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According to this video, $\varphi$ is the most irrational number, due to its continued fraction form having $1$, the smallest natural number, in the denominators.

Is it not possible to construct a "more irrational" number by using $0$?

For example,

$\iota = 1 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{0 + \cfrac{1}{1+\cdots} } } }$

Based on the argument in the video, this would appear to be more irrational.

What am I missing?

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    $\begingroup$ This number is $1+1+1+1+...$ because the recipricals cancel. $\endgroup$
    – Empy2
    Commented Sep 25, 2018 at 20:11

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No because $$\iota = 1 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{0 + \cfrac{1}{1+\cdots} } } } = 1 + \cfrac{1}{ \cfrac{1}{1 + \cfrac{1}{ \cfrac{1}{1+\cdots} } } }$$\ $$=1 + 1+\cfrac{1} {\cfrac{1}{1+\cdots}}=1+1+1+\ldots $$

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  • $\begingroup$ I don't understand how you go from form two to three, if anything, it looks like the $1/1$s would cancel, and it would be identical to the continued fraction form of $\varphi$ - which I guess also answers my question. $\endgroup$
    – RoadieRich
    Commented Sep 25, 2018 at 20:30
  • $\begingroup$ No, when two of the $1$’s cancel the rest is in the numerator, not the denominator $\endgroup$
    – Ross Millikan
    Commented Sep 25, 2018 at 20:59

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