- #1

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- TL;DR Summary
- So we know that all primes are encompassed in natural numbers, right?

But how do we show that?

w = {0,0 | 1,1 | 2,2...}

Let x = number of primes up to w+1

Let y = number of primes up to w-1

Now there's an empty prime box in the 0,0 slot not connected to anything.

So I let x = p-1 and y = p+1

p = [p0, p1, p2...]

Now p0 becomes 1,0/1

It can be either on or off.

For the sake of argument we'll turn it on.

Let x = prime numbers up to W

Let y = prime numbers up to W+1 counting 0 & 1

x = p+1

W = 0, x = 1, y = 2

W = 1, x = 2, y = 3

W = 2, x = 3, y = 4

W = 3, x = 4, y = 5

W = 4, x = 4, y = 5

W = 5, x = 5, y = 6

W = 6, x = 5, y = 6

W = 7, x = 6, y = 5

W = 8, x = 6, y = 5

W = 9, x = 6, y = 5

W = 10, x= 6, y = 5

W = 11, x = 7, y = 6

W = 12, x = 7, y = 6

W = 13, x = 8, y = 7

w = 14, x = 8, y = 7

w = 15, x = 8, y = 7

w = 16, x = 8, y = 7

w = 17, x = 9, y = 8

w = 18, x = 9, y = 8

w = 19, x = 8, y = 8

w = 20, x = 8, y = 8

w = 21, x = 8, y = 8

w = 22, x = 8, y = 9

w = 23, x = 9, y = 9

w = 24, x = 9, y = 9

w = 25, x = 9, y = 9

w = 26, x = 9, y = 9

w = 27, x = 9, y = 9

w = 28, x = 9, y = 10

w = 29, x = 10, y = 10

y = p-1

W = 0, x = 1, y = ? 1 or a 0 if you say 0 its 0/1 so i say 1 to close the loop

W = 1, x = 2, y = 1

W = 2, x = 3, y = 2

W = 3, x = 4, y = 3

W = 4, x = 4, y = 4

W = 5, x = 5, y = 4

W = 6, x = 5, y = 5

W = 7, x = 6, y = 5

W = 8, x = 6, y = 6

W = 9, x = 6, y = 6

W = 10, x= 6, y = 6

W = 11, x = 7, y = 6

W = 12, x = 7, y = 7

W = 13, x = 8, y = 7

w = 14, x = 8, y = 8

w = 15, x = 8, y = 8

w = 16, x = 8, y = 8

w = 17, x = 9, y = 8

w = 18, x = 9, y = 9

w = 19, x = 10, y = 9

w = 20, x = 10, y = 10

w = 21, x = 10,, y = 10

w = 22, x = 10, y = 10

w = 23, x = 11, y = 10

w = 24, x = 11, y = 11

w = 25, x = 11, y = 11

w = 26, x = 11, y = 11

w = 27, x = 11, y = 11

w = 28, x = 11, y = 11

Let x = number of primes up to w+1

Let y = number of primes up to w-1

Now there's an empty prime box in the 0,0 slot not connected to anything.

So I let x = p-1 and y = p+1

p = [p0, p1, p2...]

Now p0 becomes 1,0/1

It can be either on or off.

For the sake of argument we'll turn it on.

Let x = prime numbers up to W

Let y = prime numbers up to W+1 counting 0 & 1

x = p+1

W = 0, x = 1, y = 2

W = 1, x = 2, y = 3

W = 2, x = 3, y = 4

W = 3, x = 4, y = 5

W = 4, x = 4, y = 5

W = 5, x = 5, y = 6

W = 6, x = 5, y = 6

W = 7, x = 6, y = 5

W = 8, x = 6, y = 5

W = 9, x = 6, y = 5

W = 10, x= 6, y = 5

W = 11, x = 7, y = 6

W = 12, x = 7, y = 6

W = 13, x = 8, y = 7

w = 14, x = 8, y = 7

w = 15, x = 8, y = 7

w = 16, x = 8, y = 7

w = 17, x = 9, y = 8

w = 18, x = 9, y = 8

w = 19, x = 8, y = 8

w = 20, x = 8, y = 8

w = 21, x = 8, y = 8

w = 22, x = 8, y = 9

w = 23, x = 9, y = 9

w = 24, x = 9, y = 9

w = 25, x = 9, y = 9

w = 26, x = 9, y = 9

w = 27, x = 9, y = 9

w = 28, x = 9, y = 10

w = 29, x = 10, y = 10

y = p-1

W = 0, x = 1, y = ? 1 or a 0 if you say 0 its 0/1 so i say 1 to close the loop

W = 1, x = 2, y = 1

W = 2, x = 3, y = 2

W = 3, x = 4, y = 3

W = 4, x = 4, y = 4

W = 5, x = 5, y = 4

W = 6, x = 5, y = 5

W = 7, x = 6, y = 5

W = 8, x = 6, y = 6

W = 9, x = 6, y = 6

W = 10, x= 6, y = 6

W = 11, x = 7, y = 6

W = 12, x = 7, y = 7

W = 13, x = 8, y = 7

w = 14, x = 8, y = 8

w = 15, x = 8, y = 8

w = 16, x = 8, y = 8

w = 17, x = 9, y = 8

w = 18, x = 9, y = 9

w = 19, x = 10, y = 9

w = 20, x = 10, y = 10

w = 21, x = 10,, y = 10

w = 22, x = 10, y = 10

w = 23, x = 11, y = 10

w = 24, x = 11, y = 11

w = 25, x = 11, y = 11

w = 26, x = 11, y = 11

w = 27, x = 11, y = 11

w = 28, x = 11, y = 11