Unusual number bases, including Fibonacci and the irrational Phi bases, factorials and binomial coefficients bases,
rational number bases, negative bases, negative "digits", with interactive Calculators for all of these.
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Contents of this page
The icon means there is a
You Do the Maths... section of questions to start your own investigations.
The calculator icon
indicates that there is a live interactive calculator in that section.
With thanks to MikeMcl and his BigNumbers JS library
(readMe hosted on github).
It is used here in the JavaScript Calculators
for arbitrary precision number calculations.
Representing whole numbers in different bases
If you are familiar with bases 2,3,... then skip to the next section.
Otherwise begin here for basic information on bases.
We normally write numbers in base 10 so that each digit counts a power of 10 in the value:
2017 = 2×10^{2} + 0×10^{2} +1×10^{1} + 7×10^{0}
The word digit means finger
In the American and UK systems of measuring liquids in pints and gallons,
there are 8 pints to 1 gallon, so 20 pints represents 2 gallons and 4 pints.
It used to be the case that 8 gallons make 1 bushel in the UK but
the American and British systems
are now different.
This is a base 8 system where we counts in 8s:
90 pints is 1×8^{2} + 3×8 + 2: 90 in base 10 is 132 in base 8
Digits
We see that in base 10 (the decimal system), we need only 10 symbols, the digits0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
In Base 8 (octal) we need only the 8 symbols 0, 1, 2, 3, 4, 5, 6, 7
because 8 in one column will convert into 1 in the next higher column.
Similarly in base β we need only β symbols for the digits, one digit per column.
So what if the base is bigger than 10?
Computer engineers often use this since computers work in Base 2 (Binary) using binary-digits0, 1
and it was often more convenient to use base 16 (hexadecimal) using the "digits"
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F.
In base β we need only β 'digit' symbols
To show what base a number is in, we write its base as a _{suffix} after the number representation, e.g.:
90 = 132_{8} 10 = 12_{8} 7 = 12_{5}
If we omit the base then the base used is 10.
Names of Bases
base 2: binary
base 3: ternary, trinary
base 4: quaternary
base 5: quintal, quinary
base 6: sextal, senary
base 7: septal, septenary
base 8: octal
base 9: nonary
base 10: decimal, denary
base 11: undecimal
base 12: duodecimal
base 16: hexadecimal
base 20: vigesimal
base 30: trigesimal
base 40: quadragesimal
base 50: quinquagesimal
base 60: sexagesimal
base 70: septuagesimal, septagesimal
base 80: octagesimal
base 90: nonagesimal, nonogesimal
base 100: centesimal
The vigesimal system was used by the Mayan civilisation in central America and also in Britain where 20 is called a "score".
Other cultures used a base 5 system (New Hebrides) or a binary system. 12 (duodecimal) was also popular because it is easy to divide by 2, 3 and
4 and the ancient Babylonians used base 60 (sexagesimal) because it was also easy to divide by 2, 3, 4 and 5. The system of money used in Britain from the 8th century until 1971 had
12 pence (d) in a shilling (s) and 20 shillings in a pound (£).
An easy method to convert a number to and from base β
Let's look at an example.
To convert a (decimal) number to base 6, say, keep dividing the number by 6 recording any remainders. The remainders become the base 6
representation of the number.
We start with 6 dividing 5755 and work upwards , repeatedly dividing the quotients until we reach 0
0
rem = 4
6
4
rem = 2
4 = 6×0 + 4
6
26
rem = 3
26 = 6×4 + 2
6
159
rem = 5
159 = 6×26 + 3
6
959
rem = 1
959 = 6×159 + 5
6
5755
5755 = 6×959 + 1
To see the base 6 representation, write down the remainders from the top to the bottom: 5755 = 42351_{6}
This works for any numeric base and shows that the 'digits' of base β must be 0,1,2,...,β-1 because they are the remainders on
dividing by β.
Because we repeat the process on the quotients until we reach 0, we can see the 'digits' of the quotients too:
To convert a number from base β to decimal we reverse the process,
starting with the leftmost digit and continually multiplying it by the base β and adding in the next digit:
Here we convert 4235_{6} to base 10, accumulating a final sum as we go from left to right across the 'digits':
To convert 4235_{6} to base 10
'digits'
4
2
3
5
6 ×sum
0
6×4= 24
6×26= 156
6×159= 954
add 'digit'
4
26
159
959
The final sum is the base 10 value: 4235_{6} = 959
by multiplying each power of x individually by its coefficient, it is much easier and more efficient if we accumulate the sum by using Horner's rule:
P(x) = ((..( ( a_{n}) x + a_{n−1} ) x + ... + a_{1}) x + a_{0}
So to evaluate 4235_{6} = 4×6^{3}
+ 2×6^{2}
+ 3×6 + 5, we calculate instead:
((4×6 + 2)×6
+ 3)×6 + 5
which is what we are doing in the table method of the previous subsection.
The Art of Computer Programming, Vol 2:Seminumerical Algorithms D E Knuth
(first edition 1981, third edition 1998) section 4.6.4 Evaluation of Polynomials
Factorial base
Permutations
Instead of using powers of a number as the column headings, we can use other series of numbers too, for instance the factorials.
The factorial numbers count the number of permutations of n objects.
In general, whatever the objects, they will be in first place, second place and so on in the permutation list.
So we can just use their positions 1,2, ... ,n as the objects and name the objects 1,2,...,n.
So the ways to permute 3 objects (to arrange them all in some order)
are: 123, 132, 213, 231, 312, 321.
