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Here Phi = 1·6180339... = phi–1
and phi = 0·6180339... = Phi – 1 = 1/Phi = Phi–1
Phi power
phi power
A + B Phi
C + D phi
real value
Phi5
phi-5
3 + 5 Phi
8 + 5 phi
11·0901699..
Phi4
phi-4
2 + 3 Phi
5 + 3 phi
6·8541019..
Phi3
phi-3
1 + 2 Phi
3 + 2 phi
4·2360679..
Phi2
phi-2
1 + 1 Phi
2 + 1 phi
2·6180339..
Phi1
phi-1
0 + 1 Phi
1 + 1 phi
1·6180339..
Phi0
phi0
1 + 0 Phi
1 + 0 phi
1·0000000..
Phi-1
phi1
-1 + 1 Phi
0 + 1 phi
0·6180339..
Phi-2
phi2
2 - 1 Phi
1 - 1 phi
0·3819660..
Phi-3
phi3
-3 + 2 Phi
-1 + 2 phi
0·2360679..
Phi-4
phi4
5 - 3 Phi
2 - 3 phi
0·1458980..
Phi-5
phi5
-8 + 5 Phi
-3 + 5 phi
0·0901699..
We can capture these relationships precisely in two formulae:
Phin = Fib(n–1) + Fib(n) Phi
Phin = Fib(n+1) + Fib(n) phi
It is not difficult to prove (by Induction) that these formulae
are indeed correct.
They both apply to negative n as well, if we extend the Fibonacci series backwards:
..., -8, 5, -3, 2, -1, 1, 0, 1, 1,2, 3, 5, 8, ...
where we still have the Fibonacci property:
Fib(n) = Fib(n-1) + Fib(n-2)
but it now holds for all values of n, positive, zero and negative!
Another property of this extended Fibonacci series of numbers is that
In the table of powers of phi above, you will have noticed that the same multiples of Phi
occur, sometimes positive and sometimes negative. For example, 2 phi occurs in
both Phi3 = 3 + 2 phi and Phi-3 = -1 + 2 phi. If we subtract
these two powers, the multiples of phi will disappear and leave us with an integer.
Similarly, 3 phi occurs in both Phi4 = 5 + 3 phi and
Phi-4 = 2 - 3 phi. If we add these two powers, again the multiples of phi
will cancel out and leave an integer.
sums of powers of 8 (octal) and, of course, the usual way using
sums of powers of 10 (decimal)!
All the above are powers of an integer (2, 3, 8 or 10)
but the really unusual thing here
is that we are taking powers of Phi, an irrational number
and adding them to get a pure integer!
A natural question now is:
Are all integers representable as sums of powers of Phi?
The answer is Yes! The number n is just n copies of Phi0 added together!!!
So let's rephrase the question...
What we really meant to ask
was how to do this using only powers of Phi and
not repeating any power more than once in the sum
(which is what we did in the examples above).
You do the maths...
1 = Phi0 and
1 = Phi-1+ Phi-2 and
By expanding Phi-n (= phin) as
Phi-(n+1)+Phi-(n+2)
how many more ways can you find to sum powers of Phi to a total of 1 if no power of Phi can be used
more than once? e.g.
Phi-2 = Phi-3+Phi-4 so
1 = Phi-1+ Phi-2 expands to
1 = Phi-1+ Phi-3 + Phi-4
Try to express each of the following numbers as a sum of different powers of Phi
each power occurring no more than once.
You could check your answers in two ways:
(a) on your calculator to
see if you are approximately right but a more precise method is...
(b)
to use the exact values by translating all the powers of Phi into sums of integers
and multiples of Phi using the formula
Phin = Fib(n+1) + Fib(n) phi
so that you can check that all the multiples cancel out:
5 as the sum of 2 and 3
5 as the sum of 4 and 1
(use your answers to the first question
using different representations of 1)
6
6 again, but find a different answer this time
9 Find THREE different answers!
10
11
12
each of the numbers from 13 to 20
Of your representations of number 6 in the previous question,
which answer has the fewest powers of Phi?
Find a table of answers for all the values from 1 to 20 but all your answers should have
the fewest number of powers in them.
From your answers to the above questions, it may look like many numbers can be expressed in
Base Phi.
