The Golden mean (φ) can be used as a number base. It is known as the

**golden mean base**, or colloquially,

**phinary**(since the symbolism for the golden mean is the Greek letter "phi"). Any real number has a standard representation as a base-φ numeral in which only the digits 0 and 1 are used, and the digit sequence "11" is avoided. A nonstandard base-φ numeral with this digit sequence (or with other digits) can always be rewritten in standard form, relying on algebraic properties of the number φ -- most notably that φ+1 = φ

^{2}. For instance 11

_{φ}= 100

_{φ}. Despite using an irrational base, it is a remarkable fact that all integers have a unique representation as a terminating (or finite) base-φ expansion. Other numbers have standard representations base-φ, with rational numbers having recurring representations. These representations are unique, except that numbers with a terminating expansion also have a non-terminating expansion. (As they do in base 10: 2.2=2.199999...)

Examples:

Decimal | Powers of φ | Base φ |

1 | φ^{0} | 1 |

2 | φ^{1}+φ^{-2} | 10.01 |

3 | φ^{2}+φ^{-2} | 100.01 |

4 | φ^{2}+φ^{0}+φ^{-2} | 101.01 |

5 | φ^{3}+φ^{-1}+φ^{-4} | 1000.1001 |

6 | φ^{3}+φ^{1}+φ^{-4} | 1010.0001 |

7 | φ^{4}+φ^{-4} | 10000.0001 |

8 | φ^{4}+φ^{0}+φ^{-4} | 10001.0001 |

9 | φ^{4}+φ^{1}+φ^{-2}+φ^{-4}
| 10010.0101 |

## Writing a φ-base number in standard form

211.0Any 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.1_{φ}300.01_{φ}011_{φ}→ 100_{φ}1101.01_{φ}0200_{φ}→ 1001_{φ}10001.01_{φ}011_{φ}→ 100_{φ}(again) 10001.101_{φ}010_{φ}→101_{φ}10000.011_{φ}010_{φ}→101_{φ}(again) 10000.1_{φ}011_{φ}→ 100_{φ}(again)

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.

## Representing integers as Golden mean base numbers

Note that 1×1 = 1, φ × φ = 1 + φ and 1/φ = -1 + φ. Therefore, we can compute

and

- (a + bφ) × (c + dφ) = ((a × c + b × d) + (a × d + b × c + b × d)φ).

^{-1}= 1/φ.)

(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).

- To convert an integer x to a φ-base number, note that x = (x + 0φ).
- Subtract the highest power of φ, which is still smaller than the number we have, to get out new number, and record a "1" in the appropriate place in the resulting φ-base number.
- Unless our number is 0, go to step 2.
- Finished.

_{φ}= 100

_{φ}, so getting a "11" would mean we missed a "1" prior to the sequence "11".

e.g Start with integer=5, with the result so far being ...00000.00000..._{φ}

Highest power of φ ≤ 5 is φ^{3} = 1 + 2φ ≈ 4.236067977

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 **1000.1001**_{φ}.

### Non-uniqueness

- Conversion to nonstandard form: 1 = 0.11
_{φ}= 0.1011_{φ}= 0.101011_{φ}= ... = 0.10101010...._{φ} - Geometric series: 1.0101010...
_{φ}is equal to

- Difference between "shifts": φ
^{2}x - x = 10.101010..._{φ}- 0.101010..._{φ}= 10_{φ}= φ so that x = φ/(φ^{2}-1) = 1

## Representing Rationals as Golden mean base numbers

Every rational number can be represented as a recurring base φ expansion, as can any element of the field **Q**[√5] = **Q** + √5**Q**, 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):

- 1/2 = 0.010 010 010 ...
_{φ} - 1/3 = 0.00101000 00101000 00101000...
_{φ} - √5 = 10.100000
_{φ} - 2+(1/13)√5 = 10.010 1000100010101000100010000000 1000100010101000100010000000 1000100010101000100010000000 ...
_{φ}

_{φ}= 100

_{φ}/1001

_{φ}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 ------- etcThe 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.For example

- 2+3 = 10.01 + 100.01 = 110.02 = 110.1001 = 1000.1001
- 2×3 = 10.01 + 100.01 = 1000.1 + 1.0001 = 1001.1001 = 1010.0001
- 7-2 = 10000.0001 - 10.01 = 100
__1__0.0__1__01 = 11__1__0.0__1__01 = 1001.0__1__01 = 1000.1001

### 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- 2+3 = 10.01 + 100.01 = 10.01 + 100.0011 = 110.0111 = 1000.1001
- 7-2 = 10000.0001 - 10.01 = 1100.0001 - 10.01 = 1011.0001 - 10.01 = 1010.1101 - 10.01 = 1000.1001

## 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 F_{k+1}= F

_{k}+ F

_{k-1}. For example

- 30 = 1×21 + 0×13 + 1×8 + 0×5 + 0×3 + 0×2 + 1×1 + 0×1 = 10100010
_{fib}.

- 30 = 1×21 + 0×13 + 1×8 + 0×5 + 0×3 + 0×2 + 1×1 + 0×1 = 10100010