We can either consider our integer to be the (only) digit of a nonstandard base-φ numeral, and standardize it, or do the following: 1 × 1 = 1, φ × φ = 1 + φ and = −1 + φ. Therefore, we can compute : (
a +
bφ) + (
c +
dφ) = ((
a +
c) + (
b +
d)φ), : (
a +
bφ) − (
c +
dφ) = ((
a −
c) + (
b −
d)φ) and : (
a +
bφ) × (
c +
dφ) = ((
ac +
bd) + (
ad +
bc +
bd)φ). 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. (
a +
bφ) > (
c +
dφ) if and only if 2(
a −
c) − (
d −
b) > (
d −
b) × . 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 \sqrt{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 base-φ number, note that
x = (
x + 0φ). • Subtract the highest power of φ, which is still smaller than the number we have, to get our new number, and record a "1" in the appropriate place in the resulting base-φ number. • Unless our number is 0, go to step 2. • Finished. 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". Start, e.g., with integer = 5, with the result so far being ...00000.00000...φ Highest power of φ ≤ 5 is φ3 = 1 + 2φ ≈ 4.236067977 Subtracting this from 5, we have 5 − (1 + 2φ) = 4 − 2φ ≈ 0.763932023..., 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..., 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.
Non-uniqueness 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
0.999... = 1. In base-φ, the numeral 0.1010101... can be seen to be equal to 1 in several ways: • Conversion to nonstandard form: 1 = 0.11φ = 0.1011φ = 0.101011φ = ... = 0.10101010...φ •
Geometric series: 1.0101010...φ is equal to :\sum_{k=0}^\infty \varphi^{-2k}=\frac{1}{1-\varphi^{-2}} = \varphi • Difference between "shifts": φ2
x −
x = 10.101010...φ − 0.101010...φ = 10φ = φ so that
x = = 1 This non-uniqueness is a feature of the numeration system, since both 1.0000 and 0.101010... are in standard form. In general, the final 1 of any number in base-φ can be replaced with a recurring 01 without changing the value of that number. ==Representing rational numbers as golden ratio base numbers==