Fractional Bases.

Let the initial decimal integer be $k_0$ and the improper fractional base be $\dfrac pq$ , where $p$ and $q$ are positive coprime integers with $p>q$ . Then the $j$ th digit $a_j$ is given by:

$a_j = k_j \bmod p \quad \small \text{where } k_j = q \left \lfloor \frac {k_{j-1}}p \right \rfloor \text{ for }j \ge 1$

Let us use the formulas above to solve for $k_0 = 37$ , $p=4$ , and $q=3$ .

\(\begin{array} {} j= 0 & k_0 =37 & \implies a_0 = 37 \bmod 4 = 1 \\ j = 1 & k_1 = 3 \left \lfloor \dfrac {37}4 \right \rfloor = 27 & \implies a_1 = 27 \bmod 4 = 3 \\ j = 2 & k_2 = 3 \left \lfloor \dfrac {27}4 \right \rfloor = 18 & \implies a_2 = 18 \bmod 4 = 2 \\ j = 3 & k_3 = 3 \left \lfloor \dfrac {18}4 \right \rfloor = 12 & \implies a_3 = 12 \bmod 4 = 0 \\ j = 4 & k_4 = 3 \left \lfloor \dfrac {12}4 \right \rfloor = 9 & \implies a_4 = 9 \bmod 4 = 1 \\ j = 5 & k_5 = 3 \left \lfloor \dfrac 94 \right \rfloor = 6 & \implies a_5 = 6 \bmod 4 = 2 \\ j = 6 & k_6 = 3 \left \lfloor \dfrac 64 \right \rfloor = 3 & \implies a_6 = 3 \bmod 4 = 3 \\ j = 5 & k_7 = 3 \left \lfloor \dfrac 34 \right \rfloor = 0 & \implies a_7 = 0 \bmod 4 = 0 \end{array} \)

$\implies 37_{10} = 1320123_\frac 43$ , Similarly, $37_{10} = 1011012_\frac 32$ . Therefore, $\displaystyle \sum_{j=0}^6 a_j + \sum_{j=0}^6 b_j = 12+6 = \boxed{18}$ .

The above formula or algorithm can be readily implemented with a Microsoft Excel spreadsheet, using functions =INT( _ ) for $\lfloor \cdot \rfloor$ , the floor function and =MOD( _ , _ ) for $\cdot \bmod \cdot$ , the modulo operation . My spreadsheet is as follows. And $123456789_{10} = \boxed{\text{B6C23A75ACC672C6A642CC9F11D9061CFC3C6EA5B5 }}_{\frac {16}{11}}$ .

The simple algorithm below generates the conversion:

To calculate: $[M]_{\frac{a}{b}}$ , where $(a > b)$ .

Let $S_{1} = M$

Repeat

$r_{n} = S_{n} \mod a$

$q_{n} = S_{n} \:\ div \:\ a$

$S_{n + 1} = b * q_{n}$

Go until $S_{n + 1} < a$ .

Using the algorithm above:

For $(37)_{\frac{4}{3}}$ :

$\boxed{37} \rightarrow \boxed{27} \boxed{1} \rightarrow \boxed{18} \boxed{3} \boxed{1} \rightarrow \boxed{12} \boxed{2} \boxed{3} \boxed{1} \rightarrow \boxed{9} \boxed{0} \boxed{2} \boxed{3} \boxed{1} \rightarrow \boxed{6} \boxed{1} \ \boxed{0} \boxed{2} \boxed{3} \boxed{1} \rightarrow \boxed{3} \boxed{2} \boxed{1} \ \boxed{0} \boxed{2} \boxed{3} \boxed{1}$

For $(37)_{\frac{3}{2}}$ :

$\boxed{37} \rightarrow \boxed{24} \boxed{1} \rightarrow \boxed{16} \boxed{0} \boxed{1} \rightarrow \boxed{10} \boxed{1} \boxed{0} \boxed{1}\rightarrow \boxed{6} \boxed{1} \boxed{1} \boxed{0} \boxed{1} \rightarrow \boxed{4} \boxed{0} \boxed{1} \boxed{1} \boxed{0} \boxed{1} \rightarrow \boxed{2} \boxed{1} \boxed{0} \boxed{1} \boxed{1} \boxed{0} \boxed{1}$

$$

$\implies \sum_{j = 0}^{6} a_{j} + \sum_{j = 0}^{6} b_{j} = \boxed{18}$ .

$$

$$

Below is a program I wrote in Free Pascal::

program fractionalbase;

var myfile:text;

$$

procedure change(var k:string);

begin IF K = '10' THEN K:= 'A';

IF K = '11' THEN K:= 'B';

IF K = '12' THEN K:= 'C';

IF K = '13' THEN K:= 'D';

IF K = '14' THEN K:= 'E';

IF K = '15' THEN K:= 'F';

end;

$$

function fractbase(n,a,b:longint):string;

type myarray = array[1 .. 100] of longint;

var q,s,count,m:longint;

r:myarray;

strvar,strsum:string;

$$

begin

s:= n;

count:= 1;

repeat

r[count]:= s mod a;

q:= s div a;

s:= b * q;

count:= count + 1;

until(s < a);

r[count]:= s mod a;

strsum:= '';

for m:= 1 to count do

begin

str(r[m],strvar);

if r[m] > 9 then

change(strvar);

strsum:= strvar + strsum;

end;

fractbase:= strsum;

end;

$$

begin

assign(myfile,'fract.txt');

rewrite(myfile);

writeln(myfile,fractbase(123456789,16,11));

close(myfile);

readln;

end.

$$

Running the above program we obtain:

$(123456789)_{10} = \boxed{(B6C23A75ACC672C6A642CC9F11D9061CFC3C6EA5B5)_{\bf\frac{16}{11}}}$

$$

$$

Below is a program I wrote in Python:

The module below contains the majority of the program: I wrote the module so I could use the function when needed.

import math;

$$

def power(base,exponent):

product = 1;

n=1;

while n <= exponent:

product = product * base;

n = n + 1;

power = product

return power;

$$

def change(k):

if k == 10:

  k = 'A'

if k == 11:

  k = 'B'

if k == 12:

  k = 'C'

if k == 13:

  k = 'D'

if k == 14:

  k = 'E'

if k == 15:

  k = 'F'

if k == 16:

  k = 'G'

change = k

return change;

$$

import array

array.array('i')

r = array.array('i',(0 for i in range(0,1000)))

$$

$$

def fractbase(n,a,b):

s = n;

count = 1;

while s >= a:

r[count] = s % a;

q = math.floor(s/a);

s = b * q;

count = count + 1;

r[count] = s % a;

strsum = '';

for m in range(-count,0):

if r[-m] > 9:

      strsum = strsum + change(r[-m])

else:

      strsum = strsum + str(r[-m]);

fractbase = strsum;

return fractbase;

$$

Calling the function::

$$

print(mathunit.fractbase(123456789,16,11));

$$

Running the above program we obtain:

$$

$(123456789)_{10} = \boxed{(B6C23A75ACC672C6A642CC9F11D9061CFC3C6EA5B5)_{\bf\frac{16}{11}}}$