We can form sequences from an initial number using two operations " $+1=$ add 1 " and " $-i=$ minus the inverse". For example, starting with the number 3, we can form several sequences, see one of them:

$$
3 \xrightarrow{+1} 4 \xrightarrow{+1} 5 \xrightarrow{-i}-\frac{1}{5} \xrightarrow{+1} \frac{4}{5} \xrightarrow{-i}-\frac{5}{4} \xrightarrow{+1}-\frac{1}{4} \xrightarrow{+1} \frac{3}{4} \xrightarrow{-i}-\frac{4}{3} .
$$

Starting with 0, with which sequence will we obtain 0 again, using only the two operations " +1 " and " $-i$ "?

## List 3
To obtain 0, the sequence must end as:

$$
-2 \xrightarrow{+1}-1 \xrightarrow{+1} 0
$$

A requested sequence is as follows:

$$
\begin{aligned}
& 0 \xrightarrow{+1} 1 \xrightarrow{+1} 2 \xrightarrow{+1} 3 \xrightarrow{-i}-\frac{1}{3} \xrightarrow{+1} \frac{2}{3} \xrightarrow{+1} \frac{5}{3} \xrightarrow{+1} \frac{8}{3} \xrightarrow{-i}-\frac{3}{8} \xrightarrow{+1} \frac{5}{8} \xrightarrow{+1} \frac{13}{8} \xrightarrow{+1} \frac{21}{8} \\
& \xrightarrow{-i}-\frac{8}{21} \xrightarrow{+1} \frac{13}{21} \xrightarrow{-i}-\frac{21}{13} \xrightarrow{+1}-\frac{8}{13} \xrightarrow{+1} \frac{5}{13} \xrightarrow{-i}-\frac{13}{5} \xrightarrow{+1}-\frac{8}{5} \xrightarrow{+1}-\frac{3}{5} \xrightarrow{+1} \frac{2}{5} \\
& \xrightarrow{-i}-\frac{5}{2} \xrightarrow{+1}-\frac{3}{2} \xrightarrow{+1}-\frac{1}{2} \xrightarrow{+1} \frac{1}{2} \xrightarrow{-i}-2 \xrightarrow{+1}-1 \xrightarrow{+1} 0 .
\end{aligned}
$$

We have another much faster and simpler solution:

$$
0 \xrightarrow{+1} 1 \xrightarrow{-i}-1 \xrightarrow{+1} 0 .
$$

## List 3

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