Calculation: one more attempt. hopefully this isn't getting distractingly long

← Previous revision Revision as of 20:19, 27 February 2026
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==Calculation==
==Calculation==
Two non-zero quantities {{tmath|a}} and {{tmath|b}} are in the ''golden ratio'' {{tmath|\varphi}} if
Two non-zero quantities {{tmath|a}} and {{tmath|b}} are in the ''golden ratio'' {{tmath|\varphi}} if
{{bi |left=1.6 |1=
\frac{a+b}{a} = \frac{a}{b} = \varphi.
\frac{a+b}{a} = \frac{a}{b} = \varphi.
}}


To determine {{tmath|\varphi}} as a number, we use {{tmath|1= a = \varphi b}}, to write display=block>\frac{\varphi b+b}{\varphi b} = \frac{\varphi b}{b}.
To determine {{tmath|\varphi}} as a number, we can divide the numerator and denominator of the fraction on the left-hand side by {{tmath|b}},
{{bi |left=1.6 |1=\frac{\frac{a}{b}+1}{\frac{a}{b}} = \frac{a}{b},}}
[[Cancel out|Canceling out]] the appearances of {{tmath|b}} and multiplying both sides by {{tmath|\varphi}} gives
and then substitute {{tmath|1= \tfrac{a}{b} = \varphi}}, to obtain
⚫
display=block>\varphi + 1 = \varphi^2
{{bi |left=1.6 |1=\frac{\varphi + 1}{\varphi} = \varphi.}}
Multiplying both sides by {{tmath|\varphi}} gives
⚫
{{bi |left=1.6 |1=>\varphi + 1 = \varphi^2}}
which can be rearranged to
which can be rearranged to
display=block>{\varphi}^2 - \varphi - 1 = 0.
{{bi |left=1.6 |1=>{\varphi}^2 - \varphi - 1 = 0.}}


The [[quadratic formula]] yields two solutions:
The [[quadratic formula]] yields two solutions:

{{bi |left=1.6 |1=\frac{1 + \sqrt5}{2} = 1.618033\dots\ and \ \frac{1 - \sqrt5}{2} = -0.618033\dots.}}
{{bi |left=1.6 |1=\frac{1 + \sqrt5}{2} = 1.618033\dots\ and \ \frac{1 - \sqrt5}{2} = -0.618033\dots.}}


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