Does inductive reasoning lead to knowledge?
in [Logic]
|
From: Don Stockbauer on 18 Jan 2010 22:44 Does inductive reasoning lead to knowledge? Of course. That's obvious.
From: Patricia Aldoraz on 18 Jan 2010 22:49 On Jan 19, 2:44 pm, Don Stockbauer <don.stockba...(a)gmail.com> wrote: > Does inductive reasoning lead to knowledge? > > Of course. That's obvious. Another philosophically incurious usenet guy or just a naive newbie who does not really know the issues?
From: Marshall on 18 Jan 2010 23:19 On Jan 18, 7:49 pm, Patricia Aldoraz <patricia.aldo...(a)gmail.com> wrote: > On Jan 19, 2:44 pm, Don Stockbauer <don.stockba...(a)gmail.com> wrote: > > > Does inductive reasoning lead to knowledge? > > > Of course. That's obvious. > > Another philosophically incurious usenet guy or just a naive newbie > who does not really know the issues? He's sort of like a seagull. He flies in, deposits his posts on our heads, and flies away. He thinks he's funny. Marshall
From: Patricia Aldoraz on 19 Jan 2010 01:06 On Jan 19, 3:19 pm, Marshall <marshall.spi...(a)gmail.com> wrote: > On Jan 18, 7:49 pm, Patricia Aldoraz <patricia.aldo...(a)gmail.com> > wrote: > > > On Jan 19, 2:44 pm, Don Stockbauer <don.stockba...(a)gmail.com> wrote: > > > > Does inductive reasoning lead to knowledge? > > > > Of course. That's obvious. > > > Another philosophically incurious usenet guy or just a naive newbie > > who does not really know the issues? > > He's sort of like a seagull. He flies in, deposits his posts > on our heads, and flies away. He thinks he's funny. > I see <g>
From: Tim Golden BandTech.com on 19 Jan 2010 08:05
On Jan 18, 9:27 pm, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > Tim Golden BandTech.com wrote: > > On Jan 18, 1:18 pm, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > >> Tim Golden BandTech.com wrote: > >>> On Jan 16, 12:41 pm, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > >>>> Tim Golden BandTech.com wrote: > >>>>> On Jan 14, 1:19 pm, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > >>>>>> Tim Golden BandTech.com wrote: > >>>>>>> On Jan 14, 9:39 am, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > >>>>>>> J. Clarke speaks of rings above here eloquently. I wonder if you > >>>>>>> would offer your criticism on the following: > > >>>>>>> The complex number > >>>>>>> a + b i > >>>>>>> are considered to be consistent with ring terminology, with a > >>>>>>> product and sum being consistently defined and being > >>>>>>> algebraically well behaved, yet within this number form itself > >>>>>>> a + b i > >>>>>>> we see one product > >>>>>>> b i > >>>>>>> and one sum > >>>>>>> a + (bi) > >>>>>>> which are inconsistent with the group and ring definitions > >>>>>>> since a and b are real, and i is not real. Thus the very > >>>>>>> construction of the complex number via its definition is not > >>>>>>> compatible with this abstract algebraic form. > > >>>>>> Be kind enough to exactly state the definitions you are using. > >>>>>> It is difficult to follow your argument if you are not clear in > >>>>>> your definitions. And is your question whether the particular > >>>>>> examples you use violate closure or do you have some other issue > >>>>>> in mind? > > > How about > > http://en.wikipedia.org/wiki/Ring_(mathematics)#Formal_definition > > Can we agree to this as a starting point? > > The closure requirements are so far all that I have relied upon to > > make my argument. I am sorry I misuse the terminology and where I have > > used "ring quotient" I should have used "quotient ring", though I see > > no way to confuse the statement. Again, applying the closure > > principles to the two operators in the complex value > > a + b i > > we see no agreement with the ring definition. > > You are skipping steps here. You are rushing from the definition of a ring > to the complex numbers. The complex numbers constitute one ring among many. > You have not defined the set of complex numbers nor have you defined the > operations on the complex numbers. > > > It is this simple. b is > > real. i is not real. Therefore this product > > b i > > is incompatible with the ring definition's product. Further the sum > > will not resolve to a single element, where all sums will have two > > elements to operate upon. This is so simple that I cannot see how any > > confusion can creep in. > > Since you have not stated the definition of "complex number" nor have you > stated the definition of "product", there is no way to determine whether > your above statement is valid by your definitions. > > > The same concept can be reapplied to the polynomial with real > > coefficients. > > Once again you have not defined your terms. Define "polynomial with real > coefficients" and we can go from there. > > > Are the coefficients truly real? > > Does your definition require that they be? > > > Apply the operators of > > the ring > > Which operators? You have not defined any operators. > > > and we see that the elements to which these real coefficients > > apply must also be real. > > Why must they be real? > > > Thus the entire sum within such polynomials > > must be real, by the same simple closure principle. > > Again you have not defined any operations. Without defining your operations > any discussion of closure is pointless. > > > I have made these statements now several times to you and you have > > offered up that the definition of a complex value is in tuple form > > (a,b). > > If you don't like that one then offer up another one. > > > This does nothing to change my argument on the usual form > > a + b i > > and I readily admit that in the z form there can be little to argue > > over so long as we discuss in terms of > > z1 + z2, z3 z4 > > style sums and products. > > Rather than vaguely saying "the usual form", please state the definition of > "the usual form" as you understand it. > > > I got here back in time by studying the quotient ring in an attempt to > > understand some math work. > > Which particular quotient ring? There is no "the" quotient ring. > > > The work particularly relies upon > > polynomials with real coefficients. > > Which work? > > > This multiplication of a real > > valued coefficient to an X which has only vague meaning > > is beyond my ability to understand. > > So define X. > > > And in going back to the > > definition of ring I see that this construction(the polynomial itself) > > is conflicted, as I have outlined. > > Which construction? > > To create a ring you must provide certain definitions. Once you have > provided the definitions then you can discuss matters such as closure. You > seem to be assuming that there is only one kind of ring, "the" ring. That > is not the case. > > <snip> The ring definition does not itself define the actions of its operators. It does however place clear requirements upon them. One of these requirements is that they operate on the same set and have results in that same set. This is the only portion of the definition that I have relied upon, and I have stated myself clearly. One can keep asking for definitions ad nauseum, and at the bottom then what? Will you have me define the real number? Will you have me define number? I have not broken any definition and if I did then I am quite certain that by now you would have identified the conflict. Instead you've offered up several resolutions while denying the conflict. Having offered the ring definition you've merely gotten to asking for more definitions. I suggest that this behavior is indicative of the future conversation. It will be more one of stifle than of content. The reliance upon an abstract X within the polynomial, and the marrying of this completely undefined entity with a real number is ultimately the worst abuse of the abstract algebra curriculum, bolstered by abuses down lower, I believe. If I have any deeper criticism it is that modern math is off by one, but that is cryptic. Before here on this thread I have cleanly expressed a conflict. I do thank you for your patience Mr. Clarke. I have no doubt that you are a fine mathematician. I apologize for sharing an inconvenient truth. Abstract algebra is reaching, but it has been reaching up for something which lays buried beneath it. - Tim |