My task for you !

Do the math !

O

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                                                                       Sincerely yours, Svetoslav

P.S.: What is r2 and r3 after determining mass of 1 and r1=1/6th side of one square ?

Neper Epsilon vicinity to Integral

The square ABCD with side of 2 units here is "arbitrary figure" on the Ulam spiral with center the origin Oxy. This figure is characterized by a double integral down her face Natural surroundings is (Epsilon) painted triangle - it consists of three scan points of Lagrangian point on the Ulam spiral. Center coincides with the beginning of the coordinate system of the drawing falls and center of gravity of the square ABCD (intersection of diagonals). On this drawing with figure so constructed triangle area Epsilon is the smallest and the largest such area of ABCD.

In three-dimensional convex polyhedron D, will have six two-dimensional view, and its surface will be characterized by a triple integral. Six views will be orthogonal projections, which may be given on the spiral Ulamov analogous to ABCD ...

Eight-bit processor based on prime numbers

Accumulator:

Х Х Х ХХХХX —> 1 2 3 - low bits

| | | | | | | |

1713117 5 3 2 1

Register R0:

Х Х Х Х Х Х Х Х

43	41	37	31	29	23	19

__________ Primes to the processor ! (Complex for operations)

00000000 = 0 |

00000001 = 1 | “one“

00000010 = 2 | “two“

00000100 = 3 | “three“

00000110 = 4 | 3 two

00001000 = 5 | “four“

00001001 = 6 | 5 one

00010000 = 7 | “five“

00010001 = 8 | 7 one

00010010 = 9 | 7 two

00010100 = 10| 7 three

00100000 = 11| “six“

00100001 = 12| 11 one

01000000 = 13| “SEVEN“

01000001 = 14| 13 one

01000010 = 15| 13 two

01000100 = 16| 13 three

10000000 = 17| “EIGHT“

Every prime number, which is not low bit is able to be presented in more than one way.

It is always better to present one number with smaller bits.

Calculation rules

Sum:

Case А) ? no double bit:

01000100 = 16

+

00001001 = 6

=

01001101 = 22

Case B) ? the double bit is presented through lower bits:

01000100 = 16 = 13 three

+

00010100 = 10 = 7 three

=

01010100 =

+

00000100 = 00000011

=

01010111 = 26

01000000 = 13 = 13

+

01000001 = 14 = 13 one

=

01000001

+

01000000 = 00100010

=

01100011 = 27

01000001 = 14 = 13 one

+

01000010 = 15 = 13 two

=

01000011

+

01000000 = 00011000

=

00011011 = 29

We sum А and B, A>B

I. А+B=C+D , when C is a number including all significant bits no matter their repeating, and D consists only of double bits.

1. D=E+F+… ; E, F ? numbers consisted only from one digit

E ? we begin from the highest bit making it to the lowest bits

Conclusion: Since the actual discharge means multiplication of the unit with a prime number, and because the numbers are hole, transformations II are actually bringing the numbers to end chain fractiones.

Substraction:

01000100 = 16 = 00100110 = 2+3+11

-

00001001 = 6 = 00000111 = 1 + 2 + 3

=

000010110 = 10

Substraction А from B, А>B

А and B are divided to significant bits with lower rates;

А-B = D-E ,E ?

The action is repeating until:

X=00000000 (0) или X=00000001 (1)

X=0 , the result is D

X=1, analogic -> try it yourself

Multiplication:

00001000 = 5 = 00000110 = 2+3

*

00001001 = 6 = 00000111 = 1 + 2 + 3

=

11000000 = 30 = 2*3*5 = 00000010

*

00000100

*

00001000

=

(2*3)*5

2*3

=

00000010

*

00000100

=

00000100

+

00000100

=

00000111=6

(2*3)*5

=

……….

Multiplication is alike sum !

…..

General conclusion: This kind of processor would be much faster and efficient.

Conclusion: One 9-bit processor should be much more precise at the point.

X Х Х Х ХХХХX —> 1 2 3

| | | | | | | | |

19171311 7 5 3 2 1

13 17 19 highest bits !

There is geometrical-physical matter !

1 is point, and 13+17+19=7*7 =circle*circle = Sphere