**Algebra and Geometry of Phi
**

by Tom Gilmore**
**Copyright 2017

Index of all Articles

**Part I: ****Algebra of Phi**

**The Fibonacci** **Series**

Starting from 1 (and assuming a prior zero), iteratively generating the sum of the result with the previous number converges on a ratio of "phi" between successive numbers (each number in the series is equal to the sum of the previous two numbers). This iteration of sums is known as the Fibonacci Series.

0,1,1,2,3,5,8,13,21,34,55,89,144,233,377, ...

Each new number has a ratio to the previous number that gets ever closer to
the transcendent value of phi, alternating between being less than (<) and
greater than (>) phi (to seven decimals 1.6180339). The ratios are shown below in
parentheses, along with the greater or less symbol. As the fractional tail progresses into
established values, the digits are shown in **green.** Underlined digits indicate infinitely
repeating decimal strings.

1 (1 <) 1 (2 >) 2 (1.5 <) 3 (1.__6__
>) 5 (**1.6** <) 8 (1.625 >) 13
(1.__61____5384__
<) 21 (1.__61____9047__
>) 34 (1.6176471 <) **55 (1.6 18 >) 89** (1.6179775 <) 144 (1.618

The **violet**
ratio of **89/55**
is utilized in the Great Pyramid (as developed in Part II - The Geometry of
Phi).

Each ratio, when inverted, is equal to the decimal portion of the following
ratio, for example, for the ratios **8/5, **and** 13/8**,

**8/5** inverted is:

5/8 = 0.625

**13/8** = 1.625

Notice that (5/8 + 1 = 13/8) and since 8/5 is phi, 5/8 is 1/phi, this means
that (1/phi + 1 = phi).

This is the same as saying that the decimal tail of phi is equal to 1/phi.

Another way to express this relationship is that inverting a **phi fraction**
and adding 1 results in the following phi fraction.

**Iteration of
Adding 1 to the Inverted Fraction**

The general form of iteration takes a function (f) and applies it on a number (n), and repeatedly applies the function on the result.

The iteration f(n) = 1/n + 1 eventually converges on the value phi for any positive integer (n). For the purpose of this treatise, we use (n)=1.

1/n+1 (n=1)

1/1+1=**2/1**;

1/(2/1)+1=2/1+1=**3/2**;

1/(3/2)+1=2/3+1=**5/3**;

1/(5/3)+1= 3/5+1=**8/5**;

1/(8/5)+1=5/8+1=**13/8,
**1/(13/8)+1=8/13+1=

Note that this produces the identical ratio series (bold fractions) resulting from the Fibonacci numbers.

This iterative inversion, followed by adding 1, is the same functionality as Euclid's “continued fraction of 1”.

**The Continued
Fraction of 1**

*In fact there is only one number and
it is the number one (refer to the article on Number)*.

Euclid is credited with developing a "continued fraction expansion" utilizing the number 1 that is “equal” to Phi.

1 + 1 / (1 + 1 / (1 + 1 / (1 + **... …**))) = Phi.

(This continued fraction is simply a transposition of (f(n) = 1/n + 1) to (f(n) = 1 + 1/n), for n=1).

In reducing this equation from a finite portion of the continued fraction
(as detailed below), the final fractions of the 2-step reduction (shown in **bold)**
are the Fibonacci ratios. In the
2-step reduction, ratios are inverted by the fractional pattern (1/n), and in
the next step are added to 1, resulting in the next Fibonacci ratio in the
series. In order to exactly follow
the Fibonacci ratios, although the first reduction results in the number 2, it
is replaced by (2 = **2/1**).

