Welcome to JManimas web pages: (This work originally posted to the JManimas web pages in December of 2000)

Welcome to Aquarius, Volume 6, July 25, 2006

A math secret lost since 500 B.C. and re-discovered by John Manimas Medeiros on September 1, 1996.

*The Pyramid Mark (Two): The Square Root of Two *

Copyright © 2000, John Manimas Medeiros, All Rights Reserved.

__Math symbols used for this web text__: 2^{5} or 2^5 = two raised
to the fifth power.

1/ SQRT(Pi) = one over (divided by) the square root of Pi. SQRT means "the square root of."

__The question__ posed by the ancients was: "Can you construct a circle
equal in area to a given square, using only a compass and straightedge?" Or:
"Can you construct a square equal in area to a given circle, using only a
compass and straightedge?"

__An initial part of the correct answer__ (that is, a helpful path of inquiry) is the super-binary number line
system, employing the square root of two as the number base, which enables us to
construct any number, including fractions, including any irrational or
transcendental number, using only the compass and straightedge.

If the reader is not familiar with the use of a number base other than ten, the following explanation will be challenging. If the reader is familiar with the base two or binary number system, the explanation will be easy to understand and put into practice at home with an ordinary sheet of paper, pencil or pen, and a compass and straightedge. For greater control and accuracy, a sliding-bar or clip compass, with both the scriber and pivot in the vertical position, is recommended. One other item, a block of soft wood, will make the practice easier and more accurate.

When we look at the Great Pyramid from overhead, we see a square with the crossed diagonals, as shown in Figure 1 below. If the side is assigned a length of one (1), then the diagonal is the square root of two, which is an irrational number, meaning it includes an endless, non-repeating mantissa in our decimal digital place value system. Study the lines of the square and diagonals after extending the line lengths with the compass and straightedge, as in Figure 2.

To test the math, and practice using this number system, the student will need a scientific calculator.

__The Algorithm__ is best illustrated through the description of distinct
procedures which would be subroutines of a complete computer program. There is
something that occurs with D (1.414213562...) that occurs with no other number:
when it is divided in half, the result is the negative power of itself, and,
with each successive division in half, the result is the next odd negative power
of D. The even negative powers of D are the same as the negative powers of 2,
which is of course D squared. We therefore can obtain all of these calculated
number values by dividing D in half, and by dividing S in half (S = Side =
1).

Pick up your scientific calculator and play with D as the number base. First,
get D and then save it in memory. Divide D in half. The result is 0.707106781...
. Recall D again. Press 1/x to get the value of 1/D. The result is
0.707106781... . That is one over the square root of two, or D^{-1} .
Divide D^{-1} in half. The result is D^{-3} . Divide that in
half. The result is D^{-5} . Each time you divide the result in half the
next result is the next odd negative power of D. When you divide 1 in half, and
then again and then again, you get the even negative powers of D. Can we do the
same with a compass and straightedge? At first it would appear impossible,
because we cannot divide such small lines in half. But, there is a way. The
lines are proportional, and we can use a proportional multiplier to make the
lines manageable. Here the fun begins. With your calculator, recall D, then add
2. The result is 3.414213562... . Divide that result in half. The next result is
1.707106781... . Add 1. The result is 2.707106781... . Divide that in half. The
next result is 1.353553391... . Add 1. The result is then 2.353553391... .
Divide that in half. The next result is 1.176776695... . As you continue you
will see that each repetition of this simple operation yields the next odd
negative power of D plus 1 in number line value. That means we can perform these
same operations with a compass and straightedge. If we use the compass to
subtract the line length of 1, the result is the very small remaining line
length that is the odd negative power of D. We obtain the even negative powers
of D by repeatedly dividing S (1) in half. That means we can use our calculator,
or a compass and straightedge, to obtain a listing of the negative powers of D.
We can produce either by calculation, or with compass and straightedge, any
negative power of D. Those values are therefore available to us to construct
concrete number lines by summation, the same way that we construct abstract
digital numbers by summation using the negative powers of 10. For example, the
abstract digital decimal number 2,984 is:

4 * 10^{0} + 8 * 10^{1} + 9 * 10^{2} + 2 *
10^{3}

__For the first step__ in writing a simple program that repeats the
operation and prints out a list of the values, designate a variable of D value.
Create a For-Next Loop that distinguishes between the odd negative powers (using
D) and even negative powers (using S = 1). Include a counter. And then just
apply the simple operation of adding 1 and dividing in half the running value.
Print out the “next result” before the loop moves on to the next repetition. The
program print out desired is a list as follows:

D^ -1 = (D/2^ 1) +1 = 1.707106581...

D^ -2 = (S/2^ 1) +1 = 1.5

D^ -3 = (D/2^ 2) +1 = 1.35355339...

D^ -4 = (S/2^ 2) +1 = 1.25 and so forth.

This relatively simple program becomes a subroutine for holding the series of
elements, for example D^{-1} to D^{-128}, in an array to be used
later.

Link to: (Welcome) or (Geometry Alpha Index).

Next, practice with your calculator summating a number using D. What is the
value of D^{-5} plus D^{-9} ? That would be 0.176776695... plus
0.044194173... which equals the value of 0.220970869... . Our first thought
would be that we can calculate D^{-5} and save it. Then calculate
D^{-9} and add the two values. But, there is another way that we can
construct the summation in a series of repeated operations. We will divide D in
half five times in order to obtain 0.044194174... and another D in half three
times in order to obtain 0.176776695... . We perform the operation on the first
D two times, then we add the second D and continue the operations, so that we
shall have halved one D five times and the second D three times, but in a single
continuous series.

It works as follows: We recall (or construct) the diagonal. Add the value of 2. The resulting line is length 3.414213562... .

(1) Add 2, yielding 5.414213562..., and divide that in half, yielding the value 2.707106781... .

(2) We add 2, yielding 4.707106781..., and divide that in half, yielding 2.353553391... .

(3) Here we are ready to add the diagonal again, yielding 3.767766953... . Add 2, which gives us 5.767766953..., and divide in half, yielding 2.883883476... .

(4) Add 2, giving us 4.883883476..., and divide in half, yielding 2.441941738... .

(5) Add 2, giving us 4.441941738..., and divide in half, yielding 2.220970869... .

All of these line lengths have been manageable with the compass and straightedge on an ordinary sheet of paper. However, our end result is 2 plus 0.220970869... in length. We can construct only the smaller part, or at least see the smaller part, by opening the compass to a length of 2, and marking off the two, resulting in two lines: one being 2, the other being 0.220970869... . This same procedure can be used to construct any fractional number, including the irrational and transcendental numbers.

__The general rule__ which applies to the super-binary system is:

Where D = 1.414213562... for negative powers (meaning D^{-x} or
1/D^{x}):

__For even powers__ D raised to minus X power = 1 halved X/2 times.

Example: D^{(-8)} = 1/16 or .0625 and 2^{(-4)} = 1/16 (same
as 1 /2, /2, /2, /2). This is accomplished by using the side of the square, S,
which has the unit line-length of one, and dividing it in two, halving it, four
times.

__For odd powers__ D to minus X power = D halved (X+1)/2 times.

Example: D^{(-9)} = D times 2^{(-5)} or D/32, which means
1.414213562... divided by 32 = 0.044194173... . This is accomplished by using
the diagonal (D line length) and dividing it in two (halving it) five times.

__Look at Table 1__, which shows the digital decimal version of 1/Pi and
the super-binary version. Then look at Table 2, which shows the results of six
sample calculations that are also samples of constructions with the compass and
straightedge.