If we were permuting the letters of CAT we use 1 for C, 2 for A and 3 for T. Ther 6 permutations are therefore:
123=CAT, 132=CTA, 213=ACT, 231=ATC, 312=TAC, 321=TAC
Why are there exactly 6 permutations?
Because there are 3 ways to choose the position in which to put it including where it is now.
There are 2 position for the second object and then the third object goes in the remaining empty place: 3x2x1=6.
Similar reasoning shows there are 4×3×2×1 permutations
of 4 objects.
For n objects there are n! = n×(n−1)×...×3×2×1 permutations, called
factorial n, also written as n!.
n
1
2
3
4
5
6
7
8
9
10
n!
1
2
6
24
120
720
5040
40320
362880
3628800
A Factorial-base system
The factorial numbers can be used as the columns of a mixed base system,
arranging them in reverse order: ..., 5!, 4!, 3!, 2!, 1!. The digits allowed in each column depend on the column number:
the first column (1!) can have 0 or 1;
the second column (2!) can have digits 0, 1 or 2;
and in general the n-th factorial column can have 0,1, up to the column number n.
This is because n in column n means n×n! but (n+1)×n! is (n+1)!,
which "carries" to the next larger column heading (on the left).
To indicate numbers in this base, we list the "digits" and use the exclamation mark "!" as the base. For larger number using ten or more columns, the value in a column may be 10 or more.
Here we will write the values (numbers) in each factorial column as a list of numbers enclosed in { } brackets.
Easy methods of converting to and from the Factorial Base
To convert a number to base !:
We can adapt the easy method to convert a number to base β above for converting
a (decimal) number to base !. This time we start by dividing by 2 and the divisor increases each time.
Here is an example to convert 69 to base !:
2
69
3
34
rem=1
69 = 2×34 + 1
4
11
rem=1
34 = 3×11 + 1
5
2
rem=3
11 = 4×2 + 3
0
rem=2
2 = 5×0 + 2
To see the base ! representation, write down the reminders from the bottom to the top: 69 = 2311_{!}
To convert a base ! representation to decimal:
First write the multipliers above each digit by starting from the right with 2
and proceed leftwards with multipler 3 then 4 and so on.
We will not use the leftmost 'multiplier' above the leftmost digit.
Start with the most significant digit (the leftmost digit) which is the initial 'sum'
Multiply the sum by the next column multiplier on its right ...
... and then add on the 'digit' under that multiplier to make the new 'sum'
Repeat the previous two steps of multiplying and adding
until you have a 'sum' under the final (rightmost) digit which is the decimal value
For example, using the same numbers as the example above, we have:
To convert 2311_{!} to decimal:
Every factorial from n! onwards is a multiple of n.
So to test if any given factorial base representation is divisible by n, we only need to test the number represented by the
last n−1 'digits'.
For instance, to test if n_{!} is divisible by 2=2!,
look only at the last 'digit'. If it is 0, n is even, divisible by 2; if it is 1, the number is odd.
For divisibility by 3, test only the last 2 'digit's. if they are {1,1}_{!} or
{0,0}_{!} then n is a multiple of 3; otherwise it is not.
There are 6 possibilities for the final 3 'digits' to test if a value is a multiple of 4.
Show them
For larger divisors d, convert only the final d digits to base 10 and then test.
These divisibility tests are an advantage only if we are dealing with very large numbers and testing for small divisors.
An Application of Base Factorial
The 24 permutations of 4 objects can be listed in lexicographic order which means that the permutations,
as numbers, are in numerical order or dictionary or alphabetic order. For instance here is the order of all the permutations of
1,2,3 and 4 sorted into lexicographic (dictionary) order - think of 1 as "a", 2 as "b", 3 as "c" and 4 as "d":
0:
1,2,3,4
1:
1,2,4,3
2:
1,3,2,4
3:
1,3,4,2
4:
1,4,2,3
5:
1,4,3,2
6:
2,1,3,4
7:
2,1,4,3
8:
2,3,1,4
9:
2,3,4,1
10:
2,4,1,3
11:
2,4,3,1
12:
3,1,2,4
13:
3,1,4,2
14:
3,2,1,4
15:
3,2,4,1
16:
3,4,1,2
17:
3,4,2,1
18:
4,1,2,3
19:
4,1,3,2
20:
4,2,1,3
21:
4,2,3,1
22:
4,3,1,2
23:
4,3,2,1
If we want to choose a random permutation or if we want to make sure we go through all permutations, we need a way of changing the number
of the permutation in the lexicographic order to the permutation itself - and this is where base Factorial comes in.
The base factorial
representation of an index number n in the lexicographic order is easily changed into the permutation itself.
Calculator for Factorial Base and Permutations
Factorials are exact up to 500! which has 1135 digits. After that, just the number of digits is shown.
You can use this Calculator to show that the number of digits in 10^{40}! is itself a number with 42 digits(!)
Factorial Base and Permutations C A L C U L A T O R
Which is the first factorial that is a multiple of
8 ?
17 ?
100 ?
8 = 2^{3} and three 2s come from 2 and two more from 4, so 4! is the first multiple of 8
17 is prime so if first appears in 17!
100 = 2^{2} 5^{2} and these factors will come from 2, 4, 5 and 10,so 10! is the first factorial to end with 00.
2! is the first even factorial,
3! is the first facorial divisible by 3
4! is the first divisible by 4 and
5! the first that is a multiple of 5.
Which is the first divisible by 6?
