Examples:
Decimal | Powers of φ | Base φ |
1 | φ0 | 1 |
2 | φ1+φ-2 | 10.01 |
3 | φ2+φ-2 | 100.01 |
4 | φ2+φ0+φ-2 | 101.01 |
5 | φ3+φ-1+φ-4 | 1000.1001 |
6 | φ3+φ1+φ-4 | 1010.0001 |
7 | φ4+φ-4 | 10000.0001 |
8 | φ4+φ0+φ-4 | 10001.0001 |
9 | φ4+φ1+φ-2+φ-4 | 10010.0101 |
211.01φ is not a standard φ-base numeral, since it contains a "11" and a "2", which isn't a "0" or "1", and contains a 1=-1, which isn't a "0" or "1" either.
To "standardize" a numeral, we can use the following substitutions: 011φ = 100φ, 0200φ = 1001φ and 010φ = 101φ. We can apply the substitutions in any order we like, as the result is the same. Below, the substitutions used are on the right, the resulting number on the left.
Writing a φ-base number in standard form
211.01φ
300.01φ 011φ → 100φ
1101.01φ 0200φ → 1001φ
10001.01φ 011φ → 100φ (again)
10001.101φ 010φ → 101φ
10000.011φ 010φ → 101φ (again)
10000.1φ 011φ → 100φ (again)
Any positive number with a non-standard terminating base-φ representation can be uniquely standardized in this manner. If we get to a point, where all digits are "0" or "1", except for the first digit being negative, then the number is negative. This can be converted to the negative of a base-φ representation by negating every digit, standardizing the result, and then mark it as negative. For example, use a minus sign, or some other significance to denote negative numbers. If the arithmetic is being performed on a computer, an error message may be returned.
Note that when adding the digits "9" and "1", the result is a single digit "(10)", "A" or similar, as we are not working in decimal.
We can either consider our integer to be the (only) digit of a nonstandard φ-base numeral, and standardize it, or do the following:
Note that 1×1 = 1, φ × φ = 1 + φ and 1/φ = -1 + φ. Therefore, we can compute
(a + bφ) > (c + dφ) if and only if 2(a - c) - (d - b) > (d - b) × √5. If one side is negative, the other positive, the comparison is trivial. Otherwise, square both sides, to get an integer comparison, reversing the comparison direction if both sides were negative. On squaring both sides, the √5 is replaced with the integer 5.
So, using integer values only, we can also compare numbers of the form (a + bφ).
e.g Start with integer=5, with the result so far being ...00000.00000...φ
Subtracting this from 5, we have 5 - (1 + 2φ) = 4 - 2φ ≈ 0.763932023..., with the result so far being 1000.00000...φ
Highest power of φ ≤ 4 - 2φ ≈ 0.763932023... is φ-1 = -1 + 1φ ≈ 0.618033989...
Subtracting this from 4 - 2φ ≈ 0.763932023..., we have 4 - 2φ - (-1 + 1φ) = 5 - 3φ ≈ 0.145898034..., with the result so far being 1000.10000...φ
Highest power of φ ≤ 5 - 3φ ≈ 0.145898034... is φ-4 = 5 - 3φ ≈ 0.145898034...
Subtracting this from 5 - 3φ ≈ 0.145898034..., we have 5 - 3φ - (5 - 3φ) = 0 + 0φ = 0, with the final result being 1000.1001φ.
Just as with any base-n system, numbers with a terminating representation have an alternative recurring representation. In base 10, this relies on the observation that if x = 0.9999... then 10x = 9.99999... so that 9x = 9 and x=1. In base φ, the numeral 0.1010101... can be see to be equal to 1 in several ways:
Every rational number can be represented as a recurring base φ expansion, as can any element of the field Q[√5] = Q + √5Q, the field generated by the rational numbers and √5. Conversely any recurring (or terminating) base φ expansion is an element of Q[√5]. Some examples (with spaces added for emphasis):
For example
Representing integers as Golden mean base numbers
and
So, using integer values only, we can add, subtract and multiply numbers of the form (a + bφ), and even represent positive and negative integer powers of φ. (Note that φ-1 = 1/φ.)
The above procedure will never result in the sequence "11", since 11φ = 100φ, so getting a "11" would mean we missed a "1" prior to the sequence "11".
Highest power of φ ≤ 5 is φ3 = 1 + 2φ ≈ 4.236067977Non-uniqueness
This non-uniqueness is a feature of the numeration system, since both 1.0000 and 0.101010... are in standard form.Representing Rationals as Golden mean base numbers
The justification that a rational gives a recurring expansion is analogous to the equivalent proof for a base-n numeration system (n=2,3,4,...). Essentially in base-φ long division, there are only a finite number of possible remainders, and so once there must be a recurring pattern. For example with 1/2 = 1/10.01φ = 100φ/1001φ long division looks like this (note that subtraction is a bit freaky) .0 1 0 0 1
------------------------
1 0 0 1 ) 1 0 0.0 0 0 0 0 0 0 0
1 0 0 1 trade: 10000 = 1100 = 1011
------- so 10000-1001 = 1011-1001 = 10
1 0 0 0 0
1 0 0 1
-------
etc
The converse is also true, in that a number with a recurring base-φ representation is an element of the field Q[√5]. This follows from the observation that a recurring representation with period k involves a geometric series with ratio φ-k, which will sum to an element of Q[√5].Addition, subtraction, multiplication
It is possible to adapt all the standard algorithms of base-10 arithmetic to base-φ arithmetic. There are two approaches to this:
Calculate then convert to standard form
To add two φ-base numbers, add each pair of digits, without carry, and then convert the numeral to standard form. To subtract, subtract each pair of digits without borrow (borrow is a negative amount of carry), and then convert the numeral to standard form. To multiply, multiply how you normally multiply, without carry, and then convert the numeral to standard form.Avoiding digits other than 0 and 1
A more "native" approach is to avoid having to add digits 1+1 or to subtract 0-1. This is done by reorganising the operands into nonstandard form so that these combinations do not occur. For example
The subtraction seen here uses a modified form of the standard "trading" algorithm for subtraction.Division
No fractions (a/b, where a and b are integers, a not divisible by b) can be represented as a finite φ-base number, in other words, all finitely representable φ-base numbers are either integers or (more likely) an irrational in the field Q[√5]. Due to long division having only a finite number of possible remainders, a division of two integers (or other numbers with finite base-φ representation) will have a recurring expansion, as demonstrated above. A Close relation: Fibonacci Representation
A closely related numeration system is Fibonacci representation used for integers. In this system, only digits 0 and 1 are used and the place values of the digits are the Fibonacci numbers. As with base-φ, the digit sequence "11" is avoided by rearranging to a standard form, using the Fibonacci recurrence relation Fk+1 = Fk + Fk-1. For example
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