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A friend told me that the golden ratio constant $\varphi$, i.e., $\dfrac{1+\sqrt{5}}{2}$, is 'the most irrational number,' does anyone know if this is true and if so can it be proven? Thank you!

S. Maths
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Jinny Ecckle
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    https://youtu.be/CaasbfdJdJg – whpowell96 Jan 27 '20 at 01:51
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    What he meant by "the most irrational"? – S. Maths Jan 27 '20 at 01:52
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    Theorem. If $x\in \Bbb R$ then there are infinitely many $(y,z)\in \Bbb Z^2 $ such that $|x-\frac {y}{z}|<(z^2\sqrt 5)^{-1}.$ The value $\sqrt 5$ cannot be increased (for $all$ $x$) because if $x=\phi$ and $K>\sqrt 5$ then there are only finitely many $(y,z)\in \Bbb Z^2$ such that $|x-y/z|<(z^2K)^{-1}.$ – DanielWainfleet Jan 27 '20 at 02:04
  • @whpowell96, I don't intend this to be facetious, but which word do you want me to define, 'most' or 'irrational number'? – Jinny Ecckle Jan 27 '20 at 02:07

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The reason $\phi$ is sometimes called the "most irrational number" is because of its properties relating to continued fractions. A "continued fraction" is a nested fraction that goes on forever. Any number that can be expressed as a continued fraction is an irrational number. For example, the continued fraction for $\pi$ begins as $$3 + \frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{1}{292+...}}}}$$ Note that even after 2 terms, $3 + \frac{1}{7}=3.142857$ which is quite close to $\pi$. This means $\pi$ is approximated by a continued fraction quite "quickly". We may ask "what is the number that is the least well approximated by its continued fraction?" This can be twisted to essentially ask "what is the most irrational number"? The answer is $$\phi = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}$$ Note that $\phi = 1.6180339...$ and after 2 terms of our continued fraction we have $2$, and after 3 terms we have $1.6666...$. Which although it seems like it is approximating $\phi$ quite well, this is still the continued fraction that least approximates its value as we analyze more terms. Therefore some snazzy pop mathematicians like to call it the "most irrational number".

Ty Jensen
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