Concept sharing and Goldbach’s conjecture
Hide and seek is a fun game. But suppose instead of children, numbers would like to hide.
Suppose you are the number 1. Where could you hide? Numbers are so obvious. They go 1,2,3,4…etc. How could 1 hide?
But suppose I have two number 1’s. I could hide both together, by putting them together so that the two of them could not be seen, they just look like 1 number 1.
LIke so 1…..>1(x2)<……1.
Now wait you say, there aren’t two number 1’s! But there are if we have two number lines, perpendicular to each other, as shown. Then the usual number line can be thought of as the diagonal line and we are able to remove the usual 1 that’s there and replace with 2, new 1’s.
Why do we want to have two number 1’s? We can extend numbers into a number plane and see an underlying level of numbers. The symmetry of these numbers can clear up old problems, since we are looking at an underlying level to numbers.
We only have to take the usual 1 that’s there out of the way and two number 1’s can be where that 1 was.
To do this we need another level of numbers like a one story house has a basement. Then I could remove the number 1, like removing the first story, and still have the basement. Then build a two story house on top of the basement that’s left over.
The two number 1’s sharing is an example of Concept Sharing. Concept Sharing is a term used in education. But here I am giving it another definition.
As an entry into Concept Sharing let’s start with the concept of a point. In math this is the notion of an entity with no extent, or in Cartesian geometry the notion of something with position only. These are “little nothings”.
What if something of no extent could be expressed as two items of no extent, placed together. Since an item of no extent placed together with another similar item of no extent would still have no extent, yet there could still be two items here and not one.
Two little nothings still add up to nothing, yet there can be two little nothings there. The two items would just be hidden as one. Let’s hide!-they say. Who can see us? The two items would not be points, for points have a singularity to them. That is, when I combine two moving points, with the usual plane, it leads to one point, but this doesn’t have to be so with another possible entity. See the teacup shadow diagram, below. If we have another underlying plane there can be another entity which has no extent.
Consider a teacup placed on a table with two lights from above. One from the left and one from the right. See below:
Now as seen in the overlap, two different points of the shadows can take up the space of one point. This is analogous to two e’s (a new entity with no extent) sharing.
These two items could share the concept of a point (concept sharing). Let’s share the concept of nothing, the two little nothings say! This is how we can hide!
In order to do this mathematically we would have to take out the usual concept of a point and replace it with these new conceptions. There are these little entities that don’t combine to form a point, this new non-combination of two entities.
This is because a point is already there as the idea of no extent. We need some room! Kick out what’s already there!
What does it mean to take out the concept of a point? It is not the same as removing a point from a given subset of points of the plane. We wish to open up a new possibility, so we have to reject the usual concept of a point and accept this new possibility.
Yet to reject the concept of a point and replace it, it means we have to define a space where a point can be truly taken out of. We need something like another plane, a new dimension, where the usual plane is, yet co-existing with the usual plane! Places of original places. Let’s let the points have another level to them, where they can move off into another plane, coexisting with what is already there. That way it is possible to take them out, otherwise, how can we remove them? I can’t take them out unless there is something underlying!
What if a point had another type of item of no extent that it normally lived in. It is so tiny that it only needs a tiny house to house it!-we have already seen it is necessary to have other types of entities with no extent. This would necessitate another kind of plane of these items co-existing with the usual plane of points. A place where the original place can be taken out of. Another dimension to places, coexisting with existing places.
A place where the original place can be taken out of. Another level to places, coexisting with existing places. That is, a concept space. The idea is to have an infinite number of descending levels as shown below:
If you accept the notion of an entity with no extent you have to accept this new possibility as well. If you want a point, you have to have us too! The new entities say! The notion of no extent leads naturally to concept sharing!
To remove the concept of a point, we need to have a concept space, consisting of different nested levels of the concept.
This must exist because I must have this further concept of a doubled item of no extent somewhere and I need to take out the concept of a single item of no extent to have it appear somewhere. Also we must take out any other possibility of any number of items of no extent, except 2.
Furthermore, all math concepts are point-like in that they are exact ideas which have no existence in physical reality. They are mental constructions which are not available to the senses. They are in thought only. So number, set, group, function, etc. can all have concept spaces!
This, then is how the numbers move, I create a new plane, “lower” than the usual plane. See below:
The locations move off into the other dimension, coexisting with the plane that is originally there. The numbers come along with the moving points. We can have one point at 1, two points at 2, three points at 3, etc.
Now since two number 1’ s are playing hide and seek, how do we tell that it is not the case that we have the usual case of 1 number 1?
We can show that there are really 2, 1’s there by calling it (1(1) like so:
(1)(x2) = (1<….(1) =(1(1). But what are the numbers representing? The number of dots. As shown below:
The left 1 is really together with the right 1, yet I need to work with both of these so both of these are shown.
So not only would 1 like to play this game but the other numbers would like to join in. 2,3,4,5,6,…etc would also like to hide.
So we can have (2(2), (3(3), (4(4)…
Then also we can have the other games that numbers like to play like addition (+) and multiplication (x).
