Fibonacci sequence and now the AP subset sequence 0, 2, 5, 9, 14, 20, 27, 35, 44, 54, . .

[Archimedes Plutonium]

Now we all know the trick involved in writing the Fibonacci Sequence for we start with 1 and 1 and add to give 2, then we add the last two previous, of 1+2 = 3, then 2+3 = 5, then 3+5=8, giving us the sequence 2, 3, 5, 8, 13, 21, 34, 55, etc etc.

Funny how in Old Math, we have just one famous sequence like this Fibonacci, but no longer, because now we have a second famous such sequence. I am going to call it the Subset sequence.

A set of 2 members has 0 subsets, and a set of 3 members has {1,2} and {2,3} for 2 subsets, and a set of 4 members has {1,2} {2,3} {3,4} {1,2,3} {2,3,4} for a total of 5 subsets possible. Now what is the sequence pattern?

We have 2 plus 4-1 = 5, then 5 plus 5-1 = 9, then 9 plus 6-1 = 14, then 14 plus 7-1=20, etc etc

Now there is no formula for the Fibonacci sequence that I am aware of, other than to go and do the algorithm above, and likewise for Subset sequence.

AP

[red...@siu.edu]

The Fibonacci sequence isn't the only famous sequence. There its companion sequence: the Lucas sequence. There are also the Pell sequences which were more or less discovered by the Greek mathematician Theon when approximating the sqrt(2). See Theon's ladder.

Also there is a closed form for the Fibonacci sequence. It's called the Binet formula.

Your sequence, if I interpretted what you said correctly is

a_(n+1) = a_n + n+2,

for n >= 0. Thus we can find a closed form, namely

a_n = n(n + 3)/2.

Don

[Archimedes Plutonium]

On Saturday, June 20, 2015 at 1:42:12 PM UTC-5, red...@siu.edu wrote:
> On Saturday, June 20, 2015 at 1:11:48 PM UTC-5, Archimedes Plutonium wrote:
> > Now we all know the trick involved in writing the Fibonacci Sequence for we start with 1 and 1 and add to give 2, then we add the last two previous, of 1+2 = 3, then 2+3 = 5, then 3+5=8, giving us the sequence 2, 3, 5, 8, 13, 21, 34, 55, etc etc.
> >
> > Funny how in Old Math, we have just one famous sequence like this Fibonacci, but no longer, because now we have a second famous such sequence. I am going to call it the Subset sequence.
> >
> > A set of 2 members has 0 subsets, and a set of 3 members has {1,2} and {2,3} for 2 subsets, and a set of 4 members has {1,2} {2,3} {3,4} {1,2,3} {2,3,4} for a total of 5 subsets possible. Now what is the sequence pattern?
> >
> > We have 2 plus 4-1 = 5, then 5 plus 5-1 = 9, then 9 plus 6-1 = 14, then 14 plus 7-1=20, etc etc
> >
> > Now there is no formula for the Fibonacci sequence that I am aware of, other than to go and do the algorithm above, and likewise for Subset sequence.
> >
> > AP
>
> The Fibonacci sequence isn't the only famous sequence. There its companion sequence: the Lucas sequence. There are also the Pell sequences which were more or less discovered by the Greek mathematician Theon when approximating the sqrt(2). See Theon's ladder.
>
> Also there is a closed form for the Fibonacci sequence. It's called the Binet formula.
>

Alright, so looking at Wikipedia-- every sequence defined by a linear recurrence with constant coefficients, the Fibonacci numbers have a closed-form solution.

A division formula involving sqrt5, where F = (t^n -p^n) / sqrt5

Bravo

> Your sequence, if I interpretted what you said correctly is
>
> a_(n+1) = a_n + n+2,
>
> for n >= 0. Thus we can find a closed form, namely
>
> a_n = n(n + 3)/2.
>
> Don

Let me see, so 1 we get 4/2 = 2, then 2 we get 10/2 = 5, then 3 we get 18/2 = 9.