And which factorials are the first divisible by 7, by 8, by 9 and by 10?
3! is the first multiple of 6
7 is a factor first in 7!
8 in 4!
9 in 6!
10 in 5!
The sequence of n where n! is the first divisible by 2,3,4,... is: 2,3,4,5,3,7,4,6,5, ... called the Kempner Numbers A002034
How many zeroes are at the end of:
10!
20!
100!
10! = 3628800 = 2×3×4×5×6×7×8×9×10
ends with 2 zeroes
20! = 2×...5×...10×...15×...20×
ends with 4 zeroes
100! has 5 as a prime factor from 5,10,15,20,...90,95,100 and more than enough even numbers to match these.
Each of these 20 numbers has one 5 as a factor except 25, 50, 75 and 100 which have 5×5 as a factor:
20 + 4 = 24 zeroes at the end
The Binomial Representation using Pascal's Triangle
Pascal's Triangle
Here is Pascal's Triangle but pushed over ("right justified") so that the 1s at the end of the rows are aligned in a column:
r
8
7
6
5
4
3
2
1
0
n
1
0
1
1
1
1
2
1
2
1
3
3
1
3
1
4
6
4
1
4
1
5
10
10
5
1
5
1
6
15
20
15
6
1
6
1
7
21
35
35
21
7
1
7
...
⋮
Notation:
(
n
)
= Binomial(n,r)
r
=
n!
(n-r)! r!
=
n(n−1)...(n−r+1)
r(r−1)...3×2×1
Here we will use Binomial(n,r) but other notations you will see in maths books are _{n}C_{r}, ^{n}C_{r} or C_{n,r}
and it is pronounced "n choose r".
Each element in Pascal's Triangle above, which is right-justified here, is the sum of the element above
and the element to the right of that one where blank entries mean "0". The right-hand elements
in column 0 are always 1:
(
n
)
=
(
n − 1
)
+
(
n − 1
)
, n ≥ r > 0
r
r
r − 1
(
n
)
= 1;
(
n
)
= 0 otherwise
0
r
Pascal's Triangle has many interesting properties including coefficients of certain polynomials and
has many applications including probabilites.
The Binomial Representation
To use Pascal's triangle as an integer representation method,
you first decide on the number columns you want to use.
Then you choose one element from each column
and, to make the representation unique,
as we go to a smaller column number, we must choose an entry on a smaller row number too.
For instance, here is 6 represented using from 1 to 4 columns. Remember that we must choose a row for each of the N columns and as
N decreases so must the row numbers chosen. The Binomial representation is then just the list of row numbers in order of the column
numbers going down from N to 1.
All numbers in a Binomial base have their digits in decreasing order.
Notation
Once we have decided on the number of columns to use, it is the row number for each of the columns that forms the representation
as a list of the row numbers. To conform to the way other bases are shown, we list the row numbers from the largest down to 1.
Because the representation will change depending on how many columns we use,
the Base is shown as Binom.
The two examples above are therefore
Binomial(4,2) + Binomial(0,1) = 6 + 0 = 6 which is {4,0}
using 3 columns
Binomial(4,3) + Binomial(2,2) + Binomial(1,1)
= 4 + 1 + 1 = 6 which is {4,2,1}
using 4 columns
Binomial(5,4) + Binomial(3,3) + Binomial(1,2) + Binomial(0,1)
= 5 + 1 + 0 + 0 = 6 which is {5,3,1,0}
Note that all the column numbers are used in order and that the row numbers decrease as the column numbers decrease.
To convert from Binomial representation to its base 10 value:
Insert the column numbers after each 'digit', numbering the columns from 1 on the right going right to left
and then interpret these are arguments to the Binomial function: for example:
{5,3,1,0}_{Binom} has its numbers paired with columns in order, ending with column 1:
row n
5
3
1
0
col r
4
3
2
1
(
n r
)
5
1
0
0
We then interpret this as Binomial(5,4) + Binomial(3,3) + Binomial(1,2) + Binomial(0,1)
which evaluates to 5 + 1 + 0 + 0 = 6 = {5,3,1,0}_{Binom}
Using your answer to Question 1 what is {N+2, N+1, N-2, N-3, ... , 2, 1, 0}_{Binom}
where the 'digits' from N-2 descend by 1 to end at 0?
{N+2, N+1, N-2, N-3, ... , 2, 1, 0} = 2 N − 1
Express 100=10^{2} in base 9.
What is 10^{3} in base 9?
How do your answers above relate to the Binomial Theorem?
Think about 10 as 9+1 and raise it to a power.
What is 10^{3} in base 3 and how does it relate to its representation in base 9 above?
10^{2} = (9 + 1)^{2} = 1×9^{2} + 2×9 + 1 = 121_{9} = 81+18+1 = 100
10^{3} = (9 + 1)^{3} = (9+1)^{3} = 1331_{9}
In general, the 'digits' of (β+1)^{n} are the binomial coefficients of row n provided they are no bigger than the base.
Base β is related to base β^{2} by writing each base β^{2} 'digit' as two base β digits:
1000 = 1331_{9} = 01 10 10 01 _{3} because
3_{9} = 10_{3}
References
Looking into Pascal's Triangle: Combinatorics, Arithmetic, and Geometry
P Hilton, J Pedersen Mathematics Magazine vol 60 (1987), pages 305-316
JSTOR
A nice article with lots of formulae that make great investigations for students to discover for themselves
Sums of Powers J Tanton Math Horizons vol 11 (2003), pages 15-20
JSTOR
Another excellent article for teachers and others
See Eric Weisstein's MathWorld (a Wolfram Web Resource) entry on
Binomial Coefficient
The Art of Computer Programming Vol 4a: Combinatorial Algorithms Part 1 D E Knuth. Page 360
refers to the Binomial representation system using t columns as the Combinatorial Number system of degree t and the following reference...
