You may also like

problem icon

Tweedle Dum and Tweedle Dee

Two brothers were left some money, amounting to an exact number of pounds, to divide between them. DEE undertook the division. "But your heap is larger than mine!" cried DUM...

problem icon

Lower Bound

What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =

problem icon

Sum Equals Product

The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 1 [1/3]. What other numbers have the sum equal to the product and can this be so for any whole numbers?

More Twisting and Turning

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

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 