Phi's Fascinating Figures
On this page we meet some more marvellous maths about the number Phi itself, its
multiples and powers.
Contents of this Page
The icon means
there is a Things to do investigation at the end of the section.
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We return to the definition of Phi, as the positive valued solution to
P(P1)=1
P here is either of 2 values, Phi and 1Phi.
So P has the property that P^{2}=P+1.
Let's look at Phi^{2} first.
We can read the equation above as
to find P^{2}, just add 1 to P.
Things to do
 Use your calculator to evaluate Phi=(1+5)/2.
If you have a Memory on your calculator, store this value in it.
 Square it and check that it is just Phi+1 (i.e. subtract one and compare with the
Memory value).
 Use your calculator to evaluate the other value (1Phi) or (1Memory) =
(15)/2
 Again square the value just found and check that it is just the same as
adding 1 to the value you squared.
Now let's look at Phi^{3}.
Is there another way to calculate Phi^{3} apart from just Phi x Phi x Phi?
Yes  let's see how to compute it in two more ways.
We use the basic P(P1)=1 formula or, in another form,
P^{2}=P+1.
P^{3} is just P.P^{2}
= P(P+1) by our "basic formula", which expands to
P^{2}+P
...now that's interesting!...
1 + P = P^{2} AND
P + P^{2} = P^{3}
...hmmm! Is there anything in this do you think?
Question: How could this be generalized?
We'll use this result in the next subsection about a Phibonacci Brick...
but, for now, let's get back to the original equation...
and P^{2} + P = (P+1) + P after using the "basic formula" again
and this is just 2P + 1. So
Phi^{3} = 1 + 2 Phi
Notice that this needs just one multiplication rather than two if
we evaluated Phi x Phi x Phi. That's the first quick way.
The second answer was spotted by Scott Beach but it is also
in the table at the foot of this page:
since Phi=(√5+1)/2 then 2Phi is √5+1
and 1+ 2 Phi = &radic 5 + 2:
Phi^{3} = 2 + √5
[This variation was suggested by Hud Nordin of Sunnyvale, California.]
If we have a brick with sides of lengths 1, Phi=1·61803...
and phi=1/Phi=0·61803... then:
 the longest side is the sum of the other two lengths
since 1 + phi = Phi
 the largest face (area C=1 x Phi) is the sum of the other
two face's areas (area A =1 x phi and area B=phi x Phi) since
Area A + Area B
= 1 x phi + phi x Phi
= phi + 1
= Phi
= Area C
The next three interesting facts and figures on the Phibonacci brick
were first pointed out by
Donald Seitz in The Mathematics Teacher, 1986, pages 340341 in an article
entitled A Geometric Figure Relating the Golden Ratio to Pi.
 What is the surface area S of the brick?
Above we saw that
the sum of the 2 smaller face's areas equals the largest face's area,
and that this is Phi.
Since there are 2 faces with smallest area, 2 of middlesized area
(which total 2 times the largest face area, that is 2 Phi) and we also have
two other faces of the largest area (Phi) , then:
The surface area of the brick is 4 Phi
 How long is the diagonal across the brick?
Another surprise awaits us when we calculate the length of the diagonal
across the brick.
The formula is a 3dimensional analogue of Pythagoras Theorem.
For a rectangle of sides x and y, its diagonal is
(x^{2} + y^{2}).
For a 3D brick with sides x,y and z, its diagonal has length
(x^{2} + y^{2} + z^{2}).
So how long is the diagonal of our Phibonacci brick? Since its sides (x,y and z)
are 1, Phi and phi, the length of its diagonal is:
(1^{2} + Phi^{2} +
phi^{2}).
I'll leave you to check the algebra but the surprisingly simple answer is
The diagonal of the brick has length 2
 A relationship between Phi and Pi
Even more surprising is that the brick shows a simple relationship between Pi
and Phi, using the values for the diagonal D that we have just found
and the surface area S.
If we imagine the brick tightly packed into a sphere, the centre of the
sphere will be half way along diagonal D.
So the radius of the sphere will be 1 and
its surface area will be 4 Pi.
We showed above that the surface area of the brick, S, is 4 Phi.
Putting these two together we have:
The ratio of the surface areas of
the Phibonacci brick and
its surrounding sphere
is Phi : Pi
Things to do
 Suppose we have a brick with sides a, b and c. Suppose also that the sum of
the smaller sides equals the longest side (i.e. that a+b=c). Also suppose we have the
"Area sum property" too, that the smaller two areas add up to the larger, or that
ab + ac = bc. The questions here show that the only way to
have BOTH properties is when a and b as well as b and c are in the golden ratio (Phi):
Show that, if we let k stand for a/b then we must also have k = b/c .