Ernesto Pascal, Giornale di Matematiche vol 25 (1887), pages 45-49 (in Italian).
Representation of Numbers by Cascades
C C Chen, D E Daykin, Proceedings of the American Mathematical Society (Vol 59, 1976), pages 394-398.
JSTOR
Number Bases and Binomial Coefficients J M Howell, R E Horton
Maths Mag 35 (1962) pages 177-179
The Fibonacci Bases
The Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, ...
where each is the sum of the previous two numbers.
It has many interesting mathematical properties.
As a number base system, we again label the columns in order increasing to the left: 8, 5, 3, 2, 1
with just a single column labelled 1 so we only use Fibonacci numbers with index 2 or more because Fib(2)=1, Fib(3)= 2, ... .
Every whole number is a sum of different Fibonacci Numbers
that is: no Fibonacci number is used more than once.
We can use F to signify a representation using the Fibonacci numbers as column headers.
3 = 100_{F} = 2 + 1 = 11_{F}
Since Fib(1) is 1, we can simply use n Fib(1)'s to sum to any given number n.
But there is always another way to do this too for numbers bigger than 1.
If we write the number of times we need Fib(i) is column i, we have a number representation whose 'digits' are
any whole number.
The minimal and maximal Fibonacci bases
However, if we restrict ourselves to using each Fibonacci number at most once
then it is again possible to find a collection which is a set of Fibonacci numbers whose sum is any given integer.
Two in particular are of interest: since Fib(n-1) + Fib(n-2) = Fib(n), then any two consecutive 1s in a
Fibonacci representation can be "carried" into the next column. It turns out it is always possible
to find a set of Fibonacci numbers with any given sum that
either never have two consecutive 1's - so we use the smallest set of Fibonacci numbers - the minimal
Fibonacci representation or Zeckendorf representation, which we write as Fmin
or never have two consecutive 0s - we use the largest set of Fibonacci numbers - the maximal Fibonacci
representation which we write as Fmax
Here are all the Fibonacci representation for numbers 1 to 9:
Columns in the Fibonacci base as ..., 8, 5, 3, 2, 1
Columns in the Fibonacci base are ...8 5 3 2 1
n
0
1
2
3
4
5
6
7
8
9
n_{F}
0
1
10
100 11
101
1000 110
1001 111
1010
10000 1100 1011
10001 1101
To convert to and from Fibonacci bases
There are usually many Fibonacci representations of a number using only the 'digits' 0 and 1 in every column and with the column
headings (in reverse order): 1, 2, 3, 5, 8, 13, 21, ...
An easy method of converting a number to base F will give the Zeckendorf represenation which happens to have
the least number of 1s of all the Fibonacci representations for the number.
Write the Fibonacci column headings, in order from right to left, from 1,2,... until the column headings are
larger than the value to be converted
Start by writing the decimal number to be converted above the leftmost column header
If the column heading is larger than the number, put a 0 in that column and
keep the number to use on the next column on the right;
Otherwise if the column header is not larger than the number, Subtract the column header value from the number
and write a 1 in the column
repeat the step above on the same or, if you did the subtraction, the reduced decimal number
after the rightmost column you should have reached 0.
If not, you've made a mistake so check your previous steps!
Each column now has a 0 or 1 in it which is the Fibonacci (Zeckendorf) representation
For instance here we convert 52 to base Fibonacci:
A: number
52
18
18
5
5
0
0
0
B: Fib col heading
55
34
21
13
8
5
3
2
1
A-B or A
18
18
5
5
0
0
0
0
F digit
1
0
1
0
1
0
0
0
52 = 10101000_{F}
To convert from base Fibonacci, write the column headings above the digits and add the column headings above each digit '1'.
We can also express every number positive and negative using only the negative indices in the Fibonacci series:
i
...
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
...
Fib(i)
...
-21
13
-8
5
-3
2
-1
1
0
1
1
2
3
5
8
13
21
...
The advantage is that now both positive and negative whole numbers can be represented without an initial sign.
We have to choose between using the negative indices to the right of a radix point (as in integer-valued bases)
but this has the disadvantage of having 0 digits at the front.
Since the values are not fractional, listing them with indices running from -1:...-4, -3, -2, -1 makes more sense.
For instance:
i
-5
-4
-3
-2
-1
F(i)
5
-3
2
-1
1
0
0
1
1
2
1
0
0
3
1
0
1
4
1
0
0
1
0
5
1
0
0
0
0
6
1
0
0
0
1
7
1
0
1
0
0
8
1
0
1
0
1
i
-6
-5
-4
-3
-2
-1
F(i)
-8
5
-3
2
-1
1
0
0
-1
1
0
-2
1
0
0
1
-3
1
0
0
0
-4
1
0
1
0
-5
1
0
0
1
0
1
-6
1
0
0
1
0
0
-7
1
0
0
0
0
1
-8
1
0
0
0
0
0
We add up the Fibonacci numbers for each column with 'digit' 1 in it to find the value represented.
Every integer value now has a representation using 0s and 1s only.
The above are minimal forms, using the least number of 1-digits. As for minimal Fibonacci representations (Zeckendorf representations)
there will never be two 1's next to each other.