Looks like it is achieving the numbers results well, but can it be indexed so that n=1 means a set of 3 members, n=2 means a set of 4 members, etc. I see no reason why it could not be indexed. And indexing would be trivial matter.

I was hoping my sequence had irrationals, apparently not.

Does my sequence form of n(n+3) / 2 have a probability theory formula to match with? Or, has anyone seen this form elsewhere?

AP

[konyberg]

Look up the term "generating functions"

KON

[konyberg]

Look what you write: 1,2,3,4,5,6,7.8.8.,10

In 10, where did the 0 come from?

KON

[Archimedes Plutonium]

Now I need to pause and reflect here as to what I have done for Whole/Part theory and what I have done to destroy Set/subset theory.

What I have done is show that a set can only be well-defined if at the start we have a rule/formula/function that completely describes the set or Whole. That rule/formula/function can then produce the members of the set or parts of the whole.

Now can a set of 4 members whose rule is the first four Counting Numbers of Peano {1,2,3,4} have more than just 5 subsets? The subsets are {1,2}, {2,3}, {3,4}, {1,2,3}, {2,3,4} and why cannot {1} or {1,3} be a subset? The answer is that the same Rule that describes the Whole, must be the same rule used to describe the subsets, so that a singleton of {1} no longer obeys the rule of Peano of successor or math-induction, and the set {1,3} can no longer use the original rule. So a Whole must be governed by a rule/function/formula and that same rule must govern its subsets. When we take a rock ore sample and ask for its parts, the formula of its crystal nature transfers to when the ore is split into 1/3 and 2/3 sample.

So, subsets in New Math, take on a far different picture than in Old Math, and that in New Math a set can be thought of as a Rope or string where its subsets are cut pieces of that rope or string. A ore can be thought of as cut in pieces and its pieces as subsets of the original ore.

Now the implications of this are immense in that Sets are always countable, never otherwise. And the Axiom of Choice in Old Math is simply the recognition that you failed to provide a rule/formula/function at the outset, and your blatant omission is now called into duty to well define the set. In New Math, we need no Axiom of Choice, because all sets are well defined by a rule/function/formula.

In Old Math, their set theory was all focused on membership, and not focused on whether you have an object well-defined or an object that is marred in being foggy and unclear.

AP

[Archimedes Plutonium]

On Saturday, June 20, 2015 at 2:32:25 PM UTC-5, Archimedes Plutonium wrote:
> On Saturday, June 20, 2015 at 1:42:12 PM UTC-5, red...@siu.edu wrote:

(snipped)

>
> > Your sequence, if I interpretted what you said correctly is
> >
> > a_(n+1) = a_n + n+2,
> >
> > for n >= 0. Thus we can find a closed form, namely
> >
> > a_n = n(n + 3)/2.
> >
> > Don
>

Alright, looking around on the Web, I searched for n(n + 3)/2 and spotted n(n - 3)/2 which gives the very same sequence, once indexed, of 2, 5, 9, 14, 20, 27, 35, 44, 55, etc.

Surprising to me, it comes not from probability theory where Old Math's subsets are governed by 2^n that comes from n!/ k!*(n-k)!.

But rather n(n + 3)/2 indexed as n(n - 3)/2 comes from geometry as the formula of total diagonals from a given n sided polygon.

So, how do I make the diagonals associated with sides of polygon relate to number of total subsets? How do I make the idea that a Set is a rope or string, and the number of subsets of this rope or string is the number of diagonals?

Can it be that subsets are intimately related to diagonals of polygons? Seems like a stretch, and will think about it tomorrow.