 Show that k must also satisfy k^{2} = 1 + k
 ... and hence show that k must be Phi.
Golden Triangles, Rectangles and Cuboids by M Bicknell and V E Hoggatt Jr in
Fibonacci Quarterly vol 7 (1969), pages 7391
is a very readable account of some generalizations of these results.
The Golden rectangle and powers of phi
On a golden rectangle with sides of length phi and 1,
dividing at the golden ratio point gives two overlapping
squares whose sides are phi.
The gap between a square and the longer side has
length 1phi which is also phi^{2}.
So the phi by phi squares are themselves divided at their own
golden section ratios by the side of the other overlapping square.
The size of the overlap is phiphi^{2} which is just
phi^{3} (because the equation 1phi=phi^{2}
can be multiplied by phi to give the relation phiphi^{2}
= phi^{3}).
As an exercise, how would you easily construct a line of
length phi^{4} = phi^{2}phi^{3} on the diagram?
Higher powers of phi
Here is another geometrical illustration of phi^{2}, phi^{3} and
phi^{4}:
If we take a square with sides of length 1 and divide two sides at the
golden section point, we form two new squares (yellow and blue)
and two (red) rectangles.
Notice that the diagonal is also
divided at its golden section point too.
The large yellow square has sides of length phi, so its area is
phi^{2}.
The sides of a red rectangle are of length phi
and 1phi=phi^{2} so each has an area
of phi^{3}!
The small blue square has sides of length 1phi=phi^{2}
so its area is phi^{4}!!
So, the squares plus the rectangles area must add up to 1, the area of the
big square and we have yellow + 2 blue + red = 1:
1 = phi^{2} + 2 phi^{3} + phi^{4}
Things to do
 Evaluate Phi and raise it to the power 4 (hint: square it twice) on your calculator.
Save this value in a memory (or write it down).
 Check that the value just found, with (3 Phi+2) subtracted is zero.
 What is P^{5} in the form
a P + b (ie find the integers a
and b) ?
 Write P^{6} in the same form.
 Can you spot a general pattern for P^{n}?
Hint: spot the wellknown series for the values of a and b
in the successive powers of P?

Here is a cube with sides divided at the golden section points to
split it into eight pieces.
The eight pieces are of 4 different shapes:
How many of each shape are there if there are 8 pieces in total?
What is the volume of each?
( Each volume will be a power of phi!)
If the total volume is 1 (since the original cube has sides of length 1),
what relationship does this suggest between your multiples of powers of phi
and 1?

Expand (1+phi)^{3}. Since 1+phi=phi^{1} your answer will
also equal (phi^{1})^{3}=phi^{3}.
How is this answer related to the relationship you found in the last question?
There's an even more intriguing relationship between the powers of Phi than the
one you discovered for yourself in the last section. Here's how you can find it:
Things to do
 What do you notice about the difference between
Phi^{3} and Phi^{2}?
 and how does Phi^{4}Phi^{3}
compare with another power of Phi?
 Can you guess what we mustadd to Phi^{4}
to get Phi^{5}.
 What is the general pattern:
Phi^{n} = Phi^{n1} + ?
 Can you justify this using your answer to the last section?
that is, using your pattern for
Phi^{n} = a Phi + b
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We've just seen that we can use multiples of Phi to calculate its powers easily
and you might think that there's not much else we can discover about multiples of Phi.
Here we show another relationship and also explain how the
Rabbits family Tree diagram was made,
originally seen in the Fibonacci and Nature.
It's all to do with the multiples of Phi, or,
rather, the fractional parts of the multiples.
Things to do
 On a piece of paper, draw a line 10cm long near the top, or use graph paper.
This is going to be used to
plot the fractional parts of the multiples of Phi=1·618.. so it
will represent values from 0 to 1 (0·5 being at 5cm, 1 at 10cm etc).
 Draw the first point at the fractional part of Phi itself=0.618, ie
at 6·18 cm along the line. Mark it 1, since we used 1 times Phi.
 Draw another line underneath the first. On it put a point for the fractional
part of 2Phi=3.236 ie at 2·36cm. Mark the point as 2.
 Draw another line underneath the last. On it put a point for the fractional
part of 3Phi=4.854, ie at 8·54cm. Mark it as 3.
