Using Powers of Phi to represent Integers (Base Phi)
If you have already looked at the page where we showed
how to represent integers using the Fibonacci numbers,
and you have also read about the
numerical properties of powers of Phi
then this page takes you a stage further  writing the integers in base Phi!
Contents of this Page
The icon means there is a
Things to do investigation at the end of the section.
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Here is part of the
table of numerical properties of powers of Phi:
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 
Phi^{5}  phi^{5}  3 + 5 Phi  8 + 5 phi  11·0901699.. 
Phi^{4}  phi^{4}  2 + 3 Phi  5 + 3 phi  6·8541019.. 
Phi^{3}  phi^{3}  1 + 2 Phi  3 + 2 phi  4·2360679.. 
Phi^{2}  phi^{2}  1 + 1 Phi  2 + 1 phi  2·6180339.. 
Phi^{1}  phi^{1}  0 + 1 Phi  1 + 1 phi  1·6180339.. 
Phi^{0}  phi^{0}  1 + 0 Phi  1 + 0 phi  1·0000000.. 
Phi^{1}  phi^{1}  1 + 1 Phi  0 + 1 phi  0·6180339.. 
Phi^{2}  phi^{2}  2  1 Phi  1  1 phi  0·3819660.. 
Phi^{3}  phi^{3}  3 + 2 Phi  1 + 2 phi  0·2360679.. 
Phi^{4}  phi^{4}  5  3 Phi  2  3 phi  0·1458980.. 
Phi^{5}  phi^{5}  8 + 5 Phi  3 + 5 phi  0·0901699.. 
We can capture these relationships precisely in two formulae:
Phi^{n} = Fib(n–1) + Fib(n) Phi
Phi^{n} = 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(n1) + Fib(n2)
but it now holds for all values of n, positive, zero and negative!
Another property of this extended Fibonacci series of numbers is that
Fib(–n)  =  –Fib(n)  for even n 
Fib(–n)  =  Fib(n)  for odd n 
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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 Phi^{3} = 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 Phi^{4} = 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.
Here are some more examples:
Phi^{1} + Phi^{–2} = (1 + 1 phi) + (1 – 1 phi) = 2
Phi^{2} + Phi^{–2} = (2 + 1 phi) + (1 – 1 phi) = 3
Phi^{4} + Phi^{–4} = (5 + 3 phi) + (2 – 3 phi) = 7
So we have expressed the integers 2, 3 and 7 as a sum of powers of Phi.
Because Phi^{0} is just 1, we can add 1 (=Phi^{0}) to those numbers
above and so represent 3 (again), 4 and 8 as a sum of powers of Phi.
We can also add combinations of these numbers and get other ones too. In all of them, we
are writing the integer as a sum of different powers of Phi.
4 = 3 + 1 = (Phi^{2} + Phi^{–2}) + Phi^{0}
8 = 7 + 1 = (Phi^{4} + Phi^{–4}) + Phi^{0}
9 = 2 + 7 = (Phi^{1} + Phi^{–2}) + (Phi^{4} + Phi^{–4})
10 = 3 + 7 = (Phi^{2} + Phi^{–2}) + (Phi^{4} + Phi^{–4})
This reminds us of expressing numbers as :
 sums of powers of 2 (binary), or
 sums of powers of 3 (ternary), or
 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 Phi^{0} 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).
Things to do
 1 = Phi^{0} and
1 = Phi^{1}+ Phi^{2} and
By expanding Phi^{n} (= phi^{n}) as
Phi^{(n+1)}+Phi^{(n+2)}
how many more ways can you find to sum powers pf 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
Phi^{n} = 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?)
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Let's use what we learned on the
Fibonacci Bases Page
to write down our sumsofdistinctpowersofPhi 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 = Phi^{1} + Phi^{2}
so 2 in Base Phi is
powers of Phi:  ...  3  2  1  0  .  1  2  ... 
    1  0  .  0  1  
 = Phi^{1} + Phi^{–2} 
which we write as 2=10·01_{Phi} to indicate that it is a Base Phi representation.