Here we call this the Fibonacci Negative Index representation or
Fibnegi or just Fnegi.
How can we tell if a base Fibnegi value is negative?
The positive numbers start with an index of a positive Fibonacci number so are of even length.
The negative numbers start with an index of a negative Fibonacci number so are of odd length.
References
Zeckendorf Representations Using Negative Fibonacci Numbers M W Bunder
Fib Q vol 30, (1992) pages 111-115 PDF
Negative Bases
If we use base -10 (negadecimal) then the columns have the values
Column:
...
4
3
2
1
0
Meaning
...
(-10)^{4}
(-10)^{3}
(-10)^{2}
(-10)^{1}
(-10)^{0}
Value
...
10000
–1000
100
–10
1
At first it may not be obvious that
Every whole number value, both positive and negative, has a unique representation
in base -10 (negadecimal) using just the digits 0 to 9 without needing a sign
For instance,
0 to 9 are as normal
10 = 100−90 = 190_{−10}
11 to 19 are therefore
11=191_{−10},
12=192_{−10},
... ,
19=199_{−10}
What are the first 5 column headers in base −2?
Make a table of numbers from −12 to 12
in base −2.
n_{10}
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
n_{-2}
110100
110101
1010
1011
1000
1001
1110
1111
1100
1101
10
11
0
1
110
111
100
101
11010
11011
11000
11001
11110
11111
11100
Using Negative Digits
Instead of using powers of a negative number base to represent negative values, we could use negative 'digits'.
This is not as peculiar as it might at first sound. For example if the time is 2:50 then we can say "10 to 3"
which is in effect using the hour 3 and the minutes -10.
We can also see times expressed with minutes from 0 to 59 (base 60) on timetables for example and, when saying the time
the minutes go from -29 to +30 as in "29 minutes to 3" up to "3:30".
So for base β instead of using 'digits' 0 to β−1
we could just as effectively use 'digits' −β/2 to
+β/2 if β is odd (rounding the halves).
To make sure representations are unique when β is even we can decide to
use either−β/2or elseβ/2 but not both 'digits'.
Using two symbols for negative digits is inconvenient (is 3−5 two 'digits' or a subtraction?)
so
to use one single 'digit' per column and we can show negative 'digits' with the
negative sign placed above the 'digit' so that −3 is written as 3.
Base −10 using the 10 'digits': 4, 3, 2, 1, 0, 1, 2, 3, 4, 5
0 .. 5:
0_{−10} .. 5_{−10}
6 .. 15:
14_{−10} .. 15_{−10}
16 .. 25:
24_{−20} .. 25_{−20}
−4 .. −1:
4_{−10} .. 1_{−10}
−16 .. −5:
24_{−10} .. 15_{−10}
−26 .. −15:
34_{−10} .. 25_{−10}
123 is 100 −20 −3 = 77
It is easy to see if a value represented using nega-digits is positive or negative.
Show how
If the leftmost 'digit' is positive, the value is positive;
if the leftmost 'digit' is negative, the value is negative
Also, it is easy to negate a value in this system.
Show how
Change each positive 'digit' into its negative form and vice-versa.
How are -9 to 9 represented in base 3 using the 'digits' 1, 0 and 1?
n
0
1
2
3
4
5
6
7
8
9
n_{3}
0
1
11
10
11
111
110
111
101
100
m
0
-1
-2
-3
-4
-5
-6
-7
-8
-9
m_{3}
1
11
10
11
111
110
111
101
100
Rational Bases
Up to now we have only used whole numbers as our 'column headings' but always our 'digits' are whole numbers.
It is also possible to use column headings based on powers (a radix system) where the base is not a whole number but
a fraction (a rational) such as 3/2.
In base β, an integer,
when we add 1 to the units column and find a value equal to β, we take β away and add one (the carry) to the
next column to the left.
In a rational base, say 3/2, when we find 3 in any column we carry 2 to the column to the left. This applies
to any rational number β bigger than 1.
The 'digits' in rational base ^{m}/_{n} are the digits of base m, namely 0
to m−1.
1 =
1_{3/2}
1_{5/2}
1_{5/3}
2 =
2_{3/2}
2_{5/2}
2_{5/3}
3 =
20_{3/2}
3_{5/2}
3_{5/3}
4 =
21_{3/2}
4_{5/2}
4_{5/3}
5 =
22_{3/2}
20_{5/2}
30_{5/3}
6 =
210_{3/2}
21_{5/2}
31_{5/3}
7 =
211_{3/2}
22_{5/2}
32_{5/3}
8 =
212_{3/2}
23_{5/2}
33_{5/3}
9 =
2100_{3/2}
24_{5/2}
34_{5/3}
10 =
2101_{3/2}
40_{5/2}
310_{5/3}
11 =
2102_{3/2}
41_{5/2}
311_{5/3}
12 =
2120_{3/2}
42_{5/2}
312_{5/3}
Here is how we convert 11 to base 3/2:
Divide 11 by 3 noting the remainder:
3 ) 11
3 rem =
2
Multiply the quotient 3 by 2
and divide it by 3
3 ) 6
2 rem=
0
Multiply the quotient 2 by 2
and divide it by 3
3 ) 4
1 rem=
1
Multiply the quotient 1 by 2
and divide it by 3
3 ) 2
0 rem=
2
STOP because
The number is now 0 and
all future 'digits' will be 0.
