On 5/29/2012 2:22 PM, Jesse Mazer wrote:

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On Sun, May 27, 2012 at 2:51 PM, Aleksandr Lokshin<aaloks...@gmail.com <mailto:aaloks...@gmail.com>> wrote:To make the general idea more clear , suppose we are proving the well- known formula S = ah/2 for the area of a triangle. Our proof will necessarily begin as follows: “Let us consider AN ARBITRARY triangle…” Here we obviously apply the operator of the free will choice which cannot be replaced by the random choice. In fact, let us imagine that our proof begins in such a way : “Let us consider A RANDOMLY SELECTED triangle…” Surely, such a beginning will not lead us to the desired proof. The formula obtained for a randomly selected triangle is not necessarily valid for all triangles!The notion of "choosing" isn't actually important--if a proof sayssomething like "pick an arbitrary member of the set X, and you willfind it obeys Y", this is equivalent to the statement "every member ofthe set X obeys Y". In formal logic this would be expressed in termsof the upside-down A symbol that represents "universal quantification"in a given "universe of discourse" such as the set of all triangles (http://en.wikipedia.org/wiki/Universal_quantification ). In fact, inproofs like this one typically *doesn't* imagine choosing any specifictriangle, one just thinks about properties that would apply to everymember of the set and thus every "arbitrary member", like the propertyof having three sides or or the property of having its angles add upto 180 degrees in the case of a triangle obeying Euclidean axioms. Andnote that any mathematical proof can be expressed in a formal symbolicway using logical symbols/rules as well as some symbols/rules specificto the domain of mathematics under consideration (seehttp://en.wikipedia.org/wiki/Formal_proof ), and in this form theproof will often contain the universal quantification symbol, butthere is no separate symbol corresponding to the notion of "pick anarbitrary member of the set".On the other hand when proving the formula S=ab/2, obviously, it isimpossible to consider all the triangles simultaneously.Why not? One can consider the properties that all these triangles aredefined to share, and then show that these properties, along with theaxioms of geometry, can be used to derive some other properties theywill all share.Thus the operator of the free will choice must be used inevitably. More widely, let us consider a variable x which is running about a sphere of radius 1. Let us pose a question: what does x denote? Clearly, a) x does not denote an object, b) x does not denote a multitude, c) x does not denote a physical process. In my opinion, x denotes the free will choice which the reader of the mathematical text must do. So, the notion of a variable inevitably is based on the notion of the free will.If it really depended on free choice, then you would have no way ofbeing sure that just because *your* choice obeyed a certain rule,every other possible choice of examples from the same set would do soas well.Jesse --

Hi Jesse,

`Would it be correct to think of "arbitrary" as used here as meaning`

`" some y subset Y identified by some function i or mapping j that is not`

`a subset (or faithfully represented) in X, yet x => y : x /subset X"?`

`The "choice" of a basis of a linear space comes to mind. The idea is`

`that one it is not necessary to specify the method of identification ab`

`initio <http://en.wikipedia.org/wiki/Ab_initio>.`

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