On Thu, Dec 27, 2012 at 7:47 AM, Paul Tanner <firstname.lastname@example.org> wrote: > On Wed, Dec 26, 2012 at 11:38 PM, kirby urner <email@example.com> wrote: > ... >> Contradiction? >> > > In all these posts in this thread you have yet to actually state what > the contradiction is supposed to be, since you have not actually put > forth a contradiction, which is by definition a statement that is > false in all its substitution instances - that is, for instance when > doing truth tables, in the main column there would be nothing but but > F's. (A tautology is by definition a statement that is true in all its > substitution instances - when doing truth tables, in the main column > of its truth table there would be nothing but but T's.)
"Contradiction" is an English word that has survived the centuries without being co-opted by any sub-sect or religious body for purely its own purposes, although of course they're welcome to piggy-back, as is their wont.
I hope you're not so dismissive of student difficulties when they're trying to get their minds around concepts with inherent conundrums.
Picture it as a debate between two opposing sides if you like. You are the judge and need to score each debater and declare a winner.
Resolved: the limit of |360 - v| as the number n of vertexes v on a geodesic sphere increases to infinity is 0.
Debater A (affirmative): a simple epsilon-delta proof will do the job. As we all learned in calculus, if I can give you a small epsilon e such that |360 - v| < epsilon, when n (number of vertexes) > delta, and if I can show that for any epsilon, no matter how small, a corresponding delta might be found, then | 360 - v | < epsilon indeed has 0 as its limit as n -> infinity. QED.
Debater B (negation): we know conclusively and without doubt from Descartes' Deficit, that |360 - v|, no matter how small, is never 0, because the difference, however vanishing, contributes to a total of 720 and this number holds constant regardless of your delta or epsilon, so I piss on your "proof".
So there's your p & ~p where p = proposition (the resolution).
What's required to set up this debate is the conceptual process for increasing the frequency (number of edges or "spokes") from one pentagon to its neighbor to get successively higher frequency "icosa-spheres" (0 frequency is the icosahedron itself whereas 3-frequency is a soccer ball or buckyball (C60 molecule) with spokes added to the hexagons and pentagons to omni-triangulate the ball).
Here's how I imagine our judges for this debate coming down:
Judge Hansen: Debater A wins because you can never have "infinity triangles" i.e. 1/n approaches 0 as n->infinity but at no point is 1/n actually 0, just as 0.999... is never 1 if you stop with the 9s at some point. [Editor: "..." is one of the most magical of all symbols in math, given what its allowed to do for us in our imaginations]
Judge Tanner: There is no real contradiction here. Debater A wins because Debater B said "piss" which lacks decorum.
Judge Urner: I thought debater B made a stronger case by pulling in Descartes' Deficit, an unchanging 720 [Editor: 720 = number of degrees in a tetrahedron], a deft maneuver in what appeared to be a hopeless case. B rescued a more difficult position from obscurity and so wins on that basis.
> > One standard form of a contradiction is a statement of the form p & > ~p, (the conjunction of some statement p and its negation ~p), and > since you have yet to explicitly put forth any statement form that is > false in all its substitution instances - much less prove that said > statement is false in all its substitution instances, you have yet to > put forth the form p & ~p and tell us what statement p is supposed to > be. > > This is a common problem when people put forth claims of > contradiction, the sloppy argumentation of claiming that some > contradiction exists but never actually doing the careful work in > crafting an argument such that the one crafting the argument would be > able to give the actual statement of a contradiction, a statement that > is false in all its substitution instances, and prove that it is false > in all its substitution instances, where one way to do this would be > to simply put forth a statement form like p & ~p and prove that this > form necessarily derives from the argument. > > I would think that those who engage in such sloppy argumentation and > wrongly make claims of contradiction would find the error in their > thinking if only they were to actually try to put forth an argument > such that a known contradiction form like p & ~p necessarily derives > from the argument - that is, they would soon find that they cannot put