 Draw a new line and this time mark TWO points labelled 4 and
5,
so that this line will have the next
2 multiples of Phi: 4Phi and 5Phi.
 The next line will have the next THREE multiples of Phi:
6Phi, 7Phi and 8Phi.
 The next line will have the next FIVE multiples of Phi.
 The next line will have the next EIGHT multiples of Phi.
 ..you can guess the next one!
 When you're tired of drawing lines and plotting points,
add some thick vertical lines down from each point.
 Where each vertical line begins, draw a thick horizontal line
to the nearest vertical line.
We can interpret the thick lines of the Rabbit Family Tree as follows:
 The 10 cm lines across the page each represent one new generation of rabbits,
or each new month in Fibonacci's original problem.
 Each thick vertical line represents one of our rabbit pairs and..
 .. each pair is "born" when it appears for the first time as a branch from an
thick vertical line that started higher up the page (its "parents").
 Notice how each vertical line goes through
two levels (generations or months) before it starts to branch and ..
 .. then it branches on every line after that.
This shows where our rabbits must be 2 months old to start producing their own young and
then do so every month after that.
 Notice how on each generation level, the new rabbits appear equally and optimally spaced
across the page so that there is no overcrowding at one point in the tree...
 .. and also that each horizontal branching line is the same length on each
generation and ...
 .. all are on the same side of their "parents" line in each generation and...
 ..the side they appear on swops each time, from left side to right side to left.
You can probably spot some more relationships in this Tree (for instance, that the lengths
of the horizontal branches are each 0·618 times shorter than those of the previous
generation).
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A continued fraction is a fraction of the form:
a + 1 = a + 1/( b + 1/( c + 1/( d + ... )))
b + 1
c + 1
d + ...
where a, b, c etc are integers (often only positive).
Phi is the simplest form of continued fraction since all the integers a, b, c, ... are 1.
Writing such fractions in the usual form, as above, takes a lot of space on the page, so an
abbreviation often used is to just give the integers as a list: [a,b,c,...]. In this notation
then Phi=[1,1,1,1...] which we can further abbreviate to [1] where the
numbers shown like this indicate that they are repeated in the same cycle for ever.
In some books, these periodic parts may have dots over
them in the same way that we write infinite decimal fractions such as
 .   .   .   . 
1/3 =0·3333.. =0·  3 
; 1/7 = 0·  1  4285  7  = 0·142857142857142...;
1/12 = 0·0833333... = 0·08  3 
and other authors use an overline
like this so that 1/7=0·142867.
If we have more than one integer in the repeating part, such as [6,1,5,1,5,1,5,...] we write
[6,1,5].
For more see Introduction to Continued Fractions at this site.
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Phi denotes (
5+1)/2 =1·6180339... and
phi denotes (
5
1)/2 =0·6180339... .
The basic formulae are
Phi = 1/phi = phi + 1
Phi^{2} = Phi + 1 and therefore
phi^{2} = 1 – phi.
These can be generalized and give an important and remarkable pair of results:
Phi^{n} = Phi^{n1} + Phi^{n2}
phi^{n} = phi^{n+1} + phi^{n+2}
Relationships between powers of Phi, phi and the Fibonacci numbers which hold for
all integer values of n, (positive, zero or negative):
Main 
= Other 
= X + Y Phi 
= A + B phi 
Phi^{n} 
= phi^{–n} 
= F(n–1) + F(n) Phi 
= F(n+1) + F(n) phi 
phi^{n} 
= Phi^{–n} 
= (–1)^{n} { F(n+1) – F(n) Phi } 
= (–1)^{n} F(n–1) – F(n) phi 
(–phi)^{n} 
= (–Phi)^{(–n)} 
= F(n+1) – F(n) Phi 
= F(n1) – F(n) phi 
Also:

(–phi)^{n} =  L(n) – F(n)5   2 

where L(n) is the n^{th} Lucas number, defined
as L(n) = F(n+1) + F(n1).
With thanks to John McNamara for suggesting the first of these formulae. They are both just a sum and a difference of formula called
Vajda58 and Vajda59 on the
Fibonacci, Phi and Lucas numbers Formulae
page at this site.
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Summarising the above, we can make a table where we relate the powers of Phi to
its multiples  and we note the Fibonacci numbers appearing again!
We can also expand the multiples of Phi using its definition as
(5+1)/2
and find Fibonacci again. Finally, we give the continued fraction representations of these powers with the
repeating period part (if any) shown like this.
