Why can we convert a base $9$ number to a base $3$ number by simply converting each base $9$ digit into two base $3$ digits?

Why can we convert a base $9$ number to a base $3$ number by simply
converting each base $9$ digit into two base $3$ digits ?

For example $813_9$ can be converted directly to base $3$ by noting

$$\begin{array} \space 8_9&=22_3 \\ \space 1_9 &=01_3 \\ \space 3_9 &=10_3 \\ \end{array}$$

Putting the base digits together ,we get

$$813_9=220110_3$$

I know it has to do with the fact that $9=3^2$ but I am not able to understand this all by this simple fact...

Answer

Consider $N$ in base 3. For simplicity, we can assume that $N_3$ has an even number of digits: if it doesn't, just tack on a leftmost $0$. So let:

$$N_3 = t_{2n+1} t_{2n}\dotsc t_{2k+1} t_{2k} \dotsc t_1 t_0.$$
What this positional notation really means is that:
$$N = \sum_{i = 0}^{2n+1} t_i 3^i,$$
which we can rewrite as:
$$\begin{align} N &= \sum_{k = 0}^{n} (t_{2k+1} 3^{2k+1} + t_{2k} 3^{2k}) \\ &= \sum_{k = 0}^{n} (3 t_{2k+1} + t_{2k}) 3^{2k} \\ &= \sum_{k = 0}^{n} (3 t_{2k+1} + t_{2k}) 9^{k}. \\ \end{align}$$

But now, note that for each $k$, $3 t_{2k+1} + t_{2k}$ is precisely the base-9 digit corresponding to the consecutive pair of base-3 digits $t_{2k+1} t_{2k}$.