All of my posts of February 2011 should be DELETED because they contain a fundamental fallacy.
The fallacy is this: IN THE ALGEBRA OF NUMBERS, addition is distributive over multiplication. This is untrue.
In fact and this is easily proved that addition is NOT distributive over multiplication. However in the algebra of numbers, multiplication is distributive over addition.
The essential idea is that every uneven integer can be represented as a vector having a minimum number of components, depending on the said uneven integer, but any finite number of components greater than the minimum. The prime two plays a key role in the representation of any prime number that is an uneven integer. All the prime numbers with the exception of prime two are uneven integers, thus: [3, 5, 7, 11, … 7927, … to infinity ]
For example, 19 as a vector is: [16, 8, - 4, - 2, 1] and that is the minimal representation. Prime 19 is: [32, - 16, 8, - 4, - 2, 1] and
[64, - 32, - 16, 8, - 4, - 2, 1] et c. Representing prime numbers as vectors gives a different way of studying prime numbers and uneven composites using linear algebra to find linearly independent sets of vectors to form a basis spanning a given finite dimensional space.
The vectors representing; 3, 5, 7, 11 an 13 are dependent giving the linear combination: [x[1], … x[5]] = x[4].[ - 1, 1, 0, 1, -1], where x[4] can have any value. Noting that each of these vectors has five components.
However, the vectors representing 3, 5, 7, 11 and 17 are linearly independent, each vector having again five components. These linearly independent vectors span 5D space.
Finding sets of vectors representing primes that form linearly dependent sets looks like a most expeditions way to study primes.
Constructing square and rectangular matrices and solving the matrix equation AX = O, the zero vector using Gaussian elimination, looks like the best way in this context. Matrices having five rows seems to be do-able using pencil and paper.
But there is another way and that is to find sets of vectors that represent primes which span the entire space of a given dimension. The case of the primes 3, 5 and 7 spanning 3D space has been proved by solving the three equations cited below.
The case of 5D space is interesting because matrices with five rows are just do-able using pencil are paper. If such matrices have five columns, square matrices, then the five column vectors could span 5D space and thus be a basis set of linearly independent vectors. This means that if other vectors having five components are adjoined to make a rectangular matrix having five rows, then the new adjoined vectors would be some linear combination of the basis set of vectors. This implies a system of linear equations to be solved to find the linear combination.
A typical vector, here minimally representing prime 23 is: [16, 8, - 4, 2, 1]. A larger vector representing prime 23 is [32, - 16, 8, - 4, 2, 1].
It is advantageous to reclassify the primes thus:
[2] and [3, 5, 7, 11, 13, 17, … ], putting prime two into a class of its own. An integer in this context just means an uneven integer in the class:
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, … ], noting that the latter class of primes are a subclass of the integers in the above class. Composites are then integers in the class of uneven integers above. Typical composites are 9, 15, 21, 25, 27, 33, 35, … etc. The uneven integer unity, one [1] can be written as a vector having at least one component, itself, since N defined below is zero in this case, and there are [N + 1] components in a vector as a minimum. The integer one [1] can be written as an expression having at least one term, but any finite number of terms, thus:
1 = 2^{M} - 2^{M - 1} - … - 1. For example 1 = 16 - 8 - 4 - 2 - 1 giving as is seen below a 5D vector: [16, - 8, - 4, - 2, - 1].
The key fact is that when dividing an uneven integer by prime two, and having an uneven quotient in every instance, the remainders have to be either + 1 or - 1. The remainders cannot be anything else.
Formally if Q[N], an uneven integer, satisfies the inequality:
2^{N} < Q[N] < 2^{N + 1}, where N is 1, 2, 3 … then:
Q[N] = 2.Q[N - 1] + r[N - 1] and remainder r[N - 1] having the subscript [N - 1] is given by r[N - 1] = i^{1 + Q[N]} . The " i " is the imaginary number that satisfies the equation i^{2} + 1 = 0.
Counting down, Q[1] = 3 always, Q[0] = 1 and r[0] = 1. The remainder r[N] is conveniently defined as r[N] = 1, rather than r[N] = - 1, thereby fixing
Q[N + 1].