AP

[Archimedes Plutonium]

On Sunday, June 21, 2015 at 3:34:20 AM UTC-5, Archimedes Plutonium wrote:
(snipped)

>
> Alright, looking around on the Web, I searched for n(n + 3)/2 and spotted n(n - 3)/2 which gives the very same sequence, once indexed, of 2, 5, 9, 14, 20, 27, 35, 44, 55, etc.
>
> Surprising to me, it comes not from probability theory where Old Math's subsets are governed by 2^n that comes from n!/ k!*(n-k)!.
>
> But rather n(n + 3)/2 indexed as n(n - 3)/2 comes from geometry as the formula of total diagonals from a given n sided polygon.
>
> So, how do I make the diagonals associated with sides of polygon relate to number of total subsets? How do I make the idea that a Set is a rope or string, and the number of subsets of this rope or string is the number of diagonals?
>
> Can it be that subsets are intimately related to diagonals of polygons? Seems like a stretch, and will think about it tomorrow.
>

Ah yes, some light is coming through, and rather post it now than wait till tomorrow.

If we consider a set that exists, meaning well-defined by rule/formula/function, it can be considered a object. Not a rope or string with subsets as smaller pieces of the rope or string, but rather as a square or rectangle or circle or ellipse type of shape. But since the rule/formula/function makes for a sequence or math-induction upon the Set, we can consider the sides of a polygon as the terms of the sequence. And we can consider that a subset of such an object would be a cut or cleaving along two vertices-- that is a diagonal.

Now in Old Math, they loved Venn diagrams to show blobs, closed blobs for sets, in union, in intersection in compliment etc.

In New Math, our sets are not blobs that are ill defined because they have no rule/formula/function, but are well-defined since each term of the set can be found from the rule, formula, function and those terms are the sides of a polygon, and where the diagonals are thus the total possible cleaving by vertices.

AP

[Archimedes Plutonium]

On Sunday, June 21, 2015 at 3:51:26 AM UTC-5, Archimedes Plutonium wrote:
> On Sunday, June 21, 2015 at 3:34:20 AM UTC-5, Archimedes Plutonium wrote:
> (snipped)
> >
> > Alright, looking around on the Web, I searched for n(n + 3)/2 and spotted n(n - 3)/2 which gives the very same sequence, once indexed, of 2, 5, 9, 14, 20, 27, 35, 44, 55, etc.
> >
> > Surprising to me, it comes not from probability theory where Old Math's subsets are governed by 2^n that comes from n!/ k!*(n-k)!.
> >
> > But rather n(n + 3)/2 indexed as n(n - 3)/2 comes from geometry as the formula of total diagonals from a given n sided polygon.
> >

So, now, how can I justify the idea that given a ore sample in chemistry in 2nd dimension that I can pattern the ore to be a n sided polygon and then all the parts of that ore sample is a cleaving of the sample from diagonals?

Another way of saying this, is given the Counting Numbers which we usually picture as a long line, but now, picture the counting numbers folded to loop back around so the infinity borderline 1*10^604 is looped onto the 0 number, forming a circle. Now form a 1*10^604 sided polygon where each Counting Number is a Vertex. This huge polygon would have n(n-3)/2 diagonals and thus subsets.

The question is can we only cut along the diagonal one at a time to have a whole-part?

Another question, since the whole is well defined by a rule/function/formula, how unique is one diagonal with attending sides? Is each diagonal with attending sides so unique that if we tossed away the original large whole, is the remaining part of diagonal with attending sides able to uniquely reconstruct the whole? If not, then can we have a way of attaching the rule/function/formula to each part?

For example a rectangle has 2 diagonals and if we cut along one diagonal we have a right triangle, and if we threw away the other right triangle we can reconstruct the Whole from the one remaining for it is unique to the Whole. But when we get to a larger polygon and have a triangle part, it likely is not unique to reconstructing the Whole from just that part.

Now see what I have got myself tangled up in.

However, I think I may have a way out-- in the fact that all polygons are transformable into being a rectangle and finally a square. So, maybe if given the rule/formula/function we convert it into a unique square and then a triangle part taken out of the n polygon can be reconstructed back to its original whole-- not sure though.

AP

[Archimedes Plutonium]

Now here is a further interesting question, since we know all polygons can be transformed into a rectangle and finally a square. But, can we take any given polygon and transform it of same number of sides into a regular polygon?