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 +
1)))))))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (**2/1**))))))))
=

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1/2))))))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (**3/2**)))))))
=

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 2/3)))))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (**5/3**)))))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 3/5))))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (**8/5**))))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 5/8)))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (**13/8**)))) =

1 + 1 / (1 + 1 / (1 + 1 / (1 + 8 /13))) =

1 + 1 / (1 + 1 / (1 + 1 / (**21/13**))) =

1 + 1 / (1 + 1 / (1 + 13 /21))

1 + 1 / (1 + 1 / (**34/21**)) =

1 + 1 / (1 + 21/34) =

1 + 1 / (**55/34**) =

1 + 34/55 =

**89/55**
= 1.618 = Phi

The ratio **89/55**
is the Egyptian Royal Phi ratio used in constructing the Great Pyramid of Giza,
as demonstrated below.

**Part II: ****Geometry
of Phi **

**The Phi
Right-Triangle**

Two **exact**
Royal measures are given by the Nile Scholars for the G**reat Pyramid** dimensions.

**Base of
Pyramid = 440 Royal Cubits
Apothem of Pyramid = 356 Royal Cubits**

*The "apothem" of a pyramid is the line from the mid-base to the
summit*.

The measures of the pyramid in Royal Cubits are diagrammed below left.

Applying the Pythagorean formula (Hypotenuse^{2} = Base^{2}
+ Height ^{2}) to the right-triangle of height, apothem, and half-base,
(height^{2} = 356^{2} - 220^{2}), so height = ~279.8857
Royal Cubits. This is approximately
0.0004% short of the Royal approximate measure of **280 Royal Cubits. **

The relative Royal dimensions of the right triangle consisting of the
apothem, height, and half-base, are the values Phi, root-of-Phi, and 1 (illustrated
above right).

The relative lengths are found by dividing all 3 sides by 220.

Half Base = 220/220 =1

Apothem =356/220 = **89/55 **= 1.6181818 = Phi.

Height = 280/220 = 1.272727 = square-root of Phi (when Phi =1.6181818)

The measures of the apothem and half-base are 4 times the Phi ratio of **89/55**
(89x4=356 and 55x4=220).

**Approximate
Squaring of the Circle**

It is mathematically impossible to exactly
square the circle because the number Pi is transcendent.

**The ratio of base to height in the Great Pyramid squares the
circle.
**(Squaring the circle is essential to projecting global maps onto
flat surfaces).

The circumference of the square base approximately equals the circumference of
the circle with a radius of the height (see diagram above).

The height of the pyramid is ~279.8857 Royal Cubits, and for this proof is
rounded to 280.

**The circumference of a circle is (Pi x 2r).
**Taking the height of the
pyramid (280) as the radius(r) of a circle, the circumference is (Pi x 560).

The circumference of the base is 440
x 4 =** ****1760. **

**Phi as
a Rectangle**

Geometrically, the number Phi is defined as a ratio represented by a point on a line positioned such that the two segments have the property that the short segment has the same proportion to the long segment as the long segment has to the entire line.

A----------B----------------C

The formula is Phi = AC/BC = BC/AB ~ 1.618

Da Vinci named the rectangle with sides the proportion of 1 to 1.618 (AB to
BC) the Golden Section.

The phi-rectangle is shown below.

**Phi in the Pentagram**

Connecting the 5 points that are seperated by
72 degrees with all possible lines, creates a pentagram (5-pointed star) within
a pentagon, as shown below.

*At the
center of the pentagram is an inverted pentagon, and connecting the vertices of
that pentagon produces an inverted pentagram. As the symbol of magic, the outer
pentagram represents “white magic”, meaning majic used to help
others (outer), and the inner pentagram represents “black magic”,
meaning magic used for selfish purposes (inner). *

The pentagram illustrates two equations
involving Phi. The phi ratios are
integral to the proportions of the line segments.

In the diagram above, two equations are shown on the upper right, and are
illustrated (color coded) by lined-up spanned lengths between intersections of
a pentagram line, such that the sums match the equations.

**Powers of Phi**

The two equations above are specific examples
of the general equation that

Phi^{ n} = Phi^{ n-1 }+ Phi^{ n-2},** **or that each power of Phi is the sum of the
previous 2 powers of Phi.

Using n=1,

Phi^{ 1} = Phi^{ 0 }+ Phi^{ -1 }means that **Phi = 1 + 1/Phi,
**the first equation in the graphic above.