Read the remainders from bottom to top
2102_{3/2}
Rational bases less than 1
For rational base β less than 1 (but bigger than 0), reverse the digits of the representation
in base β remembering to include the radix point. This is because (β)^{n} is (1/β)^{−n}.
For example:
12 = 312_{5/3} = 2.13_{3/5}
Use the Base Converter below to convert to and from rational bases.
Rational base 2 is not the same as base 4/2
Since our algorithm to convert a number to a rational base produces 'digits' of the numerator as a base, then
base 4/2 produces a number using 'digits from the list 0,1,2,3
whereas base 2/1 produces 'digits' from 0,1!
For example 23 in base 2 is 10111_{2}.
All conversions to base 2 are the same as to base 2/1.
But in base 4/2 we have 2023_{4/2} = 23_{10}.
In base p/q where p/q is in its lowest terms every number has a unique representation.
But in a non-simple rational base p/q = b (a base that is a fraction p/q but
p is a multiple of q and so the base simplifies to the number b)
every number from b upwards has more than one representation
In base 4/2, the column's values are 8=(4/2)^{3}, 4=(4/2)^{2}, 2=4/2 and 1
and the digits we can use are 0, 1, 2, 3. So 8 has 5 representations:
8 = 3×2 + 2×1 = 32_{4/2}
8 = 1×4 + 1×2 + 2×1 = 112_{4/2}
8 = 1×4 + 2×2 = 120_{4/2}
8 = 2×4 = 200_{4/2}
8 = 1×8 = 1000_{4/2}
What we have found are the 5 partitions of 8 into bags (or multisets which are
sets but with repetition allowed) using the
parts 1,2,4 and 8 (columns of base 2, the base in simplest terms) up to a maximum of 3 times each (the remainders mod 4, the base numerator).
Some Combinatorics Of Rational Base Representations
T Edgar, H Olafson, J Van Alstine, preprint (2017)
pdf
Proof that if p and q have no common factor then every number has a unique representation in base p/q.
Negative Rational Bases
We ensure that all remainders are non-negative when dividing a number by the numerator.
The denominator of the base is always a positive number.
For example, 5 ÷ −4 is −1 with remainder 1
because 5 = −1×−4 + 1.
RULE: For all b>0 : a ÷ b = q remainder r
if and only if
a = b×q + r and 0≤r<b
Here is how we convert 5 to base −4/3 by repeatedly dividing by −4 (the numerator),
noting the reminder, then multiplying the quotient by 3 (the denominator):
Divide 5 by −4 noting the remainder:
−4 ) 5 −1 rem =
1
Multiply the quotient −1 by 3
and divide it by −4
−4 ) −3 1 rem =
1
Multiply the quotient 1 by 3
and divide it by −4
−4 ) 3 0 rem =
3
STOP because
The number is now 0 and
all future remainders will be 0.
Read the remainders from bottom to top
Convert 23 to bases 6/3 and 8/4.
Can you find a method of converting easily from one fractional base to
another whose fraction simplifies to the same number?
Make a table of the number of base 3/2 digits in numbers 0 to 30.
Find a recursive formula for the digit count.
Length is 1 for numbers 0,1,2,3
Length is 2 = length(2_{3/2})+1 for 3,4,5
Length is 3 = length(4_{3/2})+1 for 6,7,8
Length of n is length(2 Floor(n/3))+1 in general.
See A246435
The number of digits in number n is one more than the number in 2 Floor(n/3) where
for positive n, the Floor(n) is n if n is an integer otherwise it is the nearest integer below n.
For positive n, the floor of n is just n without any digits after its decimal point.
How many numbers have 1 digit in base 3/2? How many have exactly 2 digits? and length 3?
3 have 1 digit, 3 have 2, 3 have 3 ....
The list of the counts is
3, 3, 3, 6, 9, 12, 18, 27, 42, ... A081848
Which is the smallest number to have exactly n base 3/2 digits?
The smallest with 1 digit is 1 (or 0), with 2 digits is 3, with 3 digits in 6:
1, 3, 6, 9, 15, 24, 36, ... A070885
Which is the smallest even number to have n digits in base 3/2?
List them with their base 3/2 representations.
What simple pattern connects the representations?
The base 3/2 reps are
n_{3/2}
n
2
2
21
4
210
6
2101
10
21011
16
210110
24
2101100
36
21011000
54
They all are extensions of the previous one, adding one new digit on the right
A303500.
The infinite sequence of which each representation is an initial part is A304273
and the numbers themselves are in A305498
What is the largest even number that can be written with n base 3/2 digits?
What are their base 3/2 representations?
n
n_{3/2}
2
2
4
21
8
212
14
2122
22
21221
34
212211
52
2122111
80
21221112
The numbers are A305497 their representations are A304272
Each is a single digit extension of the previous one so that their representations
are the initial part of the infinite sequence
21221112212112212121.... A304274
Irrational Bases
But what if our column headings were powers not of a whole number nor even a rational number but
of an irrational number, such as Phi, the golden section number or √2? This also works
with some conditions on the base and the 'digits'.
Let's look at an example: base Phi.
Base Phi (Φ)
Phi is 1.6180339... = (√5 + 1)/2, one of the golden section numbers (or golden mean or golden ratio).
Because Phi^{2} = Phi + 1 is a definition of the positive number Phi, we only need "digits"
0 and 1, called "phigits"!
The basic Phi rule is Phi^{n+2} = Phi^{n+1} + Phi^{n}
All natural numbers are representable in Base Phi using phigits 0 and 1 but we will need negative powers
and so we need a "base point" to act like the decimal point in base 10.