I believe this is true and possible because if you can take an irregular polygon of M sides and turn it into a square, then that square should be REVERSIBLE to turn the M sides into a regular polygon.

And so the amount of diagonals remains the same in both the irregular M polygon as with the regular M polygon.

This is important because if given a triangle part from a regular N sided polygon we can always reconstruct the Whole.

Now, we can say that Regular Polygons are unique to diagonals because constant for all the sides are the same but irregular polygons are variable of sides and diagonal lengths.

So, now, the question remains as to whether a rule/formula/function defines the Whole uniquely. The Peano axioms are a rule defining the Counting Numbers and it is well known that these are a unique set.

AP

[Archimedes Plutonium]

Now there is a powerful theorem in Old Math which says that 3 points in the plane determine a unique circle. It is true and from that we can see that 3 points would determine any N sided irregular polygon inscribed inside that circle.

Further, the 3 random points determining a unique circle, can be indexed of those 3 random points to draw a Unique Regular Polygon inside that circle. What I mean by indexed is that we have the distance between points A and B and A and C and B and C and so we find a "common distance of those three points" and make this common distance be the side N of a regular polygon.

For example if the triangle produced of A,B,C were distances of 8, 20, 24 then indexed we could use 2 or 4 as the distance of the side of a Regular Polygon.

What am I driving for here? I am driving to achieve the understanding of how the diagonals of a Polygon are the subsets of the set.

AP

[Archimedes Plutonium]

Alright, I seemed to have lost my train of thought here on how diagonals of n sided polygons becomes the subset of the set involved, since the diagonals follow the sequence 2, 5, 9, 14, 20, etc etc of either n(n+3)/ 2 or indexed to n(n-3)/ 2.

What I want to know is that, since the Whole is precisely defined by a rule/function/formula, and so that very same rule/function/formula must define a snipped part of the whole. And so, it is reasonable to think that I can have a part in isolation of the original whole and by using that rule/function/formula, I should be able to assemble the original Whole?

Now, so far, I think the answer is that if the polygon is a Regular N sided Polygon with a triangle of 2 sides and the diagonal, that we can recreate back together the original whole polygon. But if the original whole polygon is irregular, seems to me, that we cannot reform what the original whole was. Unless I am missing something here, and it is too late at night, so will take this up tomorrow. The problem is that if diagonals are the geometrical analog of subsets, then I should be able to recreate the Whole given a part-- whether regular or irregular.

AP

[Archimedes Plutonium]

Nothing better than doing Examples to clear the air of confusion.

Driving Question to be made clear: If given a triangle as a part of the Whole, for which the remainder of the Whole is gone, can we recreate what the Whole was? Because in Whole/Part theory, the governing rule/function/formula governs the parts of the whole, and so a part should have the unique genetic fingerprint of the Whole. So. to test that out, let me give three Wholes with a part and see if the part can recreate the Whole.

Function y = x+1 as Successor Function in Peano Axioms giving me the Whole Set of {1,2,3,4} of domain 0, 1,2,3

Rule: Rule of Fibonacci sequence of adding prior two counting numbers to obtain next term giving the Whole Set of {2,3,5,8} of origin-points 1,1.

Formula: N(N-3)/2 (or indexed N(N+3)/2) giving me the Whole Set {2,5,9,14} of domain 4,5,6,7

IMPOSSIBLE TRIANGLE THEOREM: Given two lengths of line segments as the two sides of a triangle which when added together is equal to or greater than a third length line segment, then the three segments of the order imposed upon them, cannot form a triangle.
Proof: rather obvious, and leave it to the reader (and I suspect this is a new theorem).

Example of theorem using the Fibonacci sequence, for we have three line segments of 2,3,5 lengths and required to construct a triangle with them, but impossible because the sides 2 and 3, the first two terms in order, cannot be formed into a triangle whose long side is 5.