Using n=2,

Phi^{ 2} = Phi^{ 1 }+ Phi^{ 0} means that **Phi ^{ 2 }=
Phi + 1**, the second equation in the graphic above.

The following proof of the general equation
uses (for an example) the series numbers and ratios of

(**5**, 8/5, **8**, 13/8, **13**, 21/13, **21**).

Phi ^{1} = 8/5

Phi^{ 2} = Phi x Phi = (8/5 x 13/8) = 13/5, and

Phi^{ 3} = Phi x Phi x Phi = (8/5 x 13/8 x 21/13) = (13/5 x 21/13) =
21/5.

The product of any number of consecutive phi
ratios always reduces to a ratio with the denominator of the first ratio's
denominator and the numerator of the last ratio's numerator.

**Logical proof of the exponential equation:** Since
all consecutive powers of Phi have the same denominator, the general
exponential equation (Phi^{ n} = Phi^{ n-1 }+ Phi^{ n-2}),**
**reduces to the Fibonacci series numbers (see example below). In other words, the powers of the ratio Phi
reduce to the Fibonacci series numbers, which by definition sum the previous 2
numbers.

Example: Using Phi^{ n} = Phi^{
n-1 }+ Phi^{ n-2},** **where Phi^{ n} =21/5, Phi^{
n-1 }=13/5, and Phi^{ n-2 }=8/5,

(21/5 = 13/5 + 8/5), or (**21** = **13** + **8**).

**Phi by Compass and Ruler**

The geometric
derivation of Phi is shown below.

**It
is half the length of a unit square plus the diagonal of the half-square
rectangle**.** **

The diagonal is the hypotenuse of a right
triangle with sides of 0.5 and 1.

Thus, the diagonal is sqrt(1x1+.5x.5) = sqrt(1.25) = sqrt(5/4).

So (Phi = .5 + sqrt (5/4)) or

**Phi = (1+sqrt****5)/2**

**Part III: ****The
Transcendent Equations**

The 3 transcendent
numbers of Phi, Pi, and sqrt5 have a common transcendent root. This is established by the equations
between them.

**Phi = (1+sqrt****5)/2
**

**The (Approximate) Transcendent Equation Proof**

**Pi
= Phi ^{ 2} x 6/5**

**Proof using ****Pi ~ 22/7 and the Fibonacci series numbers ****(21, 34, 55).**

Since each number in the series increases by Phi (because Phi is the ratio
between the numbers), 2 numbers in the series increase by Phi^{2}.

This means that, using the series numbers (**21,
34, 55**), Phi = 34/21, and Phi^{2} = 55/21.

Substituting Phi^{2} = 55/21 in the formula **(X = Phi ^{ 2}
x 6/5),
**X = (55/21 x 6/5) = 330/105 =

Using higher numbers in the series increases the accuracy of both Phi and Pi in the equation, but because Pi and Phi are transcendent there is no exact proof.

**Solving for Sqrt****5 in terms of Phi**

**Demonstrating that sqrt****5 =**** ****Phi + 1/Phi**

**Phi = (1 + sqrt5) /2
**And thus:

2Phi = 1+ sqrt

sqrt

sqrt

And since

**Phi = 1 + 1/Phi**

**(Phi – 1) **= 1/Phi), and
substituting 1/Phi for **(Phi – 1)
above,**

**sqrt****5 =**** ****Phi + 1/Phi**

**Solving for** **Pi in
terms of ****Sqrt****5**

**Pi = Phi ^{2} x 6/5,
and Phi^{2} = Phi +1, **so

**Pi = (Phi + 1) x 6/5 **and** ****Phi**** = (1+sqrt****5)/2**** , so
Pi = ((1+sqrt**

= (1 + sqrt

= (1.5 + (sqrt5)/2) x 6/5

= 9/5 + (sqrt

**Pi = 9/5 +
.6 sqrt5 **

**Confirmation usimg the
following approximations:
sqrt**

**Pi = 9/5 +
.6 x ****2.236 ****= 1.8 + 1.3416 = ****3.1416**

**Part IV: ****Phi
in Earth Measures**

The **Royal
Cubit** is a Phi-Measure that was
used in building the Great Pyramid.
*Wikipedia has this wrong, and
won’t correct it.*

