Rachel thought through the problem like this:

For fractions

- \frac{n}{2}

twist until reaching

\frac{1}{2}

, then turn to get -2, and twist twice.

For fractions $\frac{2}{n}$ , start by turning once, then repeat as above.

This also works for 3 instead of 2, etc. and I think can be generalised:

for $- \frac{n}{m}$ : twist to $\frac{1}{m}$ , turn to -m, twist to 0

and for $\frac{n}{m}$ : turn to $- \frac{m}{n}$ , twist to $\frac{1}{n}$ , turn to -n, twist to 0

I think that this shows that any fraction can revert to 0.

Rachel's initial thinking regarding negative halves is absolutely correct.
Her follow-up argument is along the right lines:

if the fraction is negative, then twist (+1)

if the fraction is positive, then turn (-1/x)

However, it sometimes requires a longer sequence than the one that Rachel suggests.

You cannot guarantee that you will always arrive at a unit fraction (numerator of 1) after a sequence of twists:

e.g. starting at $- \frac{21}{13}$ , two twists take us to $- \frac{8}{13}$ and $\frac{5}{13}$ , not $\frac{1}{13}$

Oliver from Olchfa School sent us this comprehensive solution. Thank you Oliver.