# Is it possible to write a number in a base of less than 1?

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?

• Have you thought about what might go after the "decimal point"? Oct 22, 2012 at 12:03
• In the title you say less than 1, but then ask for between 0 and 1. have you considered you can have a negative base? Oct 22, 2012 at 12:34
• @DrakeClarris Hi Drake, I did, I found a post already discussing that very idea. Oct 23, 2012 at 8:29
• @MarkBennet I hadn't considered that. I'm not sure what would happen at the moment. Oct 23, 2012 at 8:31

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.

• If the base is neither an integer nor the reciprocal of an integer, the representations are either incomplete or non-unique.
– robjohn
Oct 22, 2012 at 17:20
• @robjohn: But you get non-uniqueness in any case if you want to represent more than just the integers. Oct 22, 2012 at 17:26
• 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, 2012 at 18:21

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.

• 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. Oct 23, 2012 at 8:39
• 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 Oct 23, 2012 at 9:12
• $\begin{array}{|c|c|}\hline\text{Base Ten}&\text{Base Zero}&\text{Base One}\\\hline 1&\{\}&0\\2&\{\{\}\}&00\\3&\{\{\{\}\}\}&000\\4&\{\{\{\{\}\}\}\}&0000\\\vdots&\vdots \end{array}$ Jan 3, 2013 at 15:16

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.

• Seen this? Oct 23, 2012 at 3:37
• From the link he provided it seems that he is looking for non-integer base. Oct 23, 2012 at 7:23
• Aye, it was a non-integer base, I was wondering now such a thing might be represented, if it could be. Oct 23, 2012 at 8:40