Starting from Q[1] = 2.Q[0] + r[0] and " condensing " the equations by successive elimination of Q[j]'s by multiplications by two until only Q[N] is left, an equation giving Q[N] in terms of the remainders r[1], … r[N - 1] and powers of two is arrived at. This is the minimal equation containing [N + 1] terms.
We have: Q[N] = 2^{N} + 2^{N - 1} + r[1].2^{N - 2} + … + r[N - 1].2^{0} and
2^{0} = 1, just to help the accounting.
Since 2^{N} = 2^{N + 1} - 2^{N} = 2^{N + 2} - 2^{N + 1} - 2^{N} etc. the term 2^{N} can be substituted by terms involving larger powers of prime two.
Vectors come in when the terms of the equation giving Q[N] are set in order and written as the components of a vector thus:
Q[N] as a vector is: [2^{N}, 2^{N - 1}, r[1].2^{N - 2}, … r[N - 1].2^{0}], having the minimum of [N + 1] components.
The essential matter to notice is that a vector representation of an uneven integer apart from the dimension, from a minimum dimension upwards, is unique to that uneven integer. The other matter to notice is that irrespective of dimension, subject to the minimum dimension, the algebraic sum of the components of the vector representing that uneven integer invariably reproduce that uneven integer.
A slight digression on linear independence.
The vectors that in 3D represent the primes; 3, 5 and 7 are respectively;
[4, - 2, 1], [4, 2, - 1] and [4, 2, 1]. How can it be shown that these three vectors are not dependent on one another, in other words, linearly independent ?
Ans. [a] draw a perspective picture [b] construct a model in three dimensions, a 3D model. In this case, the vertices of the vectors are the vertices of a spherical triangle of unequal sides. [c] solve the equations:
4 x[1] + 4 x[2] + 4 x[3] = 0, - 2 x[1] + 2 x[2] + 2 x[3] = 0 and
x[1] - x[2] + x[3] = 0 . The solutions are x[1] = 0, x[2] = 0 and x[3] = 0 The vectors are linearly independent.
In solving these equations, it is better to multiply the first equation by 1/4, the second equation by 1/2 and leave the third equation unaltered. This gives a new set of equations that have the same solutions as the original equations.
The proper and by far the best way to solve the simplified equations is using Gaussian elimination.
I just solve these equations in the simplest way by reverse substitutions.
In every case of solving equations involving uneven integers as vectors, it is best to remove by multiplication the descending powers of the prime two, but leave the last equation in 2^{0} terms unaltered.
Gaussian reduction works for square and rectangular systems of linear equations. In the case of rectangular systems it is usual that the number of columns, vectors representing uneven integers, exceeds the number of rows, number of components which is the dimension of the vectors.
A good way of picturing the succession of primes and composites, herein meaning uneven integers that are not primes, is to make a tableau. Using centimetre squared paper, accurately ruled, taking a sheet measuring more than 64 cm by more than 48 cm, to allow for a border, starting from x = 0 and y = 0, plotting the uneven integer along the x axis made parallel to the longer side, and plotting the components of the vector representing that particular uneven integer on the y axis that is parallel to the shorter side of the paper, then the primes from 3 to 61 can be accommodated as vectors of six dimensions. Putting N = 5 gives a six dimension vector [6D]. Four colours could be used for the primes. The greatest term 2^[5} could be one colour. The terms that are negative that exist for the smaller primes than 32 could be another colour. The positive term 2^{5 - 1}, which is 2^{4} could be another colour. The terms involving the remainders: r[1], … r[N - 1], here r[1],
r[2], r[3], and r[4], which are r[1].2^{3}, r[2].2^{2}, r[3].2^{1} and r[4].2^{0} could be the same, but another colour. If the composites are to be included, then the ordinates, the y co-ordinates could be marked using the same colour for all the composites, but different from all the other colours, of which there are at least three and at the most four colours. The tableau would then be a mosaic of dots of different colours.
There would be no point in joining any of the dots by straight lines, since this is likely to be meaningless.
Remembering that a composite here, such as 21, 45, 57 etc. is an uneven integer. Unity, one can be put on the tableau amongst the composites.
None of the above works if we put three in place of prime two, as
Q[N] = 3^{N} + etc. The result is nonsense. ONLY prime two gives the correct results.