In both the 5 forms and the continued fractions forms
we note that a new series of numbers appears 
 the Lucas numbers 2, 1, 3, 4, 7, 11, 18, 29, ... .
Phi power  phi power  A + B Phi  C + D phi  Ksqrt(5)+L  decimal value  continued fraction 
... 
Phi^{9}  phi^{9}  21 + 34 Phi  55 + 34 phi  17√5 + 38  76·0131556174..  [76] 
Phi^{8}  phi^{8}  13 + 21 Phi  34 + 21 phi  (21√5 + 47)/2  46·9787137637..  [46, 1, 45] 
Phi^{7}  phi^{7}  8 + 13 Phi  21 + 13 phi  (13√5 + 29)/2  29·0344418537..  [29] 
Phi^{6}  phi^{6}  5 + 8 Phi  13 + 8 phi  4√5 + 9  17·9442719099..  [17, 1, 16] 
Phi^{5}  phi^{5}  3 + 5 Phi  8 + 5 phi  (5√5 + 11)/2  11·0901699437..  [11] 
Phi^{4}  phi^{4}  2 + 3 Phi  5 + 3 phi  (3√5 + 7)/2  6·8541019662..  [6, 1, 5] 
Phi^{3}  phi^{3}  1 + 2 Phi  3 + 2 phi  √5 + 2  4·2360679774..  [4] 
Phi^{2}  phi^{2}  1 + 1 Phi  2 + 1 phi  (√5 + 3)/2  2·6180339887..  [2, 1] 
Phi^{1}  phi^{1}  0 + 1 Phi  1 + 1 phi  (√5 + 1)/2  1·6180339887..  [1] 
Phi^{0}  phi^{0}  1  1  1  1·0000000000..  [1] 
Phi^{1}  phi^{1}  1 + 1 Phi  0 + 1 phi  (√5  1)/2  0·6180339887..  [0, 1] 
Phi^{2}  phi^{2}  2  1 Phi  1  1 phi  (√5 + 3)/2  0·3819660112..  [0, 2, 1] 
Phi^{3}  phi^{3}  3 + 2 Phi  1 + 2 phi  √5  2  0·2360679774..  [0, 4] 
Phi^{4}  phi^{4}  5  3 Phi  2  3 phi  (3√5 + 7)/2  0·1458980337..  [0, 6, 1, 5] 
Phi^{5}  phi^{5}  8 + 5 Phi  3 + 5 phi  (5√5  11)/2  0·0901699437..  [0, 11] 
Phi^{6}  phi^{6}  13  8 Phi  5  8 phi  4√5 + 9  0·0557280900..  [0, 17, 1, 16] 
Phi^{7}  phi^{7}  21 + 13 Phi  8 + 13 phi  (13√5  29)/2  0·0344418537..  [0, 29] 
Phi^{8}  phi^{8}  34  21 Phi  13  21 phi  (21√5 + 47)/2  0·0212862362..  [0, 46, 1, 45] 
Phi^{9}  phi^{9}  55 + 34 Phi  21 + 34 phi  17√5  38  0·0131556174..  [0, 76]

... 
phi = 1/Phi so the powers of phi are the negative powers of Phi.
Note that [0,a,b,c,...] is 0 + 1/(a + 1/(b + 1/(c + ...) = 1/[a,b,c, ...].
The negative powers of Phi are the positive powers of phi and are obtained by
putting a 0 at the head of the continued fraction lists above: eg
phi = 1/Phi = [0,1]
Things to do
 Phi^{3} looks as if it is just phi^{3}+4. Perhaps this is just correct to a few
decimal places but is it really true and accurate? What other such "coincidences"
can you spot in the Table above? Almost all of them will be accurate and true
relationships, in fact.
 Try to prove some of your "coincidences" using the formulae above the table.
 What about those integers in the continued fraction forms of the powers? It looks
as if the series is
1, 3, 4, 7, 11, ..
If the next two are 18 and 29, can you spot how they are generated?
These are not the Fibonacci numbers but the pattern of generating them is similar.
The numbers are, in fact, those of a series very closely related to the Fibonacci numbers
called the Lucas numbers.
 From the table above, what do you notice about the third and fourth columns for
the following pairs of rows?
Phi and Phi^{1},
Phi^{2} and Phi^{2},
Phi^{3} and Phi^{3},
and so on?
What if we added some of these pairs to get rid of the Phi's and subtracted the
others? What series of integers results?
© 19962006 Dr Ron Knott
updated 11 June 2006