How can I convert directly between two non-power-of-ten bases?

[caters]

I can most of the time successfully convert between base 10 and another base or another base and base 10 or between 2 bases where one of them is a power of the other(like base 2 and base 4 or base 3 and base 9).

With negative bases I sometimes don't get what I want in that negative base and with fractional bases there are 2 expansions for every rational number greater than or equal to 1, They are an infinite sequence of digits to the left and a terminating or repeating decimal to the right. But how will I know if it is a repeating decimal and what repeats in that repeating decimal?

Like what if I wanted to represent 2/3 in base 1/2?

Also I don't know of an easy way to convert between 2 bases that are not powers of each other or base 10 and another base.

Here is how I convert between base 10 and another base:

1) $log_n(x) = y$ Here n is the base I want to convert to and x is the base 10 number I want to convert.

2) $y =$ some irrational number $y' =$ Just the integer and not the whole irrational

3) $\frac{x}{n^{y'}}$

4) Write down the integer part(which is equal to $z$) as the digit in the $n^{y'}$ place

5) $z*n^{y'}$

6)$x-(z*n^{y'})$

7) repeat until you reach the $n^0$ place

8) If needed extend it to decimals in base $n$.

When I convert between 2 bases that are powers of each other I start from the right and make n groups of $log_x(y)$ digits where $x$ is the current base and $y$ is a power of that base. I then take those groups as if they were individual numbers themselves, add up their values and write down their values from right to left

But how can I convert directly between 1 base and another base when it is neither one of these 2 previous scenarios like for example converting directly from base 2 to base 3?

 

[geoffrey159]

If you want to convert a positive integer from base 10 to base b (b integer greater than 1), you can do it by successive euclidean divisions :

$n = b q_0 + r_0$
$q_0 = b q_1 + r_1$
...
$q_{k-1} = b q_k + r_k$
...
until the quotient is 0.
In the end, $n = \sum_{k=0}^m r_k b^k$. So, collecting the remainders gives you the conversion from base 10 to b : $(n)_b = (r_m...r_0)$

If you want to convert from base b to c, the only option I know (and I don't know much), you can go from base b to 10 and from base 10 to c.

Assume that you now you have a positive real number in base 10 with fractional part x, and you want to convert x to base b, so that $x = \sum_{k=1}^p s_k b^{-k}$. An algorithm that gives you the $s_k$ is :

$s_1 = \lfloor bx \rfloor, x \leftarrow bx - s_1$
...
$s_k = \lfloor b x\rfloor, x \leftarrow bx - s_k$
until $x=0$

But I have the feeling you already know that, and probably better than me

 

[HallsofIvy]

For the specific case "write 2/3 in base 1/2", start by noting that 2/3 is larger than 1/2. But 2/3- 1/2= 1/6 which is less than 1/2 so the first "digit" right of the "decimal" point is 1: it starts "0.1...". Now note that (1/2)^2= 1/4 is larger than 1/6. The next place is 0: we have "0.10...". (1/2)^3= 1/8 is less than 1/6 and 1/6- 1/8= 2/48= 1/24 is less than 1/2: now we have "0.101...". (1/2)^4= 1/16 is larger than 1/24 so the next place is 0: "0.1010...". Continue like that. Of course, this will be a non-terminating "decimal".

(I put words like "digit" and "decimal" in quotes because those names are, of course, derived from the Greek "deci" for one-tenth and don't really apply to other bases.)