TRIANGLE MACHINE: this is a machine that gives exact lengths of sides as metal rods for which the end of the rods can be connected to other rods. So that if we had a rod of 2 long and another of 3 long and connected to a rod of 5 long, we cannot form a triangle, and the only configuration is that the 2+3 lies superimposed on top of the 5 long rod as a straightline segment.

So, in the Fibonacci sequence we can never form a triangle from three consecutive terms of the sequence. In the Counting Numbers sequence we cannot form a triangle with rods of 1,2,3, however we can form a triangle with the next sequence 2,3,4. For the N(N-3)/2 sequence, we cannot form a triangle from rods of length 2,5,9, nor of the next sequence 5,9,14, but we can after that of 9,14,20.

4-gon Possibility Theorem: we saw for triangles, that the 2 sides of the first two terms of sequence have to be added up and greater than the third side, third term in order to form a triangle. Would this requirement exist for the 4-gon? the 5-gon? etc etc

If we had a sequence of 1, 10, 100, 1000 of rod lengths and required to build a 4-gon, we see it is impossible. If we had the sequence 1,10,100,1000,10^4 and required to build a 5-gon, here again, that is impossible.

This post has become prematurely too long and so will continue in the next, where I examine a Part of the Whole as triangles of the Fibonacci, Counting Numbers, N(N-3)/2 sequences.

I suspect what the final answer is going to be, is that the sequences form Unique angles in the triangles, and from that unique angle, we can recreate the Whole from given just a single Part of the Whole.

AP

[konyberg]

You think this is new? You are so ignorant and got it wrong! The sum of two sides has to be greater than the third.

KON

KON

[Archimedes Plutonium]

On Monday, June 22, 2015 at 3:05:34 PM UTC-5, konyberg wrote:
> mandag 22. juni 2015 21.49.27 UTC+2 skrev Archimedes Plutonium følgende:

(snipped)

>
> "IMPOSSIBLE TRIANGLE THEOREM: Given two lengths of line segments as the two sides of a triangle which when added together is equal to or greater than a third length line segment, then the three segments of the order imposed upon them, cannot form a triangle.
> Proof: rather obvious, and leave it to the reader (and I suspect this is a new theorem)."
>
> You think this is new? You are so ignorant and got it wrong! The sum of two sides has to be greater than the third.
>
> KON

Again, the dumbest and mindless Norwegian in math, or, the finest that Norway can produce in math of Karl-Olav Nyberg. Does he have Alzheimers' disease, that he cannot comprehend what the other person writes.

I hand KON three lengths of rays 1, 2, 3. I tell him to form a triangle. KON says 2 + 3 has to be greater than 1. But, where is the triangle KON?

KON never understood that Rays come before triangles.

KON can only see a formed triangle first, and then make stupid comments on its rays.

KON cannot understand what others say or do, and is a worthless menace to sci.math.

AP

[Archimedes Plutonium]

On Monday, June 22, 2015 at 2:49:27 PM UTC-5, Archimedes Plutonium wrote:
> Nothing better than doing Examples to clear the air of confusion.
>
> Driving Question to be made clear: If given a triangle as a part of the Whole, for which the remainder of the Whole is gone, can we recreate what the Whole was? Because in Whole/Part theory, the governing rule/function/formula governs the parts of the whole, and so a part should have the unique genetic fingerprint of the Whole. So. to test that out, let me give three Wholes with a part and see if the part can recreate the Whole.
>
> Function y = x+1 as Successor Function in Peano Axioms giving me the Whole Set of {1,2,3,4} of domain 0, 1,2,3
>
> Rule: Rule of Fibonacci sequence of adding prior two counting numbers to obtain next term giving the Whole Set of {2,3,5,8} of origin-points 1,1.
>
> Formula: N(N-3)/2 (or indexed N(N+3)/2) giving me the Whole Set {2,5,9,14} of domain 4,5,6,7
>
> IMPOSSIBLE TRIANGLE THEOREM: Given two lengths of line segments as the two sides of a triangle which when added together is equal to or greater than a third length line segment, then the three segments of the order imposed upon them, cannot form a triangle.