Base Phi with no consecutive 1s:
Minimal representation (least number of 1s)
1  1 
2  10  .  01 
3  100  .  01 
4  101  .  01 
5  1000  .  1001 
6  1010  .  0001 
7  10000  .  0001 
8  10001  .  0001 
9  10010  .  0101 
10  10100  .  0101 
11  10101  .  0101 
12  100000  .  101001 
13  100010  .  001001 
14  100100  .  001001 
15  100101  .  001001 
16  101000  .  100001 
17  101010  .  000001 
18  1000000  .  000001 
19  1000001  .  000001 
20  1000010  .  010001 
21  1000100  .  010001 
22  1000101  .  010001 
23  1001000  .  100101 
24  1001010  .  000101 
25  1010000  .  000101 
26  1010001  .  000101 
27  1010010  .  010101 
28  1010100  .  010101 
29  1010101  .  010101 
30  10000000  .  10101001 
 
Base Phi with no consecutive 0s:
Maximal representation (most number of 1s)
1  1  .  
2  1  .  11 
3  11  .  01 
4  11  .  1111 
5  101  .  1111 
6  111  .  0111 
7  1010  .  1101 
8  1011  .  1101 
9  1101  .  1101 
10  1111  .  0101 
11  1111  .  111111 
12  10101  .  111111 
13  10111  .  011111 
14  11010  .  110111 
15  11011  .  110111 
16  11101  .  110111 
17  11111  .  010111 
18  101010  .  101101 
19  101011  .  101101 
20  101101  .  101101 
21  101110  .  111101 
22  101111  .  111101 
23  110101  .  111101 
24  110111  .  011101 
25  111010  .  110101 
26  111011  .  110101 
27  111101  .  110101 
28  111110  .  11111111 
29  111111  .  11111111 
30  1010101  .  11111111 

A Phigits Calculator
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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
Phi^{n} = Phi^{n1} + Phi^{n2}
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 = 10·01_{Phi} + 1·0_{Phi} = 11·01_{Phi}
but we can replace the two consecutive 1's by a 1 in the phi^{2} column:
3 = 100·01_{Phi}
Let's call this the Reducing 1's Process.
What happens if we have three 1s next to each other?
There will always be two consecutive ones that have a zero on their left,
so start with those. This will replace the two ones by zeros. We can always
start with the leftmost pair of ones and then repeat the Reducing 1's Process
on the new form if necessary.
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
This is called minimal representation since it has the least number of 1s.
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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  = 1·0_{Phi}+1·0_{Phi} Expanding the second 1·0 into 0·11: 
 = 1·0_{Phi}+0·11_{Phi} Now we can add without getting more than 1 in any column: 
 = 1·11_{Phi} and we are ready to apply the Reducing 1's process: 
 =10·01_{Phi} 
But every representation will end with a "1", which we can always expand into "011" so, if we are not careful,
all representations have an infinitely long tail!
To avoid this, we make sure no represenation in Base Phi ends with "011".
Similarly it is not difficult to show that any representation ending in
...101 is equivalent to changing the ending to
...01111 and each has the same number of 0s. For instance:
4 = 101.01 = 100.1111
10 = 1111.0101 = 1111.001111 = 1110.111111
This leads us to a decision to make about which we want to call the maximal
base Phi representation since, for instance, 27 has the following representations with no
consecutive 0s:
27 = 111011.110101 is the shortest
27 = 111011.10111111 has the least number of 0s
So here we decide that the maximalrepresentation be the shortest Base Phi representation with no
consecutive 0s (apart from an infinite number leading and trailing 0s of course).
Things to do
Check your answers with the Table shown earlier on this page
 Write 3 as 2+1 and reduce it to its minimal form (no two consecutive 1's).
 Try it for 4 = 3+1.
 Look through your answers to the earlier questions and rewrite your Table of
Base Phi representations so that all the numbers from 1 to 20 have no two consecutive 1's.
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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
Things to do
 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?
Let us call our representations of an integer n as a sum of different powers of Phi the
Base Phi representation of n.
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 GlezRegueral of Madrid for mentioning this one.]
Links and References
Some of the above
originally appeared in an article by George
Bergman, in the Mathematics Magazine 1957, Vol 31, pages 98110, where he
also gives pencilandpaper
methods of doing arithmetic in Base Phi.
C. Rousseau The Phi Number System Revisited
in Mathematics Magazine 1995, Vol 68, pages 283284.
Prof Alexey Stakhov investigates the
applications of Fibonacci and Phi number systems for representing numbers
in a computer rather than the familiar binary system. His
web site has lots more information.
© 19962007 Dr Ron Knott
updated 22 December 2007