Editing Golden ratio base (section)

==Representing integers as golden ratio base numbers==

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 {{sfrac|1|φ}} = −1 + φ. Therefore, we can compute

: (''a'' + ''b''φ) + (''c'' + ''d''φ) = ((''a'' + ''c'') + (''b'' + ''d'')φ),
: (''a'' + ''b''φ) − (''c'' + ''d''φ) = ((''a'' − ''c'') + (''b'' − ''d'')φ)

and

: (''a'' + ''b''φ) × (''c'' + ''d''φ) = ((''ac'' + ''bd'') + (''ad'' + ''bc'' + ''bd'')φ).

So, using integer values only, we can add, subtract and multiply numbers of the form (''a'' + ''b''φ), and even represent positive and negative integer [[Exponentiation|powers]] of φ.

(''a'' + ''b''φ) > (''c'' + ''d''φ) if and only if 2(''a'' − ''c'') − (''d'' − ''b'') > (''d'' − ''b'') × {{sqrt|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 [[square (algebra)|squaring]] both sides, the <math display=inline>\sqrt{5}</math> is replaced with the integer 5.

So, using integer values only, we can also compare numbers of the form (''a'' + ''b''φ).

# 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 11<sub>φ</sub> = 100<sub>φ</sub>, 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...<sub>φ</sub>

Highest power of φ ≤ 5 is φ<sup>3</sup> = 1 + 2φ ≈ 4.236067977

Subtracting this from 5, we have 5 − (1 + 2φ) = 4 − 2φ ≈ 0.763932023..., the result so far being 1000.00000...<sub>φ</sub>

Highest power of φ ≤ 4 − 2φ ≈ 0.763932023... is φ<sup>−1</sup> = −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...<sub>φ</sub>

Highest power of φ ≤ 5 − 3φ ≈ 0.145898034... is φ<sup>−4</sup> = 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'''<sub>φ</sub>.

===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...|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  = 0.11<sub>φ</sub> = 0.1011<sub>φ</sub> = 0.101011<sub>φ</sub> = ... = 0.10101010...<sub>φ</sub>
*[[Geometric series]]: 1.0101010...<sub>φ</sub> is equal to
:<math>\sum_{k=0}^\infty \varphi^{-2k}=\frac{1}{1-\varphi^{-2}} = \varphi</math>
*Difference between "shifts": φ<sup>2</sup> ''x'' − ''x'' = 10.101010...<sub>φ</sub> − 0.101010...<sub>φ</sub> = 10<sub>φ</sub> = φ so that ''x'' = {{sfrac|φ|φ<sup>2</sup> − 1}} = 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.