(snipped)

Alright, continuing where I left off.

We see that 1,2,3 cannot form a 3-gon, nor can 2,5,9, nor can 2,3,5.

We apply that theorem of the 3-gon to that of 4-gon, 5-gon etc etc, which would say that a 4-gon can only exist if the lengths of rays A, B, C when added together is greater than the length of D and the same idea for 5-gon and higher.

So, in my three sequences given of {1,2,3,4, . .} and {2,3,5,8, 13, 21, 34, . . } and {2,5,9, 14, 20, 27, 35, . .} what is the first 3-gon and first 4-gon and first 5-gon to be able to be formed if those terms each represented the length of a Ray?

Well, the first 3-gon is 2,3,4, and first 4-gon is 1,2,3,4 and first 5-gon is 1,2,3,4,5 for {1,2,3,4, . .}

The first 3-gon is none, the first 4-gon is 2,3,5,8 and first 5-gon is 2,3,5,8,13 for {2,3,5,8, 13, 21, 34, . . }

The first 3-gon is 9,14,20 and first 4-gon is 2,5,9,14 and first 5-gon is 2,5,9,14,20 for {2,5,9, 14, 20, 27, 35, . .}

Now each of those GONS of 4-gons and higher, if they exist in the sequence will have diagonals, according to the formula N(N-3)/2. And each of those 4-gons or higher with diagonals form triangles within the GONS where the sides of the GONS and the diagonal form a triangle.

And, here the question is, to be clear, is whether a GON with diagonal that forms a triangle, whether that triangle can tell us the original Sequence if all we knew were facts of that triangle?

The question, put differently, is that whether the Whole Sequence, governed by a rule/function/formula leaves enough of a genetic imprint on the triangle that it can tell us the Whole of which that triangle came from.

For example, the 4-gon of 2,5,9,14 has a triangle of 2,5, diagonal and if we had just a triangle of that 2,5,diagonal, can we find out the original Whole of {2,5,9, 14, 20, 27, 35, . .}. Can that triangle tell us the Whole was the formula N(N-3)/2 or, is it not unique the triangle?

If I were to guess, that all sequences governed by a rule/formula/function are unique and that the triangle part of the sequence would form a unique triangle angle that would come back to a uniquely specific sequence.

Sometimes we ask questions in math, and kind of have a foggy question in mind, for which we have to sort it out to make the question clear.

AP

[Port563]

"Archimedes Plutonium" <plutonium....@gmail.com> wrote in message
news:cb778aed-a3b9-4d3f...@googlegroups.com...

> On Monday, June 22, 2015 at 2:49:27 PM UTC-5, Archimedes Plutonium wrote:

(( snip Poo's poo ))

> Sometimes we ask questions in math, and kind of have a foggy question in
> mind,
> for which we have to sort it out to make the question clear.

Hahaha !! (-;

[Archimedes Plutonium]

Now I think the question of whether a triangle part can recreate the Whole is true, but whether just one part can do that is in doubt.

And I believe I can model this Whole/Part of polygons to that of Functions in Calculus. So that Set theory validity is determined by what functions exist and do not exist.

In Old Math, they allowed anyone to fill in between brackets { } and call whatever they put inside the brackets as a set. Pretty crumby way of doing things if you think math should be about "precision". So that in Old Math, to them, the set could be {0, 2, 5, 9, 14, 20, . . } and defined by N(N-3)/2 and then they would say that {0, 2, 4, 9, 14, 20, . . } was a legitimate set in Old Math, even though they had no formula/rule/function that would account for the switch of a 4 rather than a 5.

Old Math also allowed grimy, dirty tampering with functions saying that y = x is a function, but also, saying that at x=0 then y=3, and y= x elsewhere. What I called a "broken function". But is it really a function at all? In other words, if you are going to tamper with making a function by specifying the y-value at some x value, have you not then given up on the meaning of function? That a Function, if it exists can only be written as one equation. So that y=x is a function but not a tampered x=0, y=3, and elsewhere y=x. Step functions that require you to specify many x and y values rather than one equation for all the function over the domain, should not be called functions.

