Rachel thought through the problem like this:

For fractions

twist until reaching

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

For fractions
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
- n
m

: twist to
1
m

, turn to -m, twist to 0

and for
n
m

: turn to
- m
n

, twist to
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
- 21
13

, two twists take us to
- 8
13

and
5
13

, not
1
13

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