On Sun, May 27, 2012 at 2:51 PM, Aleksandr Lokshin <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
The notion of "choosing" isn't actually important--if a proof says
something like "pick an arbitrary member of the set X, and you will find it
obeys Y", this is equivalent to the statement "every member of the set X
obeys Y". In formal logic this would be expressed in terms of 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, in proofs
like this one typically *doesn't* imagine choosing any specific triangle,
one just thinks about properties that would apply to every member of the
set and thus every "arbitrary member", like the property of having three
sides or or the property of having its angles add up to 180 degrees in the
case of a triangle obeying Euclidean axioms. And note that any mathematical
proof can be expressed in a formal symbolic way using logical symbols/rules
as well as some symbols/rules specific to the domain of mathematics under
consideration (see http://en.wikipedia.org/wiki/Formal_proof ), and in this
form the proof will often contain the universal quantification symbol, but
there is no separate symbol corresponding to the notion of "pick an
arbitrary member of the set".
> On the other hand when proving the formula S=ab/2, obviously, it is
> impossible to consider all the triangles simultaneously.
Why not? One can consider the properties that all these triangles are
defined to share, and then show that these properties, along with the
axioms of geometry, can be used to derive some other properties they will
> 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?
> 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 of being
sure that just because *your* choice obeyed a certain rule, every other
possible choice of examples from the same set would do so as well.
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