__A special rule__ applies when two of the terms in the summation are an
"even pair," which then require a "combined step" in the summation procedure. An
even pair is any two consecutive exponents where the higher exponent is even and
the lower is odd. One can easily see that in this case, such as SQRT(2) to the
-10 and -9, that the construction of each term requires the same number of
steps. For the -10 value, we add 1 and then divide in half 5 times, and, for the
-9 value, we add D but also divide in half 5 times. Therefore, the "combined
step" requires that we:

Add 1, add D, add 2, and divide in half.

If the target value were 0.03125 + 0.044194173 = 0.075444173, we start with 2, and...

... 2 + 1 + D + 2) / 2 = 3.207106781 ... + 2) / 2 = 2.603553391

... + 2) / 2 = 2.301776695 ... + 2) / 2 = 2.150888348

... + 2) / 2 = 2.075444174

Thus, we have added the 1 value and the D value together and divided them both in half the same number of times, five times.

__The second step__ in the algorithm for a super-binary number program is
the number element selection step, with a precision selection command,
incorporated in a For-Next Loop or similar program subroutine. This section
begins with the operator “inputting” the target value to be summated or
constructed from the elements arrayed previously in step one. Write commands and
statements that will retrieve the elements one by one and then add them to the
running sum on a conditional basis. The condition is that if adding the next
largest element brings the running sum closer to the target value, but not
beyond it, then it is added. If the next largest element is too large, and makes
the running sum larger than the target value, it is discarded, and the loop
starts again. Incorporated in this loop you must include two other important
computer operations. There must be two command lines that select the level of
precision desired. That can be accomplished by specifying the value (for
example: 1/10^16) which, if the running sum exceeds the target value, the
element will be discarded. In a later command line, specify the value (for
example: 1/10^20) which, if the running sum is closer to the target value, the
program exits that loop and moves on to step three. The program should print out
the elements selected that sum to the target value, as shown in the Tables.

__The third step is the assembly step__ of the algorithm and is derived
from the general rule. We begin with the line that has to be divided in half the
greatest number of times.

__Look at Table 2__ and sample calculation #1, 1/ SQRT(Pi). The elements
of the series are D to:

-2 -8 -19 -24 -28 -27 -42 -46 -49 -57 -60 -66 -70 -74 -82 -88 -92 -95 -102 .

Therefore, we begin with D^{-102} and distinguish it as an even
number. The general rule tells us that if we divide 1 in half 51 times, the
result is D^{-102} . We begin our calculation, and our construction,
with S (1). Add 2, divide that number line in half, the result is 1.5, which is
just what we wanted and expected (S^{-1} + 1). The second element is
D^{-95}. The general rule tells us to distinguish that power as an odd
negative power and divide -95 in half and add one-half (round it up to the next
whole number), which gives us 48. Therefore, in order to add in our next element
in the series, we will need to add in D and divide it in half (after adding 2)
48 times. To accomplish this we divide S in half, counting backwards, from 51
(1.5) (add 2, divide in half) to 50 (add 2, divide in half) to 49 (add 2, divide
in half) to 48 (NOW add D, add 2, divide in half). There we have added in the
next element. The third element, being D^{-92} , means we add S (and 2)
at the 46th step. We add S again at the 44th step (for D^{-88}), and so
on, until we get to the final element, D^{-2}, which gives us a result
of 2.584189583, from which we can subtract the final 2, to get our target value
of 0.584189583. We shall have added 2 and divided in half 51 times, and we shall
have added in S 14 times, and added in D 5 times, adding in each S or D __at
the right step__ to achieve the correct number of divisions required to
produce each element of the summation. This same repetitious procedure works for
any fraction, including any irrational or transcendental number. For purposes of
precision, there is always a smaller element available to add that brings us
closer to the infinite decimal, but not past it, for example, D^{-500}
is a very small number, and D^{-703} is much smaller. This means that we
have exactly the same situation as with any abstract digital number system.
Irrational and transcendental numbers are infinite in the digital decimal
system, but the negative powers of D are equally infinite. We can increase our
precision indefinitely, although doing so usually has no practical purpose. I
call this method "sequential halving."