**Earth
and Time Measures ****(Base-12)**

**Earth Time
****à
Geographic **

Foot Cubit Cubit Foot Statute Foot

**360 degrees 24
hours **24** 24
hours 24
hours
60 minutes 60
minutes **60

60 seconds 60 seconds

100 feet 1000 cubits 880 6000 Roman feet 5280 Statute feet

**129,600,000 86,400,000 76,032,000 144,000,000 126,720,000**

A Cubit is 1.5 Geographic Feet **(129.6 / 86.4 = 1.5)**

A Cubit is 0.88 of a Royal Cubit **(76.032 / 86.4 = 0.88)**

A Roman Foot is 0.9 Geographic Feet **(129.6 / 144 = 0.9)**

The base of the Great Pyramid measures
__exactly__ **750 Geographic Feet**

=500 Cubits (1/2 second of time)

=440 Royal Cubits (The measure given by the Nile Scholars)

=833.33 Roman Feet (The measure given by the Roman historian Pliny)

**English Statute Foot and Freemasonry**

The English could have retained the time-based mile-measure and their foot be exactly equal to the Geographic Foot by changing the Roman Mile of 6000 feet to being 5400 feet (24x1000x5400=129,600,000), but the Masons sought to incorporate the ratio of Phi into the time measure of feet, and the predominant Masonic presence in British government allowed them to influence the British Parliament in 1593 to declare there to be 5,280 feet to a mile by statute. The Masons chose this number because it preserved the Phi-Measure of the Royal Cubit, and also closely approximated the Geographic Foot.

5,280 feet to a mile accomplishes **two major
objectives**:

**Since (126,720/129,600 =
0.9777),
**

**Since (76,032/126,720 = 0.6),
**

**The Remen ****(a
Phi-Measure)**

**The Remen is defined as the diagonal of a square Royal Cubit**. Mathematically a Royal Cubit is 1.70__45__
Geographic feet, so using the diagonal of a square Royal Cubit is **approximately** (sqrt
(2 x 1.7045^{2})) ~ **(2.4106) feet in length**. However since Phi measures are by definition
inexact, adopting approximate measures with practical applicability is
preferred, and some measures of the interior of the pyramid support the notion
that the standard measure of the Remen was 2.3885
feet (and 2 Remen was 4.7777 feet). The following are examples of the
2-Remen measure of 4.__7__ feet.

**There are 3 stone plugs 2 Remen long
lodged at the base of the ascending passage**.

The 3 plugs lodged in the passage have been estimated to be just under 4.8 feet in length, which is 2 Remen (4.777 feet).

The Grand Gallery floor is of equal width as the ascending corridor, but
there are raised shoulders on each side that are slotted at 27 intervals. These slots extend horizontally across
the shelves and vertically up the walls.
It is currently thought by many that these slots were intended for
wooden beams suspending the granite plugs for release into the ascending
passage, completely filling it. There are 27 paired slots, while the ascending
passage is 129 feet, and **(27x4.777=129**).

**All the passages, shafts, and chambers are estimated to be
offset 23.9 feet east of the centerline.**

This would be a **10 Remen** offset, since
5x4.77__7__ = 23.88__8__ feet.

If there was initially a smaller pyramid built where the Great Pyramid
stands (shown in **green** above), the opening of the descending passage would
have been outside the base of this smaller pyramid, and when the truncated
pyramid was built over it, the in-line extension of the descending passage
caused the entrance to be elevated.
The Queen's Chamber would have been centered, both north/south and
east/west in this initial pyramid.

As illustrated above, the offset is small in comparison with the length of the base. Either the offset was intentional and intended to help obscure the location of the sealed entrance, or the initial pyramid was too close to the eastern end of the plateau, and this did not allow the larger pyramid to be exactly centered east to west over it.