Just as for Fibonacci base numbers above, we need never have two 1's next to each other since they combine to give a 1
in the next column to the left using the basic Phi rule.
Here Phi = 1·6180339... = phi^{–1}
and its reciprocal (1/Phi) is also the same as Phi−1 which onthese pages is denotes
by the lower case Phi: phi = 0·6180339... = Phi – 1 = 1/Phi = Phi^{–1}
Column
Phi power
phi power
A + B Phi
C + D phi
real value
5
Phi^{5}
phi^{-5}
3 + 5 Phi
8 + 5 phi
11·0901699..
4
Phi^{4}
phi^{-4}
2 + 3 Phi
5 + 3 phi
6·8541019..
5
Phi^{3}
phi^{-3}
1 + 2 Phi
3 + 2 phi
4·2360679..
2
Phi^{2}
phi^{-2}
1 + 1 Phi
2 + 1 phi
2·6180339..
1
Phi^{1}
phi^{-1}
0 + 1 Phi
1 + 1 phi
1·6180339..
0
Phi^{0}
phi^{0}
1 + 0 Phi
1 + 0 phi
1·0000000..
-1
Phi^{-1}
phi^{1}
-1 + 1 Phi
0 + 1 phi
0·6180339..
-2
Phi^{-2}
phi^{2}
2 - 1 Phi
1 - 1 phi
0·3819660..
-3
Phi^{-3}
phi^{3}
-3 + 2 Phi
-1 + 2 phi
0·2360679..
-4
Phi^{-4}
phi^{4}
5 - 3 Phi
2 - 3 phi
0·1458980..
-5
Phi^{-5}
phi^{5}
-8 + 5 Phi
-3 + 5 phi
0·0901699..
For instance 2 is 10.01_{Phi} = Phi + Phi^{−2}.
1..13 in base Phi
1
1
.
_{Phi}
2
10
.
01_{Phi}
3
100
.
01_{Phi}
4
101
.
01_{Phi}
5
1000
.
1001_{Phi}
6
1010
.
0001_{Phi}
7
10000
.
0001_{Phi}
8
10001
.
0001_{Phi}
9
10010
.
0101_{Phi}
10
10100
.
0101_{Phi}
11
10101
.
0101_{Phi}
12
100000
.
101001_{Phi}
13
100010
.
001001_{Phi}
phi = 1/Phi = Phi-1 = 0.6180339... = (√5-1)/2 and the basic phi rule
is phi^{n} = phi^{n+1} + phi^{n+2}.
Because Phi^{−n} = phi^{n}
it is easy to see that to get a base phi representation from a base Phi representation
we merely reverse the order of the phigits
including the base point:
A Number System with Irrational Base, George
Bergman, Mathematics Magazine 1957, Vol 31, pages 98-110, where he
also gives pencil-and-paper
methods of doing arithmetic in Base Phi.
C. Rousseau The Phi Number System Revisited
in Mathematics Magazine 1995, Vol 68, pages 283-284.
Other irrational bases: √n
If n is not a square number then
the even powers of √n are just the powers of n so base √n includes all the representations
of integers in base n.
Columns for base β=√n
...
β^{6}
β^{5}
β^{4}
β^{3}
β^{2}
β^{1}
β^{0}
...
n^{3}
n^{5/2}
n^{2}
n^{3/2}
n
n^{1/2}
1
...
n^{3}
n^{2}√n
n^{2}
n√n
n
√n
1
The table shows that
the √n values are in the odd-power places
the purely integer values are in the even-power places
Because √n is itself not rational, the only values we can represent in base √n
are A + B√n for whole numbers A and B.
If we represent A in base n then shifting all its digits one place left will effectively multiply the value by the base, √n.
So any value A √n will only have non-zero digits in the even-power places and always 0s in the odd-power places.
We can then 'add' in the digits of B in base √n which similarly will have only
0s in the odd-power places and any non-zero digits will be in even-powered places.
5 = 101_{2}
which in base √2 becomes 5 = 10001_{√2} = 10001_{√2}
Similarly 7 = 111_{2}
which in base √2 becomes 7 = 10101_{√2} = 10101_{√2}
Shifting the digits of 7's representation one place to the left multiplies the value by √2: 7√2 = 10101 0_{√2}
Adding the representations of 5 and 7√2 is easy because there are never any 'carries':
Columns
4√2
2^{2}
2√2
2
√2
1
5
=
1
0
0
0
1
7 √2
=
1
0
1
0
1
0
5 + 7√2
=
1
1
1
0
1
1
5 + 7√2 = 111011_{√2}
Base Converter
About the Base Converter Calculator:
Numeric bases
Bases can be any integer, positive or negative except -1, 0 and 1
fractional bases are input as N/M with the "/" symbol separating the two whole numbers N and M.
Other bases
F which is the same as Fmin: the Fibonacci base with the minimum number of 1s known as the
Zeckendorf representation
Fmax: Fibonacci base but using the maximum number of 1s
Fnegi: Fibonacci Negative Index with the indices increasing to the right.
Phi Phi=1.618...=(√5+1)/2. The same as Phimin: base Phi with the minimum number of 1s (...11... never occurs)
Phimax base Phi but using the maximum number of 1s (...00... never occurs)
phi phi=0.618...=(√5−1)/2. The same as phimin: base phi with the minimum number of 1s (...11... never occurs)
phimax base phi but using the maximum number of 1s (...00... never occurs)
!: Factorial base
BinomN where N is a whole number. The N can be left out if converting
from a Binomial representation (as it is the length of the list of 'digits')
but is needed if converting to a Binomial base.