Do you think that ALL whole numbers can be?
If you do, how would you try to convince someone of this? If you do not, which integer do you think does NOT have a Base Phi representation? (Are you sure?)
Let's use what we learned on the
Fibonacci Bases Page
to write down our sums-of-distinct-powers-of-Phi representations of a number.
As in decimal notation, the columns represent the powers of the Base, but for us the base is Phi,
not 10. We have negative powers of Phi as well as positive ones, so, just as in decimal fractions, we need
a "point" to separate the positive powers of Phi from the negative ones.
So if 1·25 in decimal means
powers of 10:
...
3
2
1
0
.
-1
-2
...
1
.
2
5
= 1+ 2x10-1 + 5x10-2
then
2 = Phi1 + Phi-2
so 2 in Base Phi is
powers of Phi:
...
3
2
1
0
.
-1
-2
...
1
0
.
0
1
= 1 Phi1 + 1 Phi–2
which we write as 2=10·01Phi to indicate that it is a Base Phi representation.
There are Base Phi Representations for all whole numbers
You might like to convince yourself that, by successively adding 1's,
if necessary applying the Expanding 1's process, then
we can always find a way of representing ANY integer as sum of distinct powers of Phi.
By applying the Reducing 1's process as often as necessary, we can then
always
find a base Phi representation that has the minimum number of 1's and no two of them
will be next to each other.
Using the digits 0 and 1 only, we can express every integer as a sum of some
powers of Phi
You do the maths...
How unusual is this property? Could we express every integer as sum of powers
of 2? (Hint: think about even powers of
2)
What about powers of e=2.71828182...
or =3.1415926535...
or some other irrational value
like Phi that has no integer power equal to an integer?
Reducing the number of 1's in a Base Phi Representation
We haven't used much of the theory about Fibonacci numbers yet (those formulae further up this page).
There are some interesting and relevant facts in the
Formula for powers of Phi that we saw on the
Phi's Fascinating Figures page. One of these was
Phin = Phin-1 + Phin-2
This tells us that, if ever we find two consecutive 1's in a Base Phi representation,
we can replace them by an additional one in the column to the left
For instance,
3 = 2 + 1 = 10·01Phi + 1·0Phi = 11·01Phi
but we can replace the two consecutive 1's by a 1 in the phi2 column:
3 = 100·01Phi
Let's call this the Reducing 1's Process.
But what happens if we have three or more 1s next to each other?
Find the leftmost 11 and start there as there will always be
two consecutive ones that have a zero on their left.
This will replace the two ones by zeros and so any following 11 will not have a zero in front of them.
We can always
start with the leftmost pair of ones and then repeat the Reducing 1's Process
on the new form if necessary until we eliminate all "11" from our representation.
Repeatedly applying the Reducing 1's process means that we can reduce a Base Phi
representation until eventually we have no pairs of consecutive 1's
The base Phi representation of N with the least number of 1s
is called the minimal representation of N
It follows from this definition that
Every minimal base Phi representation has no consecutive 1s
The minimal representation has the least number of 1s
Every minimal representation has the greatest number of 0s ignoring any leading and trailing 0s
Every whole number has a minimal base Phi representation
Expanding the number of 1's in a Base Phi Representation
What if we get more than one of a certain power of Phi?
The solution here is to use the same formula but backwards, that is, replacing
a 1 by 1's in the two columns to the right. So that, whenever we have
...100... we can replace it by ...011...
Let's call this the Expanding 1's Process.
EG 2 = 1+1
= 1·0Phi+1·0Phi Expanding the second 1·0 into 0·11:
= 1·0Phi+0·11Phi Now we can add without getting more than 1 in any column:
= 1·11Phi and we are ready to apply the Reducing 1's process:
=10·01Phi
But every representation will end with a "1", which we can always expand into "011".
2 in base Phi is 10.01 and also 10.0011,
but we can expand the final 1 of 10.0011
to get the new form of 10.001011
and repeat on the final one again to give
10.00101011
and then 10.0010101011
and so on for ever!
All representations can be expanded to get an infinitely long tail of 010101...01011 !
To avoid this, we decide that
We will ignore base Phi representations that end ...011
But there is another case to consider too:
Any representation ending in
...101 is equivalent to ...10011 and now we can convert the initial 100 into 011 to get
...01111.