It is important to remember the unique reality of the super-binary number line system. The end results of the operations, whether performed on the calculator or on paper with the compass and straightedge, are lines. However, the lines are not abstract representations of numbers, they are the concrete numbers in proportion to the line we start with and assign the value of one. That is why we can use our line of 1/ SQRT(Pi) in length to construct a circle with the same area as the given square, subject to our chosen level of precision.

Using the super-binary number line system, we use the side 1 and D of the square that has an area of 1, to construct a line of length 1/ SQRT(Pi), and then use our constructed 1/ SQRT(Pi) (0.584189583) as the radius to construct the circle. The area of the circle is 0.000000002 less than the area of the square. A difference of 0.0000002 percent. If the side of the square were 1,000 meters, the area of the square would be 1,000,000 square meters, and the area of the circle would be 99,999.998 square meters, a difference of .002 square meters, equal to 20 square centimeters, or a rectangle of 4 centimeters by 5 centimeters within the square kilometer. That is an area smaller than one-half of a 3” by 5” index card. To appreciate this level of precision, note that if when measuring or "constructing" our square of 1,000 meters on a side, we made an error of one tenth of a centimeter short on two opposite sides, that error would reduce the area of the square by 10,000 square centimeters (1 square meter), a difference which is 500 times greater than the precision possible with the low technology of the super-binary number line system. Greater precision is possible by adding more, smaller elements.

This procedure is equally valid if the ancient question is reversed: "Can you construct a square equal in area to a given circle, using only a compass and straightedge?" In that case we begin with a circle with R = 1 and Area = Pi. We use our 1 to construct 2, the square root of two and then the square root of Pi, which is 1 +0.772453851..., for the side of our square.

End of Main Text

Bibliography: (30 math texts plus the four works listed below)

Clark, R.T. Rundle. *Myth and Symbol in Ancient Egypt*. New York: Thames
and

Hudson, 1959.

Tompkins, Peter. *Secrets of the Great Pyramid*. New York: Harper and
Row, 1971.

(Lots of math in this book.)

West, John Anthony. *Serpent in the Sky*. New York: Julian Press,
Crown

Publishing Group, 1987.

Kunzig, Robert. "A Head for Numbers." *Discover: the world of science*,
July 1997,

pp. 108-115. (This article describes research evidence that the
human

brain contains a "number line" from which we derive our concept of
number.)

Table 1: The digital decimal expression of 0.318309886... = 1/Pi

6 times 10^{-9} = 0.000000006...

+ 8 times 10^{-8} = 0.00000008

+ 8 times 10^{-7} = 0.0000008

+ 9 times 10^{-6} = 0.000009

+ 0 times 10^{-5} = 0.00000

+ 3 times 10^{-4} = 0.0003

+ 8 times 10^{-3} = 0.008

+ 1 times 10^{-2} = 0.01

+ 3 times 10^{-1} = 0.3

______________________________________

Total 0.318309886...

The super-binary expression of 0.318309886... = 1/Pi

D = diagonal or 1.414213562... (in sci notation, 10 raised to)

1 times D^{-106} = 0.0000000000000001110223... (-16)

+ 1 times D^{-101} = 0.000000000000000628037 (-16)

+ 1 times D^{-91} = 0.00000000000002009718 (-14)

+ 1 times D^{-84} = 0.0000000000002272737 (-13)

+ 1 times D^{-78} = 0.000000000001818989 (-12)

+ 1 times D^{-75} = 0.000000000005144879 (-12)

+ 1 times D^{-65} = 0.0000000001646361 (-10)

+ 1 times D^{-59} = 0.000000001317089 (-09)

+ 1 times D^{-56} = 0.00000000372529 (-09)

+ 1 times D^{-49} = 0.00000004214685 (-08)

+ 1 times D^{-46} = 0.0000001192093 (-07)

+ 1 times D^{-33} = 0.00003051758 (-05)

+ 1 times D^{-30} = 0.00003051758 (-05)

+ 1 times D^{-24} = 0.0002441406 (-04)

+ 1 times D^{-15} = 0.005524272 (-03)

+ 1 times D^{-8} = 0.0625 NA

+ 1 times D^{-4} = 0.25 NA

______________________________________

Total 0.318309886...

