Originally Posted by

**Soroban** Hello, poorna!

I'm not familiar with Cantor sets,

. . but here is a primitive conversion . . .

Suppose we want to convert $\displaystyle 0.8$ to ternary.

We assume that the ternary decimal* is of the form: .$\displaystyle 0.abcd{e}\!f\!g\hdots$

. So we have: .$\displaystyle 0.8 \;=\; 0.abc{d}e\!f\!g\hdots$ . . [Left side: base 10. .Right side: base 3]

Multiply by 3: .$\displaystyle 2.4 \;=\;a.bc{d}e\!f\!gh\hdots \quad\Rightarrow \quad \boxed{a \:=\:2}$

. So we have: .$\displaystyle 0.4 \:=\:0.bc{d}e\!f\!gh\hdots$

Multiply by 3: .$\displaystyle 1.2 \:=\:b.c{d}e\!f\!gh\hdots \quad\Rightarrow\quad \boxed{b \:=\:1}$

.So we have: .$\displaystyle 0.2\:=\:0.c{d}e\!f\!gh\hdots$

Multiply by 3: .$\displaystyle 0.6 \:=\:c.{d}e\!f\!gh\hdots \quad\Rightarrow\quad \boxed{c \:=\:0}$

.So we have: .$\displaystyle 0.6 \:=\:0.{d}e\!f\!gh\hdots$

Multiply by 3: .$\displaystyle 1.8 \:=\:d.e\!fgh\hdots \quad\Rightarrow\quad \boxed{d \:=\:1} $

. So we have: .$\displaystyle 0.8 \:=\:0.e\!fgh\hdots$ . . . . . and the cycle repeats.

Therefore: . $\displaystyle 0.8_{10} \;=\;0.\overline{2101}_3$