So here in Set theory, I run into the very same silly and hideous notion that a Set can be a specification list, as that of function with more than one equation.

So if there is a overwhelming idea that a function is only one equation over the entire domain and where a set is a set only if one equation over the entire elements, then I would have a easy proof that one triangle out of a Sequence can recreate the Whole of the polygon. For then we know that a function like y= (x^2 + 3x)/2 is unique and so is the sequence of values. And that a Sequence from N(N-3)/2 may not be the only Sequence that starts off with 0, 2, 5, but the only Sequence whose terms of 2 and 5 and then 9 and 14 create a unique angle for the diagonal involved with 2 and 5.

The trouble with the function by most people is that they only picture the function with integer x values, whereas the Sequence formula is integer N values.

So, the proof that I seek would be immensely more easy, if I could someway justify the claim that a function, a rule, a formula be just one equality and not a fabrication list of two or more equalities.

AP

[konyberg]

Can you read what I wrote. The sum of two sides have to be greater than the third. If sides equal 1, 2 and 3, then 1 + 2 = 3, not greater. So no triangle!

KON

[konyberg]

I will spell it out for you.

If the sides of a triangle is a, b and c then
a + b > c, a + c > b and b + c > a. Got it?

KON

[Archimedes Plutonium]

On Monday, June 22, 2015 at 11:45:53 PM UTC-5, Archimedes Plutonium wrote:
> Now I think the question of whether a triangle part can recreate the Whole is true, but whether just one part can do that is in doubt.
>

At the moment, I suspect that a single triangle out of a N-gon is going to tell what the original Whole was because of the angle involved with the diagonal is unique.

What I need to do is have a great justification of throwing out all these tacked-on artificial functions and Sets of artificial inclusion or exclusion of members of Old Math.

I think I found the Logical Justification which I call Double Equalities for one Whole.

> And I believe I can model this Whole/Part of polygons to that of Functions in Calculus. So that Set theory validity is determined by what functions exist and do not exist.
>
> In Old Math, they allowed anyone to fill in between brackets { } and call whatever they put inside the brackets as a set. Pretty crumby way of doing things if you think math should be about "precision". So that in Old Math, to them, the set could be {0, 2, 5, 9, 14, 20, . . } and defined by N(N-3)/2 and then they would say that {0, 2, 4, 9, 14, 20, . . } was a legitimate set in Old Math, even though they had no formula/rule/function that would account for the switch of a 4 rather than a 5.
>

So in Old Math where they had these crazy functions with multiple equality signs.

F(x) = 1 when x=0
F(x) = 0 from x>0 but less than 1
F(x) = 2 from when x=1 or greater

So, there are three equality signs in that Old Math function listed. In True Math, a function can only be created with one equality sign.

So that y = 3 is a function and y = 1/x is a function. But if you have two or more equality signs in order to describe a function, you have no function at all.

Same idea goes for Formula, that a formula such as N(N-3)/2 = number of diagonals cannot have a tacked on second or more equalities for N values.

Same thing goes for Rules, that a Rule has just one equality involved.

> Old Math also allowed grimy, dirty tampering with functions saying that y = x is a function, but also, saying that at x=0 then y=3, and y= x elsewhere. What I called a "broken function". But is it really a function at all? In other words, if you are going to tamper with making a function by specifying the y-value at some x value, have you not then given up on the meaning of function? That a Function, if it exists can only be written as one equation. So that y=x is a function but not a tampered x=0, y=3, and elsewhere y=x. Step functions that require you to specify many x and y values rather than one equation for all the function over the domain, should not be called functions.
>

Now, to show how really silly it was to tack on artificial garbage to a function, consider a Polynomial, where we have something like this x^3 -5x^2 +8 = 0 and so, in keeping with the coneheads of Calculus, suppose we say the Polynomial of:

x = 7.5, otherwise x is
x^3 -5x^2 +8 = 0

Or say in Sequence theory, what if we had the silly nonsense we find in functions of Old Math where we had.

