+118608 \(\left(\frac{1}{2}\right)_{base 10}\\ = \frac{1}{3}+\frac{1}{6}\\ = \frac{1}{3}+\frac{1.5}{9}\\ = \frac{1}{3}+\frac{1}{9}+\frac{0.5*3}{27}\\ = \frac{1}{3}+\frac{1}{3^2}+\frac{1.5}{3^3}\\ = \frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1.5}{3^4} \dots \\ =0.1111...._3\\ =0.\dot 1_{base 3}\)
there might be a better way of doing it, I am not sure about that.