So (1(1)+(1(1)=(2(2) and (2(2)x(3(3)=(6(6) for examples.
These numbers can all be on a line. The hidden number line:
But now 1 says I would like to hide with 2. Since part of 2 is 1 I could hide with part of 2. Then this could be labeled (1(2) and it would be equal to (2(1). We just have a smaller number hiding with a larger number. 1 and 2 are in the exact same place.
But where are these hidden numbers on the hidden number line?
The answer is that these numbers are on a number plane! A plane of numbers that looks like the diagram below:
On the midline we have the usual hidden numbers.
Then we can ask, how can these hidden numbers play the usual games of numbers? (addition and multiplication)
Let’s investigate!
(1(2) + (1(2) =(2(4)
But also,
(1(2) + (2(1)=(3(3). So it appears that in order for (1(2) to play (2(4)=(3(3). But there’s nothing to say this isn’t okay, since there are no rules for these partially hidden numbers-yet!
Goldbach’s Conjecture:
On June 7th, 1742 Christian Goldbach wrote a letter to Leonhard Euler In which Goldbach guessed or “Conjectured” that every even number is the sum of two prime numbers. So for example: 16=3+13.
Let’s look at the guess in terms of the hidden numbers!
We can think about new types of numbers (1(1)(1(2)), (2(1)(2(2)), (3(1)(3(2)),..These are lower numbers than the naturals, the idea of concept sharing being deeper than the usual idea of concepts. The notation is meant to show that the numbers are together in the number space, fitting into one another or hidden as one. I show a revelation of one number “peeking out” from concept sharing with another number. Shown more clearly here: Like so (a)x2….(a<……(a) leading to (a(a). For brevity I drop the (1),(2) extra designation from now on.
There are other numbers such as (2(3), (3(4), (3(5),… We can think of this as partial sharing. For (2(3) think of three dots colored red. We can have two blue dots overlapping with two red dots forming two purple dots. I say that to give the idea, numbers are not colored dots. Numbers are the exact concept of amount or position only. So that (2(3) can be thought of as two sharing with three and (3(2) can be thought of as three sharing with two. In the first case 1 is left over and in the second case 1 is an extra number. These are the same concept though so (2(3)=(3(2).
Consider the even shared numbers (4(4), (6(6),…
We can break them down into a sum of two prime shared pairs such as (4(4)=(2(2)+(2(2) if we define the other types of shared numbers with dissimilar higher numbers to fill in the rest of the possibilities of a number plane. They can form products too.
Some numbers have more than one decomposition. Such as (16(16)=(13(3)+(3(13) and (16(16)=(11(5)+(5(11).
Now think of the even shared numbers as being created from the primitives by multiplication, similar to the way composite numbers are created from primes.
I designate them (4(4) … but these numbers are not created from the usual numbers on the number lines, as these are lower numbers, these lower numbers; (4(4)…. must come from somewhere else. That is I do not choose 4 and 4 and put them together. I create (4(4) from more primitive numbers and then to get to 4, I have to replace (4(4). These even shared numbers must go on indefinitely, as to lead to the numbers 1,2,3,…at the higher level . Then the natural numbers are all at a higher level from these numbers.
For multiplication, in the usual number system we have prime numbers. We can look for the lower level of prime numbers. Suppose I have a lower level of primes by concept sharing two prime numbers. For example (7(3)=(3(7).
Prime numbers are numbers which have 2 factors, 1 and itself. 1 is not a prime number as it has only one factor, 1. It can be expressed as 1×1 or 1x1x1…etc. Any prime factorization of a composite number is unique and so does not include 1 otherwise any composite number could have any number of factors, we could just extend it indefinitely by extending 1.
If we look at (3(7) we see it can be further broken down into (3(1)(1(7). Numbers like these two can be considered the shared counterpart to primes. Yet to have it, it is not a prime factorization. Yet, (2(2)(11(1) could be a prime factorization, since I have a single 2 and 1 but I also have two prime factors. Here then 1 is considered to be a prime number. Non-prime factorization just opens the shared number up and allows for non-finite representation. Yet, so does prime factorization!
So a prime in this system might be represented as (1(p) or (p(1) where p is a prime in the higher system. Then (p(q) where p,q are primes would be a different kind of number, we could call it a binary prime shared number.
If I do include this decomposition then (3(7)=(1(3)(1(7) =(1(21) and then if I have (10(10)=(2(2)(7(3)=(7(3)+(3(7)=(2(2)(21(1)=(42(2)=(22(22). So (10(10)=(22(22). This may lead to interesting mathematics, but if we allow it it means that the representation of a shared number is not finite. We need a finite representation otherwise the sum of the shared numbers is not conserved.
We can make a further restriction to only allow shared number decompositions where the sum of the two component numbers are equal over all possible decompositions. These means the sum of original numbers is conserved. Since we are sharing numbers, this makes sense that the amount we are sharing one way or another can vary but the sum of the numbers must remain the same.
This works out with a factorization where one of the factors is (2(2) as we can switch the two numbers in the other factor. This makes the sum the same. If we allow the sum to be different, there is no finite representation.