'Digits':
'Digits' in between { and } brackets can be any integers, including negative values
for example {12,-3,0} . The { } may not be omitted.
A negative number using { } is written with the − before the { bracket
for example -{12,3} -123 is acceptable and means the same as
-{1,2,3} but is not the same as
{-1,2,3}
Radix point:
A 'decimal point', called a radix point in bases other than 10,
if using { } notation is inserted just as it is, for example {1,2,.,3,4}
in Fibonacci bases, we use only the columns indexed from 2 upwards corresponding to Fibonacci numbers ..8,5,3,2,1.
If a radix point is used, 'digits' after the radix point refer to Fibonacci numbers with indices -2,-3,-4,...
and the radix point separates the positive indexed digits from the negative:
i
5
4
3
2
1
0
-1
-2
-3
-4
-5
●
Fib(i)
5
3
2
1
1
0
1
-1
2
-3
5
Different answers?
There is no check on the validity of 'digits' when converting to base 10:
321_{2} means 3×2^{2} + 2×2^{1} + 1 = 17
and similarly for the other bases.'Digits' can be negative or any whole number.
However on converting from base 10 to another base, the 'digits' will be valid for that base.
Converting 17 to base 2 will give 10001_{2}
and similarly for the other bases
Make a table
To make a table of conversions of a number to a range of bases, put the number in the from box
and list the bases separated by a comma.
Click on
to make a table of a numbers converted to one base, fill in the from and to input boxes and
put the base in the converted to these bases box.
Click on
You can also specify a range of numbers and a list of bases to convert them all to.
Need more space for numbers?
Resize the Result box and the input boxes for conversion to and from base 10
by grabbing a corner with the mouse.
On devices without this facility:
Use the
and
buttons to
expand or decrease the input box height
We can again extend our possible bases in a radix system to complex numbers and find a unique way to
represent integers and other values.
...... more coming soon .....
The Chinese Remainder Theorem Number Base
This is a system which has many computational advantages, particularly for computers than can run parts of
a computation in parallel
since each digit is independent of the others.
.... coming soon .....
Monotonic numbers
For a given base, Monotonic numbers are those that have all their base digits already in order, from least to most, left to right,
repetitions allowed. This is sometimes called non-decreasing order since any digit
must be less than or equal to the digit in the next
column to its right. In this section we investigate the monotonic numbers in base 10:
13 and 112 are monotonic but 103 and 192 are not monotonic.
You Do The Maths...
How many monotonic numbers are there between 1 and 10?
How many monotonic numbers are there between 10 and 100?
How many monotonic numbers are there between 100 and 1000?
1 to 9: all are monotonic: 9 numbers
10 to 100:
11 to 19, 22 to 29, ... 88-89, 99: 9+8+7+6+5+4+3+2+1 = 45 numbers
13 and 112 are monotonic numbers. 13×112= 1456 is also monotonic
33366667 and 333667 are monotonic and their product is 33366667×333667 = 11122255677889, also monotonic.
Here is a method to find pairs of monotonic numbers whose product is also monotonic:
Take two numbers each of which is either composed solely of 3s or has only 6s as its digits or with some 3s first followed by some 6s.
Add 1 to both numbers
Their product is monotonic
For example: (33+1)×(366+1) = 34×367 = 12478 a monotonic number.
Can you find any other examples that are not formed by the above method?
All are of the form above: see the Blecksmith and Nicol Reference below
5, 35 and 335 are monotonic.
Their squares are also monotonic: 5^{2} = 25, 35^{2} = 1225 and 335^{2} = 112225.
What is the pattern here?
5, 35, 335, 3335, 33335, ... any number of 3s followed by 5
have a monotonic square.
See the Blecksmith and Nicol Reference below
There is another similar pattern starting with 17. What is it?
17, 167, 1667, 16667, ... begin with 1, then any number of 6s and a final 7
all have monotonic squares.
This series and the one above are the only numbers with monotonic squares.
See the Blecksmith and Nicol Reference below
References
Monotonic numbers R Blecksmith, C Nicol,
Maths Mag vol 66 (1993) pages 257-262
Links and References
An Introduction to Computational Combinatorics
E S Page, L B Wilson (1979 Cambridge)
see chapter 5 on several ways of ordering permutations as well as the factorial representation method of
generating them in lexicographic order.
Some theorems on Completeness V E Hoggatt, B Chow Fib Quarterly, vol 10 (1972)
pages 551-554,
560.
How can we decide if a given sequence of numbers could be used as the column headings of a base to
represent every whole number? This paper gives two theorems.
Systems of Enumeration A Fraenkel, Amer. Math. Monthly vol 92 (1985), pages 105-114
JSTOR
A proof of the existence and uniqueness of representations of positive integers in any system of "column headers"
described by a recurrence relation. This includes base β (powers of β) and mixed radix, factorial base, our Pascal's
triangle combinatorial system above as well as various systems based on the Fibonacci numbers.
Powers of rationals modulo 1 and rational base number systems
S Akiyama, C Frougny, J Sakarovitch
Israel J Math vol 168 (2008), pages 53-91
pdf
Some Combinatorics of Factorial Base Representations
T Ball, P Dalenberg, J Beckford, T Edgar, T Rajabi
Journal of Integer Sequences Vol. 23 (2020), Article 20.3.3
pdf