Is 2/9 = 0 in 2-adic?

Question

If my math checks out we have that the 2-adic integers:

$$...0101010101 = 1 + 4 + 16 + ... = 1/(1-4) = -1/3$$

and

$$...1010101010 = 2 + 8 + 32 + ... = 2/(1-4) = -2/3$$

(which makes sense to me because multiplying by 2 "shifts" a 2-adic number left.)

Since p-adic multiplication is done component-wise, I expect:

$$...1010101010 × ...0101010101 = ...0000000000$$

Analogously we would have that:

$$-1/3 × -2/3 = 2/9 = 0$$

I tried to check my work by writing out the expansion of $2/9$, but:

$$\frac{2}{9} = \frac{2}{(1-(-8))} = 2 - 16 + 128 - 1024 + ... =$$ equal to what? I don't know how to deal with the negative terms.

Is $\frac{2}{9} = 0$ in $2$-adic?

Answer 1

The product ...0101 times ...1010 should be computed in column, in the usual way:

...0101 * 0 =  ...0000000
...0101 * 1 =  ...010101 
...0101 * 0 =  ...00000  
...0101 * 1 =  ...0101   
            .            
            .            
-------------------------
            =  ...1010010

the result, if carefully computed, should be $\frac 2 9$.

Answer 2

As a way to check the direct multiplication given in Emanuele Paolini's answer, you can compute

$$ -\frac19 = \frac{7}{1-64} = \ldots000111000111 $$

We can subtract this from $0$ to get

$$ \frac19 = 0 - \left(-\frac19\right) = \ldots111000111001 $$

and then shift to double, yielding

$$ \frac29 = 2\left(\frac19\right) = \ldots1110001110010 $$

which is indeed the result of the direct multiplication given in Emanuele's answer. Note that multiplication by $1001_2 = 9$ gives just $10_2 = 2$.