The digits of 1/2 (base 2.5)

Posted by: Gary Ernest Davis on: April 27, 2010

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The unusual title for this post comes from a tweet from my Twitter colleague Peter Flom.

Peter tweeted that he was thinking about writing numbers in base 2.5

I recalled that some years ago when I was teaching a graduate mathematics education class at Rutgers, we looked at an algorithm for generating the expansion of a number in a given whole number base, and one of the students asked what would happen if we chose a fractional base – like $2\frac{1}{2}$ .

We implemented the algorithm in Excel, so we just changed the base from a whole number – 2 – to a fractional number – $2\frac{1}{2}$ – and bingo! the digits of the expansion appeared.

Writing numbers in different whole number bases

How do we go about writing a number such as $\frac{1}{3}$ as a sequence of digits base 2?

First we have to agree on what the digits base 2 are. The usual convention is that they are the digits 0 and 1.

So we are looking for a representation of $\frac{1}{3}$ in the form $\frac{1}{3}=0.b_1b_2 b_3\ldots$ Â (base 2) where the $b_n$ are digits 0 or 1.

By this we mean that $\frac{1}{3}=\frac{b_1}{2}+\frac{b_2}{2^2}+\frac{b_3}{2^3}+\ldots$ .

How can we find the sequence of digits $b_n$ ?

The key is to begin by multiplyingÂ $\frac{1}{3}$ by the base 2. Â This gives us the representation $\frac{2}{3}=b_1+0.b_2b_3b_4\ldots$ .

The digit $b_1$ is then the integer part – or floor – of $\frac{2}{3}$ , so $b_1=0$ and $\frac {2}{3}=0.b_2b_3b_4\ldots$

Now we multiply by 2 again to get the repesentation $\frac {4}{3}=b_2+0.b_3b_4b_4\ldots$ .

This means that $b_2$ is the floor of $\frac{4}{3}$ , namely 1, and $\frac{1}{3}=\frac{4}{3}-1=0.b_3b_4b_5$ .

Now we are in luck, because we have a representation for $\frac{1}{3}$ again, so we can take $0=b_1=b_3=b_5\ldots$ and $1=b_2=b_4=b_6\ldots$ .

So the base 2 representation for $\frac{1}{3}$ is $0.01010101\ldots$

Recall this means that $\frac{1}{3}=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\ldots$

Writing the algorithm in Excel

The recursive algorithm for finding the digits in the base 2 expansion of a number is so simple that we can code it in Excel (of course feel free to use Python or R).

Here’s how the Excel spreadsheet looks:

base =	2	n	b(n)	a(n)
number =	1/3	1	0	1/3
		2	1	2/3
		3	0	1/3
		4	1	2/3
		5	0	1/3
		6	1	2/3
		7	0	1/3
		8	1	2/3
		9	0	1/3
		10	1	2/3
		11	0	1/3
		12	1	2/3
		13	0	1/3
		14	1	2/3
		15	0	1/3
		16	1	2/3
		17	0	1/3
		18	1	2/3
		19	0	1/3
		20	1	2/3

with the entries in the a(n) column set to be fractions (rather than decimal expansions).

The simple Excel code to generate this table is as follows:

base =	2	n	b(n)	a(n)
number =	=1/3	1	=INT(E2*$B$1)	=B2
		=C2+1	=INT(E3*$B$1)	=E2*B$1-D2
		=C3+1	=INT(E4*$B$1)	=E3*B$1-D3
		=C4+1	=INT(E5*$B$1)	=E4*B$1-D4
		=C5+1	=INT(E6*$B$1)	=E5*B$1-D5
		=C6+1	=INT(E7*$B$1)	=E6*B$1-D6
		=C7+1	=INT(E8*$B$1)	=E7*B$1-D7
		=C8+1	=INT(E9*$B$1)	=E8*B$1-D8
		=C9+1	=INT(E10*$B$1)	=E9*B$1-D9
		=C10+1	=INT(E11*$B$1)	=E10*B$1-D10
		=C11+1	=INT(E12*$B$1)	=E11*B$1-D11
		=C12+1	=INT(E13*$B$1)	=E12*B$1-D12
		=C13+1	=INT(E14*$B$1)	=E13*B$1-D13
		=C14+1	=INT(E15*$B$1)	=E14*B$1-D14
		=C15+1	=INT(E16*$B$1)	=E15*B$1-D15
		=C16+1	=INT(E17*$B$1)	=E16*B$1-D16
		=C17+1	=INT(E18*$B$1)	=E17*B$1-D17
		=C18+1	=INT(E19*$B$1)	=E18*B$1-D18
		=C19+1	=INT(E20*$B$1)	=E19*B$1-D19
		=C20+1	=INT(E21*$B$1)	=E20*B$1-D20

The key is to start with $a(1)$ as the number whose base 2 representation we want to find. Then $b(n)=$ integer part (=floor) of $a(n-1)*base$ Â (in this case the base is 2), and $a(n)=a(n-1)*base-b(n-1)$ .

Having set up the Excel code this way we can change the number whose representation we want to find from $\frac{1}{3}$ to $\frac{1}{4},\frac{1}{5}$ or even $\pi$ .

Changing to another whole number base

We can also change the base. If we change the base to 3, so using the digits $0, 1, 2$ in our representations, we get, for example, $\frac{1}{4}=0.02020202\ldots$ (base 3).

Changing to a Â fractional base

But nothing stops us from choosing a fractional base, such as $2\frac{1}{2}$ .

In this case the digits will be 0, 1 or 2 Â – the non-negative integers less than $2\frac{1}{2}$ .

Then, we get, for example,Â $\frac{1}{2}=0.10111000011012100011111210001002000100120100111110\ldots$ (base 2.5).

Changing to arbitrary real number bases

We can even choose transcendental bases such as $\pi$ and to within the numerical accuracy of Excel, we get $\frac{1}{2}=0.11211202102012230001010010211022022021112301100001\ldots$ (base $\pi$ ).

Changing to complex integer bases

We can even calculate the base b expansion of a number where b is a complex integer – that is a complex number of the form $m+ni$ where m and n are integers. Â This is because there is a nice floor function for complex integers, which is $\textrm{Floor}(m+ni)=\textrm{Floor}(m)+\textrm{Floor}(n)i$ .

We can use the built-in complex arithmetic of Excel to carry out the calculations as before (but you might want to use Mathematica, MATLAB, R or Python instead).

Two examples are:

$\frac{1}{2} = 0.00i$ (base 1+i ) and $\frac{1}{3 }= 0.00-1(-1+i)0i -1i$ repeated (base 1+i).

Postscript

Because of the problem of non-uniqueness of representation in weird bases, one needs to think a little about what are appropriate digits for a base b representation, where b is something other than a positive integer, as pointed out in the comment by John Lame, below.

A lovely account of representation in weird bases appears in James Tanton’s Exploding Dots.

Republic of Mathematics blog

The digits of 1/2 (base 2.5)