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. The icon means there is an
interactive calculator in this 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.
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?
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.
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
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".
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:
4 = 101.01 = 100.1111
10 = 1111.0101 = 1111.001111 = 1110.111111
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) = Phi^{n} + (–phi)^{n} = Phi^{n} + (–Phi)^{–n} = Phi^{n} + (–1)^{n} Phi^{–n}
For instance
L(2) = 3 = Phi^{2} + Phi^{–2} and so is 100.01 in Base Phi.
L(4) = 7 = Phi^{4} + Phi^{–4} and so is 10000.0001 in Base Phi.
But for L(0) = 2, we get Phi^{0} + Phi^{0} so the two powers of Phi are the same and 2 is 2 Phi^{0}.
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 viceversa.
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All forms of Base Phi representation
By converting "100" to "011" (or viceversa) 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:
1  2  3  4  5  6  7  8  9  10  11 
1
 10.01 1.11
 100.01 11.01 10.1111
 101.01 100.1111 11.1111
 1000.1001 1000.0111 110.1001 110.0111 101.1111
 1010.0001 1001.0111 1001.1001 111.1001 111.0111
 10000.0001 1100.0001 1011.0001 1010.1101 1010.101111
 10001.0001 10000.1101 10000.101111 1101.0001 1100.1101 1100.101111 1011.1101 1011.101111
 10010.0101 10010.001111 10001.1101 10001.101111 1110.0101 1110.001111 1101.1101 1101.101111
 10100.0101 10100.001111 10011.0101 10011.001111 10010.111111 1111.0101 1111.001111 1110.111111
 10101.0101 10101.001111 10100.111111 10011.111111 1111.111111

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
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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 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.
© 19962015 Dr Ron Knott
updated 22 June 2015