Table 2: Six sample calculations, or constructions, of irrational or transcendental numbers:

1) 0.5641895835477562 = 1/ SQRT(Pi) = the sum of D raised to the following:

-2 -8 -19 -24 -28 -37 -42 -46 -49 -56 -62 -69 -73 -82 -89 -96 -100

2) 0.7724538509055160 = [SQRT(Pi)] - 1 = the sum of D raised to the:

-1 -8 -17 -28 -31 -38 -43 -48 -52 -62 -72 -77 -84 -90 -98 -102

3) 0.1415926535897932 = Pi -3 = the sum of D raised to the:

-6 -12 -21 -24 -30 -38 -42 -48 -52 -57 -67 -73 -78 -82 -94 -99 -103

4) 0.6180339887498948 = Phi - 1 = 1/ Phi = the sum of D raised to the:

-2 -7 -11 -15 -18 -28 -34 -38 -42 -47 -51 -57 -61 -68 -75 -79 -83 -87 -92 -95 -101

5) 0.8090169943749474 = Phi / 2, = sine of 54 degrees = the sum of D raised to the:

-1 -7 -13 -18 -22 -30 -39 -49 -55 -58 -64 -69 -76 -79 -102

6) 0.7320508075688772 = [SQRT(3)] - 1 = the sum of D raised to the:

-1 -11 -17 -28 -31 -38 -44 -49 -59 -70 -77 -81 -87 -97 -101

Figure 1: A square. If side S = 1, then diagonal D = SQRT(2)

Figure 2: The pyramid mark - top view of a regular, four sided pyramid, which provides for (x's left to right) construction of 1, SQRT(2), and 2, the three values used to construct all other numbers. The end points of the lines can be pressed into a block of wood to provide stable guides for each time the compass is opened to these values.

__To order the author’s versions__ of the computer programs that
illustrate the Super-Binary Number Line System, send $50.00 U.S. to cover the
cost of producing the 3.5" diskette (or diskettes), suitable packaging, the
postage to mail first class from my address (John M. Medeiros, P.O. Box 536,
Bellows Falls, VT 05101) to yours. The author will not use any diskette received
in the mail and will not accept responsibility for its return. If you send a
check or money order, please make it payable to "John Manimas." The diskette
contents will include text instructions and the best programs available from the
author to date.

We begin every construction with a square and the diagonal. The square provides the three line lengths that are needed to construct any irrational fraction expressed as a line length. The side S is one. The diagonal D is SQRT(2). Twice S is two. To help keep these line lengths the same each time the compass is used, you can make tiny dimples or impressions in a piece of wood to mark off each length. Below, I have used the space bar of the keyboard to create the lines. One is ten spaces, D is fourteen spaces to approximate 1.414213562... , and two is twenty spaces.

line length of one: __________

line length of D: ______________

line length of two: ____________________

Our example from pyr 1, p. 2: construct 0.220970869... (0.176776695... + 0.044194173... ) is the same as D^{-5} plus D^{-9} and we obtain this sum by sequential halving.

Step 1 ( 2 + D ): ___________________+______________

Our line length is: 3.414213562...

Step 2 ( add 2): ___________________+______________+____________________

Our line length is: 5.414213562...

Step 3 (halve): ___________________________

Our line length is: 2.707106781...

Step 4 (add 2 ): ___________________________+____________________

Our line length is: 4.707106781...

Step 5 (halve): _______________________

Our line length is: 2.353553391...

Step 6 (add D ): _______________________+_____________

Our line length is: 3.767766953...