S= 2, 4, 6, 8, 10, 12, 13, 16, 18 . . .

of the Rule 2x where x is a counting number, except for when x = 7 we have 13.

Can the reader detect how phony such tack ons were in Old Math.

> So here in Set theory, I run into the very same silly and hideous notion that a Set can be a specification list, as that of function with more than one equation.
>

So, finally, Set theory of Old Math was rife with mischief of phony tackons, and of deletions.

> So if there is a overwhelming idea that a function is only one equation over the entire domain and where a set is a set only if one equation over the entire elements, then I would have a easy proof that one triangle out of a Sequence can recreate the Whole of the polygon. For then we know that a function like y= (x^2 + 3x)/2 is unique and so is the sequence of values. And that a Sequence from N(N-3)/2 may not be the only Sequence that starts off with 0, 2, 5, but the only Sequence whose terms of 2 and 5 and then 9 and 14 create a unique angle for the diagonal involved with 2 and 5.
>
> The trouble with the function by most people is that they only picture the function with integer x values, whereas the Sequence formula is integer N values.
>
> So, the proof that I seek would be immensely more easy, if I could someway justify the claim that a function, a rule, a formula be just one equality and not a fabrication list of two or more equalities.
>

So, I have the justification I need to rid math of artificial tackons, because a tackon requires a second or more equality signs. A Whole entity requires a single rule/function/formula with one Equality sign to describe it in full.

AP

[konyberg]

In "Old math" we talk about relations. A function is a well ordered relation. That is one value for each input. y = x^2 is a well ordered relation, therefor a function. y^2 = x is not well ordered, therefor not a function. It splits into two functions.

KON

[Port563]

"Archimedes Plutonium" <plutonium....@gmail.com> wrote in message

news:1e92a2d5-cb34-4da9...@googlegroups.com...

> I hand KON three lengths of rays 1, 2, 3. I tell him to form a triangle.
> KON says 2 + 3 has to be greater than 1. But, where is the triangle KON?
> KON never understood that Rays come before triangles.
> KON can only see a formed triangle first, and then make stupid comments on
> its rays.

"Rays"??

Hahaha!

The gibbering loon means "line segments", perhaps? Or even "lines"? Who
knows?


> AP cannot understand what others say or do, and is a worthless menace to
> sci.math.


I corrected your mistake in the above sentence.

[Port563]

"Archimedes Plutonium" <plutonium....@gmail.com> wrote in message

news:814fb7f2-d970-4d0c...@googlegroups.com...

On Monday, June 22, 2015 at 11:45:53 PM UTC-5, Archimedes Plutonium wrote:
>> Now I think the question of whether a triangle part can recreate the
>> Whole is true,
>> but whether just one part can do that is in doubt.
>
> At the moment, I suspect that a single triangle out of a N-gon is going to
> tell what
> the original Whole was because of the angle involved with the diagonal is
> unique.

Hahaha!!

For once - just once - Archie Poo made a statement (in his earlier post)
that is both:

1) Vaguely math-related, and

2) Comprehensible.

Of course it is also utterly trivial, but that went without saying.

It is absurd to even consider that one can deduce what was the original, not
necessarily regular, N-gon from one "sectorial" (I justify that term from
the limiting case of the regular N-gon as N-->oo) triangle.

But in 814fb7f2-d970-4d0c...@googlegroups.com the moron
concludes that it is possible, hahaha.

> What I need to do is have a great justification of throwing out all these
> tacked-on artificial functions and Sets of artificial inclusion or
> exclusion
> of members of Old Math.

> I think I found the Logical Justification which I call Double Equalities
> for one Whole.

Hahaha!

The above is quite a good strategy, Poehlmann.

Be quite as insane as that - I mean, you are nearly in Gabriel territory
there, and then it is true....

... that no asylum will have you.

So you may be permitted to be at large in the Land of the Free for a while
longer.