3. The hypercuboid hotel
So let’s imagine
the entire number chain laid out as a hotel in the shape of this hypercuboid.
From one angle it
looks like this.
2n
|
2n+1
|
2
|
1
|
4
|
3
|
6
|
5
|
8
|
7
|
10
|
9
|
where n is zero or
any positive integer *
(You could imagine this as the start of two infinite
series of numbered rooms - odd numbers and even numbers– you walk through
room 1 to get to room 3, 5, 7 etc.)
Turned through a
right angle it looks like this
3n
|
3n+1
|
3n+2
|
3
|
1
|
2
|
6
|
4
|
5
|
9
|
7
|
8
|
12
|
10
|
11
|
Turned through
another right angle it looks like this
5n
|
5n+1
|
5n+2
|
5n+3
|
5n+4
|
Each dimension is
laid out like this for a higher prime, with the layers arranged by modulo
residue. Within the 5n plane, all the rooms are composite, except for the first
one, number 5. The numbers in the other four corridors are not multiples of 5.
Within the hotel,
we can define any number by an infinite set of co-ordinates of the modulo
residues with respect to each prime.
For instance the
infinite co-ordinates for room 4 can be expressed thus:
2n, 3n+1, 5n + 4,
7n + 4 and pn + 4 for every prime thereafter.
The co-ordinates
for room 5 can be expressed thus:
2n+1, 3n+2, 5n, 7n
+ 5 and pn + 5 for every prime thereafter.
How do we “find”
prime numbers in the hotel?
A first naïve
thought is that you can simply “walk down” a non-zero residue corridor in every
dimension. For instance you could walk down the pn + 1 corridor on every plane.
This actually does create the set of co-ordinates for the number 1, as 1 has a
modulo residue of 1 for every other number. However this is the only way you
can choose a non-zero residue corridor in every dimension and create an
admissible set of co-ordinates.
This is because
every number has a zero residue in its own modulo plane, for instance, in the
5n plane, the number 5 has a zero residue even though it is a prime number. 35
has a zero residue for 5 and 7 and so on.
If a number is
prime, it has only one zero residue co-ordinate. (Though we need to be aware
that powers of primes also only have one zero residue co-ordinate, for instance
25 and 125 only have a zero residue at 5n - this is a problem we will deal with later in this proof.)
If a number has
two or more zero residue co-ordinates it is definitely not a prime.
4. The hypercuboid and the Menger Sponge
At this point, the
imaginary hypercuboid can be seen to resemble a multi-dimensional
Menger Sponge, the fractal object which can be compared in turn to a Sierpinski
Carpet taken up a dimension. All we are doing here is extending that analysis
to higher dimensions. Just to clarify this – imagine a slightly different way
of visualizing the hotel.
3n+1
|
3n
|
3n+2
|
||
10
|
30
|
20
|
5n
|
|
4
|
24
|
14
|
5n+4
|
|
2n
|
28
|
18
|
8
|
5n+3
|
22
|
12
|
2
|
5n+2
|
|
16
|
6
|
26
|
5n+1
|
|
1
|
21
|
11
|
5n+1
|
|
7
|
27
|
17
|
5n+2
|
|
2n+1
|
25
|
15
|
5
|
5n
|
13
|
3
|
23
|
5n+3
|
|
19
|
9
|
29
|
5n+4
|
5n+1
|
5n+2
|
5n
|
5n+3
|
5n+4
|
|
7n+1
|
1
|
127
|
85
|
43
|
169
|
7n+2
|
121
|
37
|
205
|
163
|
79
|
7n+3
|
31
|
157
|
115
|
73
|
199
|
7n
|
91
|
7
|
175
|
133
|
49
|
7n+4
|
151
|
67
|
25
|
193
|
109
|
7n+5
|
61
|
187
|
145
|
103
|
19
|
7n+6
|
181
|
97
|
55
|
13
|
139
|
And here is that
same section extended into the 11n dimension.
7n+1
|
7n+2
|
7n+3
|
7n
|
7n+4
|
7n+5
|
7n+6
|
|
11n+1
|
1
|
331
|
661
|
1981
|
991
|
1321
|
1651
|
11n+2
|
211
|
541
|
871
|
2191
|
1201
|
1531
|
1861
|
11n+3
|
421
|
751
|
1081
|
91
|
1411
|
1741
|
2071
|
11n+4
|
631
|
961
|
1291
|
301
|
1621
|
1951
|
2281
|
11n+5
|
841
|
1171
|
1501
|
511
|
1831
|
2161
|
181
|
11n
|
2101
|
121
|
451
|
1771
|
781
|
1111
|
1441
|
11n+6
|
1051
|
1381
|
1711
|
721
|
2041
|
61
|
391
|
11n+7
|
1261
|
1591
|
1921
|
931
|
2251
|
271
|
601
|
11n+8
|
1471
|
1701
|
2131
|
1141
|
151
|
481
|
811
|
11n+9
|
1681
|
2011
|
31
|
1351
|
361
|
691
|
1021
|
11n+10
|
1891
|
2211
|
241
|
1561
|
571
|
901
|
1231
|
We can arrange the
layers of the hotel so that in each dimension we are taking a symmetrical
“slice” through the centre. Of course this doesn’t produce a symmetrical
pattern in the number chain, as we have had to rearrange the number chain to
fit this visualization. But we can see more clearly how the construction of
this hypercuboid is similar to the construction of a Menger Sponge or a Cantor
Set.
