# 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!

The 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... and phi = 0·6180339... = Phi-1 = 1/Phi
Phi
power
A + B Phireal
value
Phi5 8 + 5 phi 11·090169..
Phi4 5 + 3 phi 6·8541019..
Phi3 3 + 2 phi 4·2360679..
Phi2 2 + 1 phi 2·6180339..
Phi1 1 + 1 phi 1·6180339..
Phi0 1 + 0 phi 1·0000000..
Phi-1 0 + 1 phi 0·6180339..
Phi-2 1 - 1 phi 0·3819660..
Phi-3 -1 + 2 phi 0·2360679..
Phi-4 2 - 3 phi 0·1458980..
Phi-5 -3 + 5 phi 0·0901699..
We can capture these relationships precisely in a formula:
Phin = Fib(n+1) + Fib(n) phi
It is not difficult to prove (by Induction) that this formula is indeed true. This formula applies 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 and = 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 :
• 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 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.
(a) on your calculator to see if you are approximately right but a better way (that is, more precise) 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 -3 ... 1 0 . 0 1
which we write as 2=10·01Phi to indicate that it is a Base Phi representation.
Base Phi with no consecutive 1s:
 1 2 3 1 10 . 01 100 . 01 101 . 01 1000 . 1001 1010 . 0001 10000 . 0001 10001 . 0001 10010 . 0101 10100 . 0101 10101 . 0101 100000 . 101001 100010 . 001001 100100 . 001001 100101 . 001001 101000 . 100001 101010 . 000001 1000000 . 000001 1000001 . 000001 1000010 . 010001 1000100 . 010001 1000101 . 010001 1001000 . 100101 1001010 . 000101 1010000 . 000101 1010001 . 000101 1010010 . 010101 1010100 . 010101 1010101 . 010101 10000000 . 10101001
Base Phi with no consecutive 0s:
 1 2 3 1 . 1 . 11 11 . 01 11 . 1111 101 . 1111 111 . 0111 1010 . 1101 1011 . 1101 1101 . 1101 1111 . 0101 1111 . 111111 10101 . 111111 10111 . 011111 11010 . 110111 11011 . 110111 11101 . 110111 11111 . 010111 101010 . 10101111 101011 . 10101111 101101 . 10101111 101110 . 11101111 101111 . 11101111 110101 . 11101111 110111 . 01101111 111010 . 10111111 111011 . 10111111 111101 . 10111111 111110 . 11111111 111111 . 11111111 1010101 . 11111111

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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

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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

#### Things to do

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 2? (The answer is easy if you think about even powers of 2)
2. What about powers of e or 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
• 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.]