ABSTRACT We are all aware of the concept of integers 1, 2, 3, 4, 5, 6, ⋯. Which all have the nice property of a unique representation of a product of powers of primes. We can write numbers by their divisibility by a prime, or just the powers of the primes in this product for example 2=1\\ 3=0,1\\ 4=2,0\\ 5=0,0,1\\ 6=1,1\\ 7=0,0,0,1 where we see each new digit is the next prime. We can have powers greater than one etc. Any expression we can write with any number of commas, and any powers will then be a real, tangible integer. Now, instead of prime factors, let each new digit in such an expression be a positive divisor, and the “power” (it’s not really a power any more) is the number of times that divisor can be applied 1=1\\ 2=1,1\\ 3=1,0,1\\ 4=1,2,0,1\\ 5=1,0,0,0,1\\ 6=1,1,1,0,0,1 this creates the standard divisor map pattern but there is a non-standard 2 in the representation of 4 for example. Now, we can realised that only _certain_ expressions in this format constitute actual integers. If we write 1, 0, 0, 1, this would then imply a number which can be divided by 4 once, and by 1, but not by 2 at all! Obviously this would be a contradiction if it were an integer, so it is a fictional integer that does not exist. However, we can now explore the concepts behind some of these numbers. MAIN Let a string of integers seperated by commmas as above be called an “aggregate”. We can tabulate the first few examples of aggregates below with some arbitrary labels a=1\\ b=0,1\\ c=1,1\\ d=0,0,1\\ e=1,2\\ f=1,3 we can see that a = 1, the normal unity we are used to, b however, is a fictional integer, it is divisible by 2 but not by 1. c is an integer and c = 2. d is a fictional form of 3 which is not divisible by 1, e is a fictional form of 2, which is divisible by 2 twice, but not by 4. f is divisible by 2 three times but not by 4 nor by 8. We can define a product, of aggregates, which is simply a sum of constituents. for example 0,1 \cdot 1,1,0,1 = 1,2,0,1 = 4 above we can see that two fictional integers, can form an integer under this product. One question is the validity of a digit to represent 1. Fictional integers which can not be divided by 1 are in a way more strange than fictional integers that can. We can define a span of an aggregate, as the number of digits written, which is also the largest number that aggregate is divisible by \; a,b,c,\cdots,z = 24 we can also define the fill of an aggregate, which is the sum of all the entries \; a,b,c,d = a+b+c+d we can write any ’positive’ aggregate, where positive is that each digit is positive or zero, as the product of one or more fill-1 aggregates, for example [a,b,c,d] = [a] \cdot [0,b] \cdot [0,0,c] \cdot [0,0,0,d] Theorem 1, there is only one fill-1, span-n aggregate, for each n. Proof, there is only one way to arrange m − 1 zeroes and one 1 such that the 1 is in the mth position. Theorem 2, all prime numbers are the product of the unit aggregate and the fictional fill-1 prime aggregate. Proof trivial by above theorem. for example 