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The
golden ratio (
golden section,
golden mean) is the positive root
of the
quadratic equation
 $x^{2}x1=0,$
which has roots
 $\phi ={\frac {1+{\sqrt {5}}}{2}},\ \varphi ={\frac {1{\sqrt {5}}}{2}}.$
Note that
 $\phi +\varphi =1,$
 $\phi \,\varphi =1.$
Decimal expansion of the golden ratio
The decimal expansion of the golden ratio (A001622) is
 $\phi =1.6180339887498948482045868343656381177203091798057628621\ldots$
and the decimal expansion of the conjugate root of the golden ratio is
 $\varphi =0.6180339887498948482045868343656381177203091798057628621\ldots$
Since
 $x\,(x1)=1,$
the
multiplicative inverse of the root
is
(same
fractional part), and since
 $x+[(x1)]=1,$
the root
added with the
additive inverse of its
multiplicative inverse also gives
.
Powers of ϕ and Fibonacci numbers
 $\phi ^{n}={\bigg (}{\frac {1+{\sqrt {5}}}{2}}{\bigg )}^{n}=F_{n1}+F_{n}\,\phi ,$
where
is the golden ratio and
is the
th Fibonacci number.
Powers of



6


18

5



4


7

3



2


3

1



0


2

−1



−2


3

−3



−4


7

−5



−6


18

Continued fraction and nested radicals expansions
The golden ratio has the simplest continued fraction expansion (the all ones sequence A000012)
 $\phi =1\,+\,{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}=1+\left[1+\left[1+\left[1+\left[1+\left[1+\cdots \right]^{1}\right]^{1}\right]^{1}\right]^{1}\right]^{1},$
since
 $\phi 1={\frac {1}{\phi }},$
and also the simplest nested radicals expansion (again, the all one's sequence)
 $\phi ={\sqrt {1+{\sqrt {1+{\sqrt {1+{\sqrt {1+{\sqrt {1+\cdots }}}}}}}}}}=\left[1+\left[1+\left[1+\left[1+\left[1+\cdots \right]^{\frac {1}{2}}\right]^{\frac {1}{2}}\right]^{\frac {1}{2}}\right]^{\frac {1}{2}}\right]^{\frac {1}{2}},$
since
 $\phi ^{2}1=\phi .$
Approximations
 $e{\frac {11}{10}}=1.61828182845904\ldots (1.000153173364\ldots \times \phi ),$
where
is
Euler's number.
 ${\sqrt {\frac {5\pi }{6}}}=1.6180215937964\ldots (0.999992339\ldots \times \phi ).$
As an infinite series
 $\phi =\sum _{k=0}^{\infty }\left({\frac {3{\sqrt {5}}}{2}}\right)^{k}=\sum _{k=0}^{\infty }\left(1+\varphi \right)^{k}={\frac {1}{1(1+\varphi )}}={\frac {1}{\varphi }}=\phi .$
 $\varphi =\sum _{k=0}^{\infty }\phi ^{2k}=\sum _{k=0}^{\infty }\left({\frac {1}{\phi ^{2}}}\right)^{k}=\sum _{k=0}^{\infty }\left({\frac {1}{\phi +1}}\right)^{k}={\frac {1}{1(\phi +1)}}={\frac {1}{\phi }}=\varphi .$
See also
 {{Fibonacci}} (mathematical function template)