In particular, a fraction that, in a certain base, will be recurring - such as $\frac12$ in base 3. (I used a base changing calculator to find it out)
I have tried using repeated multiplation as described by this site: http://www.mathpath.org/concepts/Num/frac.htm
However I don't quite understand the process, please describe it.
Your figure out fractions in another base just as you figure them out in base 10.
To figure out $a/b$ in base 8 you divide be into $a$ and take the remainder, multiply it by the base and divide $b$ into it and repeat.
Example $3/5$ in base $8$ is...
$5$ goes into $3$, $0$ times with $3$ remainder. So $0.rem3$. So we multiply the $3$ by $8$ to get $24$.
$5$ goes into $24$, $4$ times with $4$ remainder. So $0.4rem4$. So we multiply the $4$ by $8$ to get $32$.
$5$ goes into $32$, $6$ times with $2$ remainder. So $0.46rem2$. So we multiply the $2$ by $8$ to get $16$.
$5$ goes into $16$, $3$ times with $1$ remainder. So $0.463rem1$. So we multiply the $1$ by $8$ to get $8$.
$5$ goes into $8$, $1$ time with $3$ remainder. So $0.4631rem3$. But we started with 3 so this is just going to repeat indefinately.
So $3/5 = 0.\overline{4631}_8$.
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Why?
Because $3/5 = 0 + 3*8/5*8 = 0 + [5*4 + 4]/5*8=$
$0 + 4/8 + 4/5*8 = 0 + 4/8 + 4*8/5*8^2 = 0 + 4/8 + [5*6 + 2]/5*8^2 = 0 + 4/8 + 6/8^2 + 2/5*8^2=$
$0 + 4/8 + 6/8^2 + 2*8/5*8^3 = 0 + 4/8 + 6/8^2 + [5*3 + 1]/5*8^3=$
$0 + 4/8 + 6/8^2 + 3/8^3 + 1*8/5*8^4 = 0 + 4/8 + 6/8^2 + 3/8^3 + 1/8^4 + 3/5*8^5$
Just as $.1 = 1/10$ in base 10. $.1 = 1/8$ in base 8 so...
$ 0 + 4/8 + 6/8^2 + 3/8^3 + 1/8^4 + 3/5*8^5 = 0.4631 + 3/5*8^5 = 0.4631 + 0.00004631 + 0.000000004631 + .... = 0.\overline{4631}$.
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Furthermore $0.4631.... *8^4 = 4631.4631....$
So $(8^4 - 1)0.4631.... = 4631 = 4*8^3 + 6*8^2 + 3*8 + 1$
So $0.4631 = (4*8^3 + 6*8^2 + 3*8 + 1)/(8^4 -1)= (4*512 + 6*64 + 3*8 + 1)/(4096-1)= 2457/4095=3/5$.