Step 7 (add 2 ): _______________________+_____________+___________________

Our line length is: 5.767766953...

Step 8 (halve): _____________________________

Our line length is: 2.883883476...

Step 9 (add 2 ): _____________________________+___________________

Our line length is: 4.883883476...

Step 10 (halve): ________________________

Our line length is: 2.441941738...

Step 11 (add 2): ________________________+___________________

Our line length is: 4.441941738...

Step 12 (halve): ______________________

Our line length is: 2.220970869...

We can subtract the 2 to yield a very short line that is 0.220970869... in length. This is a sort of beginners' example. It demonstrates that if we divide D in half 5 times and also divide D in half 3 times in a continuous series of steps, we have accomplished the sequential halving that yields our target value. This same procedure can be used to produce a summation of any set of values that represent D to specified negative powers. By use of a computer program that completes the time consuming calculations, it has been demonstrated that any irrational number can be given expression as a line length using this method, and the operator can achieve a target level of precision. The necessity or "limitation" that the operator chooses the level of precision is in fact a reality that applies to any and all constructions. There is no method to construct *exactly *any irrational number, because an irrational is an endless, non-repeating decimal. Although any geometer can construct any square root as a line length, using the Pythagorean Theorem, all such line lengths are exact only in theory. For concrete lines, it is not possible to construct a second line, using only the compass and straightedge, and prove mathematically that the two lines are the exact irrational and also perfectly equal. Neither the mathematician nor the geometer can demonstrate that two lines are equal to the infinite decimal place.

__Advanced fraction
constructions__ using the base SqRt(2) and sequential halving.

Originally uploaded 3/11/01. Reviewed 11/10/01.

You always begin with a square and diagonal providing line lengths of 1, D, and 2.

The number line construction begins with a line length of 2.

#1 construction: 1 over the square root of Pi:

* Your Target Value is: 0.564189583547756

--> Enter decimal level of precision (1 to 15):

