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Following on from this question:

https://math.stackexchange.com/a/217112/45127

If we take base 10 as an example, the granularity is 1. I.e. we increment the digits in an increment of 1 until we reach 9 and then start a new column.

If the base is less than 1, then would any number we write require an infinite number of columns or is there a way of writing a number in a base between 0 and 1?

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BanksySan
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3 Answers3

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HINT: $123.456_{\text{ten}}=6543.21_{\text{one-tenth}}$:

$$\begin{align*} 1\cdot10^2&+2\cdot10^1+3\cdot10^0+4\cdot10^{-1}+5\cdot10^{-2}+6\cdot10^{-3}\\ &=6\left(\frac1{10}\right)^3+5\left(\frac1{10}\right)^2+4\left(\frac1{10}\right)^1+3\left(\frac1{10}\right)^0+2\left(\frac1{10}\right)^{-1}+1\left(\frac1{10}\right)^{-2} \end{align*}$$

Added: Of course if you insist that the ‘digits’ base $b$ must lie in the interval $[0,b)$, then you will have a hard time constructing a reasonable base $b$ notation. A more interesting question might be:

If $0<b<1$, what sets of ‘digits’ can be used to give every real number a base $b$ representation.

If $b=\frac1n$ for some integer $n\ge2$, the digits $0,1,\dots,n-1$ will serve, as in the example above.

Brian M. Scott
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    If the base is neither an integer nor the reciprocal of an integer, the representations are either incomplete or non-unique. – robjohn Oct 22 '12 at 17:20
  • @robjohn: But you get non-uniqueness in any case if you want to represent more than just the integers. – Brian M. Scott Oct 22 '12 at 17:26
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    True. However for an integer base, there are a countable set of numbers which have double representations. In base $2.5$, if we only use digits ${0,1}$ we cannot represent any number in $\left(\frac23,1\right)$. If we use digits ${0,1,2}$ all numbers have an uncountable number of representations. Non-integer bases are just not as nice. – robjohn Oct 22 '12 at 18:21
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If you give up the condition that digits should be from the set $\{0,1,\ldots,b-1\}$ (which makes no sense unless $b$ is a non-negative integer, gives the empty set for $b=0$ and for $b=1$ only gives you the digit $0$ which will not generate much else) but from some other set, possibly accompanied with additional restrictions on the sequences of digits, then number systems are possible with fractional, negative or even complex bases. You can even relax the condition that place values are all powers of some "base" in the first place, giving mixed-radix systems, or other things like the Fibonnacci number system. There are really too many ideas possible (and tried) to mention them all here.

Marc van Leeuwen
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  • Thanks, I hadn't imagined a complex base, or a variable radix base. It's got awesome possibilities, few of which I am clever enough to figure out. I think I have a new "Maths Fad" for a few weeks. i really like the idea of a Taylor series-esk base for using in wave filtering. – BanksySan Oct 23 '12 at 08:39
  • I might have added a reference for these things. You can find many discussions about weird number systems in Concrete Mathematics, and I'm fairly sure there is some discussion in The Art of Computer Programming, but I forgot which volume; since it is quite basic volume 1 would be a guess, also maybe the latest volume 4A – Marc van Leeuwen Oct 23 '12 at 09:12
  • $\begin{array}{|c|c|}\hline\text{Base Ten}&\text{Base Zero}&\text{Base One}\\hline 1&{}&0\2&{{}}&00\3&{{{}}}&000\4&{{{{}}}}&0000\\vdots&\vdots \end{array}$ – Elements In Space Jan 03 '13 at 15:16
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Yipes! I just tried base -10.

$$12345.678=\begin{aligned} 2&\times\;^{_-}10^{4}\\ +\;8&\times\;^{_-}10^{3}\\ +\;4&\times\;^{_-}10^{2}\\ +\;6&\times\;^{_-}10^{1}\\ +\;6&\times\;^{_-}10^{0}\\ +\;4&\times\;^{_-}10^{^{_-}1}\\ +\;8&\times\;^{_-}10^{^{_-}2}\\ +\;2&\times\;^{_-}10^{^{_-}3} \end{aligned}$$ $$12345.678_{10} = 28466.482_{^{_-}10}$$

Although, that's using positive digits with a neative base. You may need to use negative digits and negate the whole number.

Hand-E-Food
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