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 Things To Do icon means there is a Things to do investigation at the end of the section.

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Powers of Phi

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 phireal
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

Fib(–n) = Fib(n) for even n
Fib(–n) = Fib(n) for odd n

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Integers as sums of powers of Phi

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.

Here are some more examples:

Phi1 + Phi–2 = (1 + 1 phi) + (1 – 1 phi) = 2
Phi2 + Phi–2 = (2 + 1 phi) + (1 – 1 phi) = 3
Phi4 + 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 Phi0 is just 1, we can add 1 (=Phi0) 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 = (Phi2 + Phi–2) + Phi0
8 = 7 + 1 = (Phi4 + Phi–4) + Phi0
9 = 2 + 7 = (Phi1 + Phi–2) + (Phi4 + Phi–4)
10 = 3 + 7 = (Phi2 + Phi–2) + (Phi4 + Phi–4)
This reminds us of expressing numbers as :

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).

/ Things to do /

  1. 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 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
  2. 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:
    1. 5 as the sum of 2 and 3
    2. 5 as the sum of 4 and 1
         (use your answers to the first question using different representations of 1)
    3. 6
    4. 6 again, but find a different answer this time
    5. 9 Find THREE different answers!
    6. 10
    7. 11
    8. 12
    9. each of the numbers from 13 to 20
  3. Of your representations of number 6 in the previous question, which answer has the fewest powers of Phi?
  4. 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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Base Phi Representations

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
= Phi1 + Phi–2
which we write as 2=10·01Phi to indicate that it is a Base Phi representation.
Base Phi with no consecutive 1s:
Minimal representation (least number of 1s)
11
210.01
3100.01
4101.01
51000.1001
61010.0001
710000.0001
810001.0001
910010.0101
1010100.0101
1110101.0101
12100000.101001
13100010.001001
14100100.001001
15100101.001001
16101000.100001
17101010.000001
181000000.000001
191000001.000001
201000010.010001
211000100.010001
221000101.010001
231001000.100101
241001010.000101
251010000.000101
261010001.000101
271010010.010101
281010100.010101
291010101.010101
3010000000.10101001
  Base Phi with no consecutive 0s:
Maximal representation (most number of 1s)
11.
21.11
311.01
411.1111
5101.1111
6111.0111
71010.1101
81011.1101
91101.1101
101111.0101
111111.111111
1210101.111111
1310111.011111
1411010.110111
1511011.110111
1611101.110111
1711111.010111
18101010.101101
19101011.101101
20101101.101101
21101110.111101
22101111.111101
23110101.111101
24110111.011101
25111010.110101
26111011.110101
27111101.110101
28111110.11111111
29111111.11111111
301010101.11111111

A Phigits Calculator

C A L C U L A T O R
up to


R E S U L T S

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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.

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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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" 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
  1. Write 3 as 2+1 and reduce it to its minimal form (no two consecutive 1's).
  2. Try it for 4 = 3+1.
  3. Look through your answers to the earlier questions and re-write 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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Minimal Base Phi Representations

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 /

  1. How unusual is this property? Could we express every integer as sum of powers of sqrt2? (Hint: think about even powers of sqrt2)
  2. What about powers of e=2.71828182... or  pi =3.1415926535... or some other irrational value like Phi that has no integer power equal to an integer?

Other names for Base Phi

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

Links and References

Article: Some of the above originally appeared in an article by George Bergman, in the Mathematics Magazine 1957, Vol 31, pages 98-110, where he also gives pencil-and-paper methods of doing arithmetic in Base Phi.

Article: C. Rousseau The Phi Number System Revisited in Mathematics Magazine 1995, Vol 68, pages 283-284.

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.

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updated 22 December 2007