User:Numberslogicacquisition

Each of the following steps(lemmas) belong to the overall proof of polyhedrons as proposed by Cauchy:

Step 1:

Let us imagine the polyhedron to be hollow, with a surface made of thin rubber. If we cut out one of the faces, we can stretch the remaining surface flat on the blackboard, without tearing it. The faces and edges will be deformed, the edge become curved, but V and E will not alter, so that if and only if V-E+F=1 for this flat network -remember that we have removed one face.

Step 2:

Now we triangulate our map - it does indeed look like a geographical map. We draw(possibly curvilinear)diagonals in those(possibly curvilinear) polygons which are not already (possibly curvilinear) triangles. By drawing each diagonal we increase both E and F by one, so that the total V-E+F will not be altered.

Step 3:

From the triangulated network we now remove the triangles one by one. To remove a triangle we either remove an edge-upon which one face and one edge disappear, or we remove two edges and a vertex - upon which one face, two edges and one vertex disappear.

The lemmas they're referring to here are the individual steps above comprising the whole proof. Remember, that the original conjecture is Euler's characteristic: V-E+F=1.

First subconjecture:

I wonder, I see that this experiment can be performed for a cube or for a tetrahedron, but how am I to know that it can be performed for any polyhedron? For instance, are you sure, Sir, that any polyhedron,after having a face removed, can be stretched flat on the blackboard?

Second subconjecture:

Are you sure that in triangulating the map one will always get a new face for any new edge?

Third subconjecture:

Are you sure that there are only two alternatives - the disappearance. of one edge or else of two edges and a vertex - when one drops the triangles one by one? Are you even sure that one is left with a single triangle at the end of this process?

Criticism of the Proof by Counterexamples which are Local but not Global:

- given on page 10 it is being brought up to just remove an "interior face" of the network(there is an interior face and an external face which surrounds the network on the other side)leaving all edges and vertexes intact disproving the third lemma.

-The rebuttal or refute:

This first counter-example is one against the proof but still isn't a counter-example to the main conjecture:V-E+F=2. So it's stated: "You have shown the poverty of the argument-the proof-but not the falsity of our conjecture.

Local counterexample - an example which refutes a lemma (without necessarily refuting the main conjecture).

Global counterexample - an example which refutes the main conjecture itself.

The counterexample given above is local and is therefore a criticism of the proof but not the conjecture.

Refutation of Local Counterexample

The refute made by the Teacher by Lakatos on page 11 is kind of silly actually. Normally, when you look up the Euler Characteristic the steps state clearly to remove the boundary triangle when removing faces from the network first.Under Step 3 for the proof in this book,there is no mention of boundary. Unless Lakatos is trying to give an example of a refute by assuming an observation and putting this detail in later in hindsight for the purpose of this book. -An added note though, at the bottom of the page it is mentioned that Euler(even though this proof is by Cauchy)couldn't make the observation needed to correct a proof for another problem.

Criticism of the Conjecture by Global Counterexamples

The counterexample provided by Alpha on page 13 is a convex polyhedron with the faces of the two cubes being shared. Not only does this contradict the proof by Cauchy since obviously you can't stretch both cubes to form a plane it also contradicts the original conjecture since now you have double of everything: V-E+F=4.

The following is the rejection of the counterexample brought under: Criticism of the Conjecture by Global Counterexamples(use of a convex polyhedron).

(b) Rejection of the counterexample. The method of monster-barring/pathological - this is the elimination round to rule out all nasty counterexamples and verify the original conjecture.

At this point, Alpha and Gamma are providing the counterexamples against the original conjecture and Delta is providing the definitions on the behalf of proving the original conjecture. Notice that Delta is redefining in order to refute the counterexamples. Also like to note that the proceeding of the counterexamples and definitions that follow are with the introduction of the counterexample first and definition of the original conjecture which is explained in a way that refutes the counterexample.

Counterexample against the original conjecture

6(a) - Take two tetrahedra which have an edge in common. 6(b) - Take two tetrahedra which have a vertex in common.

Definition supporting the original conjecture

D3.(1) - A system of polygons arranged in a such a way that exactly two polygons meet at every edge and (2)- it is possible to get from the inside of any polygon to the inside of any other polygon by a route which never crosses any edge at a vertex.

Pause:

Take two tetrahedra which have an edge in common. -This one only has that one edge that it shares with another tetrahedra. In the definition it states that two polygons meet at every edge. I think a more clear definition would be:

A polyhedron(plural:polyhedra)is a three dimensional figure enclosed by a finite number of polygons arranged in such a way that (a)exactly two polygons meet(at any angle)at every edge.

(b)it is possible to get from every polygon to every other polygon by crossing edges of the polyhedron.

- This means that any two polygons meet at angle at every side and not that they meet at every side with each other for one specific pair. It just means that if you look let's say at part of the polyhedron you'll see two polygons attached at their sides for that portion only. If travel around that same polygon it will have another of it's side being shared with another polygon and so on...

Counterexample:(Use of a star-polyhedron in this particular counterexample to disprove the main conjecture):

(1) - exactly two polygons meet at every edge, and (2) - it is possible to get from every polygon to every other polygon without ever crossing a vertex of the polyhedron.

Definition 4(these are all refutations) A polygon is a system of edges arranged in such a way that (1) - exactly two edges meet at every vertex and (2) - the edges have no points in common except the vertices.

Typically polygons are associated with convexity with their edges fairly even and straight to one another at their vertices. The Platonic Solids follow this kind of form. With stars you have the reaching out of it's edges from the center and then coming back in again. The variation in this kind of pattern is great enough so that many would argue about it's being a true convex polygon. This is known as the farthest-pair problem with each space in-between every successive vertex being a convex hull. Finding the roots for each of these points is where the problem comes in. The portion where Gamma describes :"You are misled by your embedding the polygon in a plane - you should let it's limbs stretch out into space!" Is analogous to taking the standard definition of a polygon(Platonic Solid) with it's convexed sides and pulling them outward as opposed to thinking that your embedding the form inward instead.