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Let's say we have two functions, $f$ and $g$. $f:\mathbb{R}\mapsto [0,1]$ where $0,1$ denote true, false respectively. $f(x)=1$ when $x$ contains any zeroes as a digit; $f(x)=0$ otherwise.

Now let's suppose that $g(x)$ essentially removed all numbers preceding the decimal point and the decimal itself from $x$ (that is, $g(4.5)=5$, $g(e)=71828182\ldots$ etc.).

Would $f(g(\pi))=1$? That is, does the $g(\pi)$ contain any zeroes as a digit? Do you have any evidence to support your claim?

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  • $\begingroup$ What are you asking? You need to clarify this question, especially your definiton of the function f.. $\endgroup$ – Quality Dec 3 '14 at 2:41
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    $\begingroup$ This would have been a fabulous question in 1630! $\endgroup$ – Mark McClure Dec 3 '14 at 2:43
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    $\begingroup$ @MarkMcClure People say I'm rather old school for my age ;) Thanks for the reference! $\endgroup$ – Conor O'Brien Dec 3 '14 at 2:52
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    $\begingroup$ See also oeis.org/A050279, which shows the start of the positions of 0, 00, 000, ... in the decimal expansion of pi. $\endgroup$ – Charles Dec 3 '14 at 2:52
  • $\begingroup$ $\pi$, e, and other famous irrationals are conjectured to be normal numbers, at least in base $10$. $\endgroup$ – Lucian Dec 3 '14 at 4:07
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Yes, of course there is a 0 in the decimal expansion of $\pi=3.1415926535897932384626433832795\underline{0}2884197...$.

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    $\begingroup$ It was touch and go for a while though, wasn't it? $\endgroup$ – TonyK Dec 5 '14 at 0:41

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