User talk: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 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 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.

Initial questions are being asked in regard to each of the steps of the proof above. This helps to establish the counterexamples that follow.

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?

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

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:

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 vertices intact disproving the third lemma.

This first counter-example is one against the proof(Cauchy) but the not the conjecture(V-E+F=2). This counterexample is an example of a local counterexample which refutes a lemma(-part of the proof from one of the steps provided above)so as a result the whole proof suffers.

At this point we haven't gotten to a Global counterexample which then would bring about the validity of the conjecture itself.

The refutations are coming from both directions: the counterexamples debunking either the proof or conjecture or the definitions or redefinitions upholding the original proof and conjecture and debunking the counterexamples. As with the above counterexample then follows the refutation to this local counterexample in this instance being made by the teacher. The refutation being made by the TEACHER(Lakatos pg. 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 the 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

This counterexample provided by Alpha on pg 13 is a convex polyhedron with the faces of 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 you now have double of everything:V-E+F=4.

The following is the rejection of this counterexample(again a refutation).

(b)Rejection of the counterexample. This method of monster-barring is to eliminate this counterexample and all subsequent counterexamples and in process verifying the original conjecture.

It's at this point, that 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 making redefinitions in order to refute the counterexamples. What follows are the counterexamples and redefinitions for and against the original conjecture.

Counterexamples 6 a and b

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

Definition 3

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

Pause:

Take two tetrahedral which have an edge in common. -This one only has that one edge that is 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 an 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 you travel around that same polygon it will have another of it's sides being shared with another polygon and so on....

Counterexample 7 This one uses a star-polygon

(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 polygon.

Definition 4 A poly 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.

An explanation of star polygons and the controversy that surrounds them

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 a star form of polygon you have this 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. That part where Gamma states:"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.

Goldbach's Conjecture:

Any even number may be expressed as the sum of two primes.

Proof:

Steps in making the proof:

Step 1: calculating the probabilities for prime and non-prime meetings:

Only three combinations: prime+prime, prime+non-prime, non-prime+non-prime - abbreviations P+P,P+N,and N+N.

Step 2: Redefine and transform probability fractions into densities,allowing to develop a proof without probabilities.

We are given that some fraction of all integers is prime:

- fraction immediately in this case means the portion of all the available numbers to choose from.

Example of the combinations given:

20=

1+19 2+18  Using 20 to be able to be able to start the pairs from 1. 3+17 4+16 5+15 6+14 7+13 8+12 9+11 10+10

We are given that some fraction of all integers is prime. If that fraction were > 1/4, then we would need no further proof of GC, since in that case the number of primes would be greater than non-prime odds: The non-prime odds are the N+N combinations. From the following table on page 45 Critical Thinking there are exactly 25 primes out of 100

so primes = only 1/4 and no more:

2   3      5    7      1   13   17     19   23     29   31   37     41   43     47   53   59     61   67     71   73   79     83   89     97

If the number of primes were >1/4 it would overwhelm the combination pairs in the table above. A prime would have to meet a prime and since the count in the above table only goes up to 20 the prime combinations are frequent since most of the low numbers are prime. But as the count goes higher the primes go down some. As the even numbers get larger and larger the fraction of primes gets smaller and smaller.

Again,the portion of primes was lowered to 1/6 in order to work the combination of pairs table going up to 20. As it stands since we're at the lower numbers there are more primes involved and keeping the primes at 1/4 would be too much to work with. Lowering the fraction to 1/6 gives more of an example of what it would be like if more integers beyond 20 were to be used.

Just a note:It's mentioned that there are complicated proofs that show the limit to when the separation or difference of even numbers compared to primes comes to a point but they're not shown here.

Using a fraction that is less than 1/4 let's say 1/6

The probabillity of a P meeeting a P are 1/36 where you mutliply 1/6 * 1/6 to arrive at 1/36.You take the previous example of 1/6 * 1/6 for P+P and the remaining of the whole which would be 6/6 is 5/6 since 1/6 is already taken by P+P. And since you have to take both instances of P+N and N+P then you have to multiply by 2. N meeting N becomes 5/6 * 5/6=25/36.

In the conjecture any even number may be expressed as the sum of two primes. Now it's being stated that a lot of those N's are even and if you look in the list of of the above sums going up to 20 you'll notice that it happens that all of evens are matching up with other evens(and odds are matching up with odds). -So the probability that an even will meet an even in this list is 1. It is 100%. It's on this presumption that all of the evens are eliminated and odd numbers are involved so that there is a variance in order for probabilities to exist. If the original sum of all integers is 1/6 and you removed the evens you have just increased the proportion of prime instances within the remaining odds up to 1/3. Then P+P = 1/9. Don't get P+P confused with P meeting P. One is added and the other is multiplied. The probability of a P meeting an N are 2(1/3 * 2/3)= 4/9. Again since 1/3 is the amount of primes this leaves the remaining 2/3 to be non-primes for a complete 3/3=1. Since N=2/3 then N

But there is a serious problem with this proof. At no point does the author use any properties of the prime numbers, except that they are odd. If the proof were valid, replacing the word prime in the proof with any word that signifies a set of odd numbers would again give a valid proof and thus every even integer would be representable as the sum of two such numbers. That is certainly not the case for all sets of odd numbers, showing that this proof is not correct.

The following is an extraction from the proof and could be considered a subconjecture. The author of the proof is stating it based on the example that he gave for the combination of pairs going up to 20:

Lemma or subconjecture

We are given that some fraction of all integers is prime. If that fraction were > 1/4, then we would need no further proof of GC, since in that case the number of primes would be greater than non-prime odds: The non-prime odds are the N+N combinations.

Local Counterexample

- Try to find a 20 pair combination in possibly the same up/down arrangement that he gave or a variation of this picking at random any 20 pair combination within a particular set of 20 numbers anywhere for all and any integers.