7=[1,0,0,0,0,0,1] =[1]\cdot[0,0,0,0,0,0,1] Division is the inverse of the product, so one may think ⋅, 6/3=[1,1,1,0,0,1]/[1,0,1]=[0,1,0,0,0,1] and we see the aggregate division of two integers can yield a fictional integer. This poses a problem. In the original notation we have placed a 1 in the 1’s column of the aggregate, but all numbers are divisible by 1 as many times as needed. So really they should have an ugly ∞ in the 1’s column. This mans when dividing we should always re-add a one if one was present in the first place, or altogether ignore the 1′s column of the aggregates. MORE OPERATIONS We may create a number of different operations for such numbers, for example substitution/composition [1,0,2,3]\#[1,0]\to[[1,0],[0,0],[2,0],[3,0]]=[1,0,0,0,2,0,3,0] a number which is divisible by 1, 5 twice and by 7 three times, this may seem useless, however, note that [0,1]\#[0,1]=[0,0,0,1] [0,0,1]\#[0,0,1]=[0,0,0,0,0,0,0,0,1] [0,1]\#[0,0,1]=[0,0,0,0,0,1] [0,0,1]\#[0,1]=[0,0,0,0,0,1] so we see, we may write composite m, fill-1, span-m aggregates as # substitutions of their factors which appear to commute. Then we may write 4=[1,2,0,1]=[1]\cdot[0,1]\cdot[0,1]\cdot[0,0,0,1] but [0,0,0,1]=[0,1]\#[0,1] 4=[1,2,0,1]=[1]\cdot[0,1]\cdot[0,1]\cdot([0,1]\#[0,1]) 2=[1,1]=[1]\cdot[0,1]\\ 2\times2=([1]\cdot[0,1])\times([1]\cdot[0,1])=[1]\cdot[0,1]\cdot[0,1]\cdot([0,1]\#[0,1]) if we accept the rules [1]\cdot[1]=[1]\\ [1]\#[1]=[1] then this reads a\times a= a\cdot a \cdot(a\# a) for 9 we may try 3\times 3= ([1]\cdot[0,0,1])\times([1]\cdot[0,0,1])= [1]\cdot[0,0,1]\cdot[0,0,1]\cdot([0,0,1]\#[0,0,1])=[1,0,2,0,0,0,0,0,1]=9 so we may represent the squaring of integers under this method, in fact it appears to work for composites with two factors 3\times 2= ([1][0,0,1])\times([1][0,1])= [1][0,1][0,0,1]\cdot([0,1]\#[0,0,1])=[1][0,1][0,0,1][0,0,0,0,0,1]=6 long winded, but consistent on the priviso we write the numbers explicity in their prime-product form! Then try 2\times6 = ([1][0,1])\times([1][0,1][0,0,1][0,0,0,0,0,1])= [1][0,1][0,1][0,0,1]([0,1]\#[0,1])([0,1]\#[0,0,1])([0,1]\#[0,0,0,0,0,1]) which makes less sense, it would appear we take all but the last divisor in the sequence as a product with [1] and then take the # product with all divisors. As a final check we try two composites in a product 6\times10= [1][0,1][0,0,1][0,0,0,0,0,1]\times[1][0,1][0,0,0,0,1] = [1\#1][1\#2][2\#1][3\#1][2\#2][1\#5][6\#1][2\#5][6\#2][3\#5][2\#2\#5][6\#5][6\#2\#5] MAGNITUDE We can also define a magnitude, which is a weighted fill. For an aggregate [a,b,c,d]= a+2b+3c+4d this gives us a way to rank aggregates, we can then rank first by magnitude, then by span, then by fill. We can tabulate some below & & & & & \\ (0) & 0 & 0 & 0 & y & n\\ (1) & 1 & 1 & 1 & y & n\\ (2) & 2 & 1 & 2 \\ (0,1) & 2 & 2 & 1 \\ (3) & 3 & 1 & 3 \\ (1,1) & 3 & 2 & 2 & y & y\\ (0,0,1) & 3 & 3 & 1 \\ (4) & 4 & 1 & 4 \\ (0,2) & 4 & 2 & 2 \\ (2,1) & 4 & 2 & 3 \\ (1,0,1) & 4 & 3 & 2 & y & y\\ (0,0,0,1) & 4 & 4 & 1 \\ (5) & 5 & 1 & 5 \\ (1,2) & 5 & 2 & 3 \\ (3,1) & 5 & 2 & 4 \\ (0,1,1) & 5 & 3 & 2 \\ (2,0,1) & 5 & 3 & 3 \\ (1,0,0,1) & 5 & 4 & 2 \\ (0,0,0,0,1) & 5 & 5 & 1 \\ (6) &6 &1 &6 \\ (0,3) &6 &2 &3 \\ (2,2) &6 &2 &4 \\ (4,1) &6 &2 &5 \\ (0,0,2) &6 &3 &2 \\ (1,1,1) &6 &3 &3 \\ (3,0,1) &6 &3 &4 \\ (0,1,0,1) &6 &4 &2 \\ (2,0,0,1) &6 &4 &3 \\ (1,0,0,0,1) &6 &5 &2 & y & y\\ (0,0,0,0,0,1) &6 &6 &1 \\ (7) & 7 &1 &7\\ (1,3) & 7 &2 &4\\ (3,2) & 7 &2 &5\\ (5,1) & 7 &2 &6\\ (0,2,1) & 7 &3 &3\\ (1,0,2) & 7 &3 &3\\ (2,1,1) & 7 &3 &4\\ (4,0,1) & 7 &3 &5\\ (0,0,1,1) & 7 &4 &2\\ (1,1,0,1) & 7 &4 &3\\ (3,0,0,1) & 7 &4 &4\\ (0,1,0,0,1) & 7 &5 &2\\ (2,0,0,0,1) & 7 &5 &3\\ (1,0,0,0,0,1) & 7 &6 &2\\ (0,0,0,0,0,0,1) & 7 &7 &1 \\ (8) &8 &1 &8 \\ (0,4) &8 &2 &4\\ (2,3) &8 &2 &5\\ (4,2) &8 &2 &6\\ (6,1) &8 &2 &7\\ (0,1,2) &8 &3 &3\\ (1,2,1) &8 &3 &4\\ (2,0,2) &8 &3 &4\\ (3,1,1) &8 &3 &5\\ (5,0,1) &8 &3 &6\\ (0,0,0,2) &8 &4 &2\\ (0,2,0,1) &8 &4 &3\\ (1,0,1,1) &8 &4 &3\\ (2,1,0,1) &8 &4 &4\\ (4,0,0,1) &8 &4 &5\\ (0,0,1,0,1) &8 &5 &2\\ (1,1,0,0,1) &8 &5 &3\\ (3,0,0,0,1) &8 &5 &4\\ (0,1,0,0,0,1) &8 &6 &2\\ (2,0,0,0,0,1) &8 &6 &3\\ (1,0,0,0,0,0,1) &8 &7 &2 & y & y\\ (0,0,0,0,0,0,0,1) &8 &8 &1\\ (9) & 9 & 1 & 9 \\ (1,4) & 9 & 2 & 5 \\ (3,3) & 9 & 2 & 6 \\ (5,2) & 9 & 2 & 7 \\ (7,1) & 9 & 2 & 8 \\ (0,0,3) & 9 & 3 & 3 \\ (0,3,1) & 9 & 3 & 4 \\ (1,1,2) & 9 & 3 & 4 \\ (2,2,1) & 9 & 3 & 5 \\ (3,0,2) & 9 & 3 & 5 \\ (4,1,1) & 9 & 3 & 6 \\ (6,0,1) & 9 & 3 & 7 \\ (0,1,1,1) & 9 & 4 & 3 \\ (1,0,0,2) & 9 & 4 & 3 \\ (1,2,0,1) & 9 & 4 & 4 & y & n \\ (2,0,1,1) & 9 & 4 & 4 \\ (3,1,0,1) & 9 & 4 & 5 \\ (5,0,0,1) & 9 & 4 & 6 \\ (0,0,0,1,1) & 9 & 5 & 2 \\ (0,2,0,0,1) & 9 & 5 & 3 \\ (1,0,1,0,1) & 9 & 5 & 3 \\ (2,1,0,0,1) & 9 & 5 & 4 \\ (4,0,0,0,1) & 9 & 5 & 5 \\ (0,0,1,0,0,1) & 9 & 6 & 2 \\ (1,1,0,0,0,1) & 9 & 6 & 3 \\ (3,0,0,0,0,1) & 9 & 6 & 4 \\ (0,1,0,0,0,0,1) & 9 & 7 & 2 \\ (2,0,0,0,0,0,1) & 9 & 7 & 3 \\ (1,0,0,0,0,0,0,1) & 9 & 8 & 2 \\ (0,0,0,0,0,0,0,0,1) & 9 & 9 & 1 \\ (1,1,1,0,0,1) & 12 & 6 & 4 & y & n \\ (1,3,0,2,0,0,0,1) & 23 & 8 & 7 & y & n\\ we can see that there are likely p(n) aggregates of magnitude n, where p(n) is the nth partition number, this makes sense from the definition of magnitude, this table gives us a convenient way of ranking the entries. Question, can two different entries have exactly the same magnitude, span and fill. Answer, no. If one aggregate has a set of digits, to conserve fill, exactly the same number of digits must be used, but for a different configuration one or more must be moved, thus changing the magnitude. We see that an integer prime p will always have a magnitude p + 1 a span p and a fill 2. We say an integer prime p, as we may be able to extend the notion of primality to the fictional integers above, we can declare any aggregate with fill 2 a prime aggregate, and check their density. We can then tabulate our ’fictional primes’ (2)\\ (1,1)\\ (0,2)\\ (1,0,1)\\ (0,1,1)\\ (1,0,0,1)\\ (0,0,2)\\ (0,1,0,1)\\ (1,0,0,0,1)\\ (0,0,1,1)\\ (0,1,0,0,1)\\ (1,0,0,0,0,1) we can then realise a few (aggregate) products of the above list (1,1)\cdot(0,1,0,1)=(1,2,0,1)=4\\ (0,1,1)\cdot(1,0,0,0,0,1)=(1,1,1,0,0,1)=6\\ (0,0,2)\cdot(1,0,0,0,0,0,0,0,1)=9\\ (0,2)\cdot(0,1,0,1)\cdot(1,0,0,0,0,0,0,1)=8\\ (0,0,1,0,1)\cdot(1,\cdots,1)=15 we then can categorize the natural numbers, how many prime aggregates are needed to construct the numbern, P(n), how many different ways can the number be written in prime aggregates, w(n). For example 4=(1,1)\cdot(0,1,0,1)=(0,2)\cdot(1,0,0,1) so w(4)=2. This means w(1) is zero, w(2)=1, w(3)=1, w(5)=1. In fact any prime p, w(p)=1, we have 6=(0,1,1)\cdot(1,0,0,0,0,1)=(1,1)\cdot(0,0,1,0,0,1)=(1,0,1)\cdot(0,1,0,0,0,1) so w(6)=3. Some questions, are all natural numbers n > 1 able to be created by a finite number of prime aggregates? If the fill of the aggregate representation of n is even then this is true for all n. Is the set of fictional numbers creatable by prime aggregates smaller than the set fictional integers? This is true, as we can create fictional integers with non-even fills. INFERRED RULES If 4 = (1, 2, 0, 1), and 2 = (1, 1), then (0, 1, 0, 1)=2*, a special version of two. However, if 6 = (1, 1, 1, 0, 0, 1) and 3 = (1, 0, 1), then 2* = (0, 1, 0, 0, 0, 1) which is not consistent. Really we should write 4/2 = (0, 1, 0, 1), and 6/3 = (0, 1, 0, 0, 0, 1) where / is aggregate division. These are both prime aggregates. We can then write 8/(4/2)=(1, 2, 0, 0, 0, 0, 0, 1), and (8/(4/2))/2 = (0, 1, 0, 0, 0, 0, 0, 1). In some sense then, these prime aggregates can be named based on the integers. One naming concept is the shifting of prime aggregates, for example 2=(1,1)\\ 2>>1=(0,1,1)\\ This can provide names for stranger aggregates 4/((6/2)<<2)=(0,2) and we gain amazing strings of formulae like 4=2\cdot(3>>1)\\ 6=3\cdot(5>>1)\\ 8=4\cdot(7>>1)\\ 10=5\cdot(9>>1) implying that 2m = m((2m − 1)> > 1). However... 6\cdot(11>>1)=(1,1,1,0,0,1)\cdot(0,1,0,0,0,0,0,0,0,0,0,1)=(1,2,1,0,0,1,0,0,0,0,0,1)\ne12 we are missing a factor of (0, 0, 0, 1)! and we would require 6 ⋅ ((1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1)> > 1) to keep the trend running. This brings us to the observation that 12 cannot be made by a finite number of prime aggregates. \;12=\;(1,2,1,1,0,1,0,0,0,0,0,1)=7 not even. We then see that 14=7\cdot(13>>1)\\ but that 16 is not constructable. STANDARD PRODUCT So far we have missed out the standard numeric product between these aggregates (1,1)\times(1,1)=(1,2,0,1)\\ (1,1)\times(1,0,1)=(1,1,1,0,0,1) which is clearly done by the standard method of pairing each term and adding to the relevant box in the new aggregate. That is C_k= A_iB_j we can then extend this form of product to our prime aggregates above, (1,1)\times(0,2)=(0,1,0,1)\cdot(0,1,0,1)\\ (0,1,1)\times(1,0,0,0,0,1)=(0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1)\\ (1,1)\times(1,0,0,1)=8/(0,2) This product is consistent with multiplication of integers, we would also have the concept of addition from normal integers which makes less sense (1,1)+(1)=(1,0,1)\\ (1,0,1)+(1)=(1,2,0,1)\\ (1,1)+(1,1)=(1,2,0,1)\\ (1,1)+(1,0,1)=(1,0,0,0,1)