It is accepted
that, taken to infinity, the Menger Sponge approaches zero volume but still has
an infinite surface area.
However the
geometry here is slightly different – there aren’t any “holes” because every
number is on its own modulo zero plane. So we need to think slightly
differently about the geometry of the hypercuboid.
When we look for
primes, we are looking for the points that are only intersected by one zero
residue plane. The more zero residue planes a point is intersected by, the more
different factors that number has.
Next, in order
to deal with the problem of powers of primes it is necessary to use a slightly
more complicated version of the hypercuboid. Then we can (finally) move on to
the twin prime proof.
(nb C.R.Greathouse, for whom I have a lot of respect, thinks the Menger Sponge analogy is weak, which is fair enough - certainly on its own it doesn't prove anything though I don't think anything else that follows depends on it, it's just a way I thought of visualising the hypercuboid).
5. The hypercuboid arranged
by squares of primes
In order to make sure we can’t confuse primes with powers of primes
we are going to rearrange the hypercuboid so that in each dimension we are
considering the modulo residues of squares of primes, instead of primes.
For instance, rather than dividing the numbers into 2n and 2n+1, we
divide them into 4n, 4n+1, 4n+2, 4n+3.
2 is a 4n+2 number. All powers of 2 (4, 8, 16 etc) are 4n numbers.
For any prime p, p is now in a different plane to p^2 (p squared), p^3 etc. In general, p will be in the (p^2)n+p plane, and not in any other pn plane.
Within this hypercuboid each number still has a unique set of
co-ordinates. We can now say without exception that a prime number is one that
is only intersected by one (p^2)n+p plane.
Here is a colour-coded visual representation of this, for residues
of 4 and 9.
4n
|
4n+1
|
4n+2
|
4n+3
|
|
9n
|
36
|
9
|
18
|
27
|
9n+1
|
28
|
1
|
10
|
19
|
9n+2
|
20
|
29
|
2
|
11
|
9n+3
|
12
|
21
|
30
|
3
|
9n+4
|
4
|
13
|
22
|
31
|
9n+5
|
32
|
5
|
14
|
23
|
9n+6
|
24
|
33
|
6
|
15
|
9n+7
|
16
|
25
|
34
|
7
|
9n+8
|
8
|
17
|
26
|
35
|
·
Green = numbers that are
co-prime to 2 and 3
·
Red = p^2n+p = prime, if only
intersected by one red plane across all dimensions, otherwise not prime (for instance in the grid above, 2 and 3 are red but won't be intersected by any other red planes)
·
Black = all composite numbers
·
Blue = intersection of two red
planes, therefore a composite.
·
When we go up to the next dimension, we will have 25 slices coloured
thus:
·
Black: 25n, 25n+10, 25n+15,
25n+20
·
Coloured the same as above: All
other planes except for 25n+5
·
The 25n + 5 plane will be
coloured blue where it coincides with the red squares above, and red where it
coincides with green squares above (as in the image below).
4n
|
4n+1
|
4n+2
|
4n+3
|
|
9n
|
180
|
405
|
630
|
855
|
9n+1
|
280
|
505
|
730
|
55
|
9n+2
|
380
|
605
|
830
|
155
|
9n+3
|
480
|
705
|
30
|
255
|
9n+4
|
580
|
805
|
130
|
355
|
9n+5
|
680
|
5
|
230
|
455
|
9n+6
|
780
|
105
|
330
|
555
|
9n+7
|
880
|
205
|
430
|
655
|
9n+8
|
80
|
305
|
530
|
755
|
5 is the only remaining number on this section that will not be
intersected by another red plane (and turned blue) in other dimensions.
Since we are marking numbers as composite if they are on the
intersection of two p^2n+p (red) planes, and we are not mixing up powers of
primes with primes, we are not missing any composite numbers in this sieving
process.
All numbers will end up coloured red, blue or black. If we imagine
each number as a room in the infinite hotel, they all start off green, then get
painted black if an all-composite plane intersects them, red if an (p^2)n+p plane
intersects them, and blue if two or more (p^2)n+p planes intersect them. Numbers
that remain red must be primes.
We haven't at this point proved anything about prime distribution. However we know (from Euclid and other prime proofs) that there remains an infinitude of numbers in the
hypercuboid that are only intersected by one (red) (p^2)n+p plane, because we
know there is an infinitude of primes.
So far this has just been an elaborate way of visualising something
we already knew about, but the interesting thing is how we can extend this
analysis to twin primes.
No comments:
Post a Comment