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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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    @MarkBennet I hadn't $c$onsidered that. I'm not sure what would happen at the moment.2012-10-23

3 Answers 3

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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, 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.

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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.2012-10-22
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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.

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    \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}2013-01-03
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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.

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    Aye, it was a non-integer base, I was wondering now such a thing might be represented, if it could be.2012-10-23