Here is an example: (2(2)=(3(1)
If we move by one up or down from (3(3), for example we can obtain (2(4) and ask is (2(4) okay? But (2(4)=(2(1)(1(4) (since 4 is composite this is an allowed decomposition)=(2(1)(1(2)(1(2)=(1(2)(1(2)(1(2)=(1(8) and 1+8=9 and not 6. We seek a prime decomposition which maintains addition of shared numbers.
Then for (16(16) we get (16(16)=(2(2)(p(q) and (13(3)=(11(5). No other binary numbers work as I can further decompose them and by switching lead to two different sums. Also I can’t decompose (13(3) and (11(5) further as this will not be prime factorization.
Think of (12(12)=(2(2)(6(6). So if we are at (6(6) I can move it to (7(5) and this works out because (7(5) has no prime factors of either 7 or 5. This is the way it will work. I must have for example (16(16)=(2(2)(11(5) the second factor has to be where both numbers have to be primes. This is then the definition of these shared even numbers, greater than or equal to (4(4), since they are conserving the sum of the numbers.
Then starting at (4(4), we can ask, how is (4(4) created? (4(4)=(2(2)(2(2). (4(4) is an even shared number so there is a division by (2(2) possible. This can be the definition of an even shared number. When I divide both sides by (2(2) I have a shared number on both sides.
Then also (6(6)=(2(2)(3(3). Also (6(6) =(2(2)(5(1) also as (5(1)+(5(1)=(5(1)+(1(5)=(6(6).
This is a factorization, one is considered a prime number.
In general (2n(2n)=(2(2)(p(q).
So in the shared number system we need a new definition of prime factorization. Let’s look at some more examples.
(6(6)=(2(2)(4(2) but (4(2)=(4(1)(1(2) (4 is composite so it is okay to express the decomposition this way as the shared product is that of a composite)=(2(1)(2(1)(1(2) which does not work as I can break it down further into (6(6)=(2(2)(2(1)(2(1)(1(2)= (2(1)(2(1)(1(2) +(1(2)(1(2)(1(2)=(4(2)+(1(8)=(5(10). 5+10=15 not 12.
Also (8(8) is not equal to (2(2)(4(4) as (2(2)(4(4)=(2(2)(4(2)(1(2)=(2(2)(2(2)(2(1)(1(2). This can be further broken down to (2(2)(2(1)(1(2)+(2(2)(1(2)(1(2)=(4(4)+(2(8)=(6(12). This is not allowed. We find (8(8)=(5(3)+(3(5)=(2(2)*(3(5)=(2(2)*(5(3)=(10(6), and 10+6=16=8+8.
And so on. Each even shared number must be a multiple of (2(2) and another prime shared number.
Every even shared number can be divided by (2(2). The other factor must be a prime shared number as we need this to work in the shared number system. For example (12(12)=(2(2)(6(6). (6(6) must separate into two prime numbers, the left number moving up by one and the right number moving down by one.
We also can’t have multiple factors like (12(12)=(2(2)(2(2)(3(3) because I can have then (12(12)=(2(2)(6(2)(1(3) and then (12(12)=(2(2)(2(6)(1(3)=(2(2)(2(18) and it won’t add up. Having multiple factors again opens the shared number up for a non-finite representation. There must be some creation of (12(12) which has only finite representation. It must exist since I need to have 12. There must be a stable way of defining it. You see, if we allow (12(12)=(4(36) I can further extend (4(36) indefinitely.
I can now take (12(12) out and replace it by 12. So we must have a decomposition with (2(2)(a(b) with a, b both being ordinary prime numbers.
You see as we now have (4(4), (6(6), (8(8),…. we can now go on to create the usual natural numbers. The picture below, would now be in reverse. We would replace (1(1) by 1 and (2(2) by 2, etc.
Divide (4(4) by (2(2) to get (2(2) and divide (2(2) again by (2(2) to get (1(1). The we can go forward by dividing (6(6) by (2(2) to get (3(3) and (8(8) by (2(2) to get (4(4) etc. Then all this sequence can lead to the natural numbers. I can replace (1(1) by 1, (2(2) by 2, (3(3) by 3 etc.
This also demonstrates the equality of every primitive decomposition as well (which the lower level was hinting at) since for any even shared number there may be more than one decomposition. We go back to form (16(16) and then can separate once again to find another decomposition. Here we find two, (2(2)(13(3) and (2(2)(11(5).
Also, since we include 1 as a prime in some cases, it can be seen that there must always be at least two of these different representations of any even shared number greater than (4(4). So if we have one of them being (p(1) another one must be (r(s) where p,r,s are primes. For example with (10(10) we have (5(5) and (7(3). Since once I form the number 10, I need at least two of these to feedback to the other, original number line. See the picture below. We might also have more than two representations, In which case there are extra added dimensions.
Then there must be at least two of these decompositions for each even shared number. We might have one (p(1) but then I have another one (r(s). If we look at one half of these binary decompositions, that is, for example, look at 16(16)=(13(3)+(3(13) and see 16 =13+3 we can see Goldbach’s conjecture is true. It was just a part of a deeper understanding of numbers.