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 |

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.

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).

(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.

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}.

- 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

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}

.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

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

- 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

- 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