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==Representing rational numbers as golden ratio base numbers== Every non-negative rational number can be represented as a recurring base-Ο expansion, as can any non-negative element of the [[field (mathematics)|field]] '''Q'''[{{sqrt|5}}] = '''Q''' + {{sqrt|5}}'''Q''', the field generated by the [[rational number]]s and [[square root of 5|<math display=inline>\sqrt{5}</math>]]. Conversely any recurring (or terminating) base-Ο expansion is a non-negative element of '''Q'''[{{sqrt|5}}]. For recurring decimals, the recurring part has been overlined: *{{sfrac|1|2}} = 0.<span style="text-decoration:overline;>010</span><sub>Ο</sub> *{{sfrac|1|3}} = 0.<span style="text-decoration:overline;>00101000</span><sub>Ο</sub> *{{sfrac|1|4}} = 0.<span style="text-decoration:overline;>001000</span><sub>Ο</sub> *{{sfrac|1|5}} = 0.<span style="text-decoration:overline;>001001010100100100</span><sub>Ο</sub> *{{sfrac|1|10}} = 0.<span style="text-decoration:overline;>000010000100010100001010001010101000100101000001001000100000</span><sub>Ο</sub> 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-Ο [[long division]] there are only a finite number of possible remainders, and so once there must be a recurring pattern. For example, with {{sfrac|1|2}} = {{sfrac|1|10.01<sub>Ο</sub>}} = {{sfrac|100<sub>Ο</sub>|1001<sub>Ο</sub>}} long division looks like this (note that base-Ο subtraction may be hard to follow at first): <pre> .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 β 1001 = 1011 β 1001 = 10 1 0 0 0 0 1 0 0 1 ------- etc. </pre> The converse is also true, in that a number with a recurring base-Ο; representation is an element of the field '''Q'''[{{sqrt|5}}]. This follows from the observation that a recurring representation with period k involves a [[geometric series]] with ratio Ο<sup>βk</sup>, which will sum to an element of '''Q'''[{{sqrt|5}}].
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