan 20, 7:04 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jan 20, 8:00 pm, Marshall <marshall.spi...(a)gmail.com> wrote: > > > On Jan 20, 3:25 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > > wrote: > > > > Regardless of how many zeros you'd like to enforce you'll still have a > > > term > > > b i > > > which is a product and which is inconsistent with the closure > > > requirement of the ring product because b is real and i is not real. > > > You are hallucinating: b is complex. You also are apparently > > Here I take the opportunity to falsify your statement above. > b is not complex. b is real. > Please falsify my own statements in such a direct manner and you will > have something. You're on: There does not exist a number that is a member of the set of real numbers that is not also a member of the set of complex numbers. Marshall
From: jbriggs444 on 21 Jan 2010 08:33
On Jan 20, 10:24 pm, Marshall <marshall.spi...(a)gmail.com> wrote: [...] > You're on: > > There does not exist a number that is a member of > the set of real numbers that is not also a member of the > set of complex numbers. You've overstated the case significantly here. _IF_ we define the complex numbers as ordered pairs of reals under the obvious cartesian coordinate method _THEN_ there is no real number that _is_ also a complex number. [My background is real analysis. This is the obvious construction] _IF_ we define the complex numbers as the closure of the real numbers plus i under the obvious rules for how addition and multiplication treat imaginary numbers _THEN_ there is no real number that _is not_ also a complex number. [I've never been exposed to the foundations of the complex numbers from an algebraic point of view, but I expect that this is the kind of basis you might want to put under them] _IF_ we define the "foobar numbers" as ordered pairs of reals under the obvious cartesian coordinate method and then define the "complex numbers" as the isomorphic set produced by replacing each (x,0) pair in the "foobar numbers" with the real number x _THEN there is no real number that _is not_ also a complex number. [This is the obvious foundation an analyst could put under the complex numbers if somebody wants to get bitchy about subnet relations] To an analyst, all three statements are obviously true. (*) And the distinction is irrelevant. Whether there is a sub-ring of the complex numbers that _is_ the real numbers or whether the sub-ring is merely isomorphic to the real numbers is of little consequence. I was trained as a Dedekind cut guy. But I feel no need to declare Jihad against the infidel Cauchy sequence dudes. |