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If a number with more than one digit occurs in the fraction, it should be expanded to as many digits in the expansion. I will be even more impressed, however, if the fraction consists entirely of 1-digit terms. No integers allowed.

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  • $\begingroup$ By the last sentence I imagine that you allow a continued fraction of the form $[a_0,a_1,..]$ in which the $a_i$ may not be $<10$ but then you concatenate the digits in the decimal expansion as $\overline{a_0,a_1a_2...}$. Is that what you mean? $\endgroup$ – OR. Aug 1 '13 at 22:33
  • $\begingroup$ @RGB That is correct, except that it would most likely not be a repeating decimal, and I assume you mean > 10. $\endgroup$ – Lee Sleek Aug 1 '13 at 22:56
  • $\begingroup$ Bizarre question! $\endgroup$ – user43208 Sep 8 '13 at 13:22
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There is no solution with single digits. You clearly have to start with a $3$ to get $0.3 \approx [0;3]=\frac 13$ Now whatever you put next the CF is too small.

You have the amusing $\frac 13=0.3333333\ldots=[0;3,333333333\ldots]$ In a bit of playing, this "attractive fixed point" is hard to avoid. You need a big second number to get anywhere close. As you work at it, the $3$'s keep coming. I haven't made it into a proof that there isn't one.

I presume you don't like $1=[1;]$ or any other integer.

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  • $\begingroup$ How about irrational numbers? This is not yet a complete answer so I won't upvote or downvote it for now. $\endgroup$ – Lee Sleek Aug 2 '13 at 1:37
  • $\begingroup$ I don't think you will get an irrational. After starting with $0.3$, you need a large second term to make them equal. Everything I tried lead to the infinite series of $3$'s-otherwise the value of the CF was below the fraction. But I agree it isn't complete. I hoped this would prompt some thought. $\endgroup$ – Ross Millikan Aug 2 '13 at 1:42
  • $\begingroup$ It just occurred to me that $0.1 = [0; 10]$ is a solution (trailing zeroes should be omitted in case that wasn't clear). $\endgroup$ – Lee Sleek Aug 7 '13 at 16:22
  • $\begingroup$ But the digits of $0.1$ are 0 and 1, not 0 and 10. If you allow 10 as a digit, it must have the value of 10, which means that 0.(10) = 1, not $\frac1{10}$. $\endgroup$ – MJD Jan 8 '14 at 3:42

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