* Your precision level is: 0.0000000000000001

__Term, power, value, LVS = line value summation thus far__

1: -2 --> +0.500000000000000 LVS= 0.500000000000000

2: -8 --> +0.062500000000000 LVS= 0.562500000000000

3: -19 --> +0.001381067932005 LVS= 0.563881067932005

4: -24 --> +0.000244140625000 LVS= 0.564125208557005

5: -28 --> +0.000061035156250 LVS= 0.564186243713255

6: -37 --> +0.000002697398305 LVS= 0.564188941111560

7: -42 --> +0.000000476837158 LVS= 0.564189417948718

8: -46 --> +0.000000119209290 LVS= 0.564189537158008

9: -49 --> +0.000000042146849 LVS= 0.564189579304856

10: -56 --> +0.000000003725290 LVS= 0.564189583030146

11: -62 --> +0.000000000465661 LVS= 0.564189583495808

12: -69 --> +0.000000000041159 LVS= 0.564189583536967

13: -73 --> +0.000000000010290 LVS= 0.564189583547257

14: -82 --> +0.000000000000455 LVS= 0.564189583547711

15: -89 --> +0.000000000000040 LVS= 0.564189583547752

16: -96 --> +0.000000000000004 LVS= 0.564189583547755

17: -100 --> +0.000000000000001 LVS= 0.564189583547756

Constructed number line summation to precision

level 15 is: 0.564189583547756

Target fraction was: 0.564189583547756

Number of terms is: 17. Last exponent is: -100

__Step, term, (power), RLV = running line value__

50, 17 (-100) : (RLV = RLV +1 +2) halved = 2.500000000000000

49, *** *** : (RLV = RLV +0 +2) halved = 2.250000000000000

48, 16 (-96) : (RLV = RLV +1 +2) halved = 2.625000000000000

47, *** *** : (RLV = RLV +0 +2) halved = 2.312500000000000

46, *** *** : (RLV = RLV +0 +2) halved = 2.156250000000000

45, 15 (-89) : (RLV = RLV +D +2) halved = 2.785231781186548

44, *** *** : (RLV = RLV +0 +2) halved = 2.392615890593274

43, *** *** : (RLV = RLV +0 +2) halved = 2.196307945296637

42, *** *** : (RLV = RLV +0 +2) halved = 2.098153972648318

41, 14 (-82) : (RLV = RLV +1 +2) halved = 2.549076986324159

40, *** *** : (RLV = RLV +0 +2) halved = 2.274538493162080

39, *** *** : (RLV = RLV +0 +2) halved = 2.137269246581040

38, *** *** : (RLV = RLV +0 +2) halved = 2.068634623290520

37, 13 (-73) : (RLV = RLV +D +2) halved = 2.741424092831807

36, *** *** : (RLV = RLV +0 +2) halved = 2.370712046415904

35, 12 (-69) : (RLV = RLV +D +2) halved = 2.892462804394500

34, *** *** : (RLV = RLV +0 +2) halved = 2.446231402197250

33, *** *** : (RLV = RLV +0 +2) halved = 2.223115701098625

32, *** *** : (RLV = RLV +0 +2) halved = 2.111557850549312

31, 11 (-62) : (RLV = RLV +1 +2) halved = 2.555778925274656

30, *** *** : (RLV = RLV +0 +2) halved = 2.277889462637328

29, *** *** : (RLV = RLV +0 +2) halved = 2.138944731318664

28, 10 (-56) : (RLV = RLV +1 +2) halved = 2.569472365659332

27, *** *** : (RLV = RLV +0 +2) halved = 2.284736182829666

26, *** *** : (RLV = RLV +0 +2) halved = 2.142368091414833

25, 9 (-49) : (RLV = RLV +D +2) halved = 2.778290826893964

24, *** *** : (RLV = RLV +0 +2) halved = 2.389145413446982

23, 8 (-46) : (RLV = RLV +1 +2) halved = 2.694572706723491

22, *** *** : (RLV = RLV +0 +2) halved = 2.347286353361746

21, 7 (-42) : (RLV = RLV +1 +2) halved = 2.673643176680873

20, *** *** : (RLV = RLV +0 +2) halved = 2.336821588340436

19, 6 (-37) : (RLV = RLV +D +2) halved = 2.875517575356766

18, *** *** : (RLV = RLV +0 +2) halved = 2.437758787678383

17, *** *** : (RLV = RLV +0 +2) halved = 2.218879393839191

16, *** *** : (RLV = RLV +0 +2) halved = 2.109439696919596

15, *** *** : (RLV = RLV +0 +2) halved = 2.054719848459798

14, 5 (-28) : (RLV = RLV +1 +2) halved = 2.527359924229899

13, *** *** : (RLV = RLV +0 +2) halved = 2.263679962114949

12, 4 (-24) : (RLV = RLV +1 +2) halved = 2.631839981057475

11, *** *** : (RLV = RLV +0 +2) halved = 2.315919990528737

10, 3 (-19) : (RLV = RLV +D +2) halved = 2.865066776450916

9, *** *** : (RLV = RLV +0 +2) halved = 2.432533388225458

8, *** *** : (RLV = RLV +0 +2) halved = 2.216266694112729

7, *** *** : (RLV = RLV +0 +2) halved = 2.108133347056365

6, *** *** : (RLV = RLV +0 +2) halved = 2.054066673528182

5, *** *** : (RLV = RLV +0 +2) halved = 2.027033336764091

4, 2 (-8) : (RLV = RLV +1 +2) halved = 2.513516668382046

3, *** *** : (RLV = RLV +0 +2) halved = 2.256758334191023

2, *** *** : (RLV = RLV +0 +2) halved = 2.128379167095511

1, 1 (-2) : (RLV = RLV +1 +2) halved = 2.564189583547756

0 step, subtract 2, final Line Value is: 0.564189583547756

to precision level 0.0000000000000001

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