Note that the original and the final forms here have the same number of 0s.
For instance:
This leads us to a decision to make about the base Phi representation for n which contains the most 1s:
which we want to call the maximal
base Phi representation.
For instance, 27 has the following representations with no
consecutive 0s:-
27 = 111011.110101 is the shortest base Phi representation of 27.
Now we use the two expansions just explained to convert an ending of 101 into 01111:
27 = 111011.11001111 but now another "100" appears which we can convert to reduce the number of 0s:
27 = 111011.10111111 and this now has the least number of 0s in any base Phi representation of 27.
So here we define the maximal representation as follows
The base Phi representation of N that does not end in 011 and
with the greatest number of 1s
is called the maximal base Phi representation of N.
Several things follow from this definition:
Every maximal representation has no two consecutive 0s
Every maximal representation has the greatest number of 1s
Every maximal representation has the least number of 0s
Every whole number has a maximal base Phi representation
Note that if we had use "the least number of 0s" in the definition and not "the greatest number of 1s"
then we would not have got a unique maximal representation:
for instance, 7 has representations 1010.1101 and 10101.01111 both of which contain the least number of 0s (three)
but only the latter has the greatest number of 1s (seven) amongst all the base Phi reps of 7.
Comparing the Minimal amd Maximal Representations
Here is a table of the minimal and maximal base Phi representations of 1 up to 30:
N
Minimal rep
no 11s
fewest 1s
Maximal rep
no 00s
most 1s
1
1
.
1
.
2
10
.
01
1
.
11
3
100
.
01
10
.
1111
4
101
.
01
11
.
1111
5
1000
.
1001
101
.
1111
6
1010
.
0001
111
.
0111
7
10000
.
0001
1010
.
101111
8
10001
.
0001
1011
.
101111
9
10010
.
0101
1101
.
101111
10
10100
.
0101
1110
.
111111
11
10101
.
0101
1111
.
111111
12
100000
.
101001
10101
.
111111
13
100010
.
001001
10111
.
011111
14
100100
.
001001
11010
.
110111
15
100101
.
001001
11011
.
110111
16
101000
.
100001
11101
.
110111
17
101010
.
000001
11111
.
010111
18
1000000
.
000001
101010
.
10101111
19
1000001
.
000001
101011
.
10101111
20
1000010
.
010001
101101
.
10101111
21
1000100
.
010001
101110
.
11101111
22
1000101
.
010001
101111
.
11101111
23
1001000
.
100101
110101
.
11101111
24
1001010
.
000101
110111
.
01101111
25
1010000
.
000101
111010
.
10111111
26
1010001
.
000101
111011
.
10111111
27
1010010
.
010101
111101
.
10111111
28
1010100
.
010101
111110
.
11111111
29
1010101
.
010101
111111
.
11111111
30
10000000
.
10101001
1010101
.
11111111
1
1.
2
10.01
1.11
3
100.01
4
11.1111
7
10000.0001
11
1111.111111
18
1000000.000001
29
111111.11111111
Some patterns are visible in the table above and shown on the left here:
2,1,3,4,7,11,18,29 are formed in the same way as the Fibonacci numbers, by adding the latest two to get the next,
but instead of starting with 0 and 1 as we do for the Fibonacci Numbers, we start with 2 and 1.
They are called the Lucas numbers, denoted L(n), and almost always appear where the Fibonacci numbers do!
Sum the two Fibonacci numbers on either side of each Fibonacci number and you generate each Lucas number:
L(n) = F(n–1) + F(n+1)
The formula for the Lucas numbers involves Phi and phi too:
L(n) = Phin + (–phi)n = Phin + (–Phi)–n = Phin + (–1)n Phi–n
which tells us that L(n) is a sum of just two powers of Phi when n is even. For instance
L(2) = 3 = Phi2 + Phi–2 and so is 100.01 in Base Phi.
L(4) = 7 = Phi4 + Phi–4 and so is 10000.0001 in Base Phi.
But for L(0) = 2, we get Phi0 + Phi-0 so the two powers of Phi are the same and 2 is 2 Phi0.
The significance of the maximal representation is seen in the following table of sums all with a total of eleven:
1+
1.
2+
10.01
3+
100.01
4+
101.01
10
1110.111111
9
1101.101111
8
1011.101111
7
1010.101111
11
1111.111111
11
1111.111111
11
1111.111111
11
1111.111111
Since 11, the fifth Lucas number, L(5), has a maximal representation of 1111.111111 with no zeros then
the 1s of the top number in the sum are the 0s in the bottom number.
Sometimes we need to change the final 1 (of either number) into 011 to make the number of phigits after the (phigital-)point the same as in 5+6=11 :
5+
1000.1001 =
1000.100011+
6
111.0111
111.0111
11
1111.111111
The next number with a maximal representation with no zeros is L(7), 29, and you will notice the same patterns between
1 and 28, 2 and 27, 3 and 26, etc where the "h0les" in the larger numbers are "f1lled" in the smaller
and vice-versa.
By converting "100" to "011" (or vice-versa) in any Base Phi representation, we get another valid base Phi form.
As we saw above when converting to the maximal form (no consecutive 0s) we can always convert the final
"1" to "011" and so continue the expansion of N in Base Phi for ever. So we will only count
finite Base Phi forms which do not end in "011" to prevent this.
Here is a table of equivalent Base Phi forms for the values 1 to 11:
The counts here are 1, 2, 3, 3, 5, 5, 5, 8, 8, 8, 5 all of which should look familiar to you by now!
However the pattern does not continue and is not even solely Fibonacci numbers.
A Phigits Calculator
C A L C U L A T O R
Natural numbers up to
A Phi + B, A and B integers
Phi +
up to
Phi +
Phi
R E S U L T S
Key: Lucas Minima, Fibonacci, Fibonacci Pair.
The Calculator reveals some interesting relationships in this plot of the numbers of base Phi representations for
numbers 1 to 2000:
The Lucas numbers (3,4,7,11,18,29, ...) have the least number of representations:
Lucas
3
4
7
11
18
29
47
76
...
#Reps
3
3
5
5
7
7
9
9
...
The Fibonacci numbers 1,2,3,5,8,13,21, ... all have the amazing property that
the number of base Phi representations is the number itself!
Fibonacci
1
2
3
5
8
13
21
34
55
89
...
#Reps
1
2
3
5
8
13
21
34
55
89
...
They also make a pair with another number a little way ahead that also has the same number of
representations. How far ahead? The next with F(n) representations after F(n) is after a gap of F(n−3),
at F( n + F(n−3) ):
13 and 13+3=16 (13 forms), 21 and 21+5=26 (21 forms), 34 and 34+8=42, and so on.
Fib+gap
5+1=6
8+2=10
13+3=16
21+5=26
34+8=42
...
#Reps
5
8
13
21
34
...
The plot of the number of representations has other nice fractal properties, so that later sections
exhibit the same patterns but with more detail.
The plots on the right are arranged so that you can see this pattern when one plot
merges into another and the pattern of red, green and magenta lines is maintained as is the symmetry
of the points between the Lucas minima.
Representations of an integer n as a sum of different powers of Phi are aclled the
Base Phi representation of n on this page.
Other names that have been suggested are
Phigital: compare with digital for Base Ten;
Phinary: compare with Binary
since we are also using just the digits 0 and 1 but to base Phi
[with thanks to Marijke van Gans for this term];
expressing a number in Phigits
[With thanks to Prof Jose Glez-Regueral of Madrid for mentioning this one.]
Links and References
Some of the above
originally appeared A Number System with Irrational Base, George
Bergman, Mathematics Magazine 1957, Vol 31, pages 98-110, where he
also gives pencil-and-paper
methods of doing arithmetic in Base Phi.
C. Rousseau The Phi Number System Revisited
in Mathematics Magazine 1995, Vol 68, pages 283-284.
icon means there is a
You do the maths... section of questions to start your own investigations.
The 
calculator icon
indicates that there is a live interactive calculator in that section.
2? (Hint: think about even powers of
=3.1415926535...
or some other irrational value
like Phi that has no integer power equal to an integer?
The Mathematical Magic of the Fibonacci Numbers
Fibonacci Home Page
The Golden String
© 1996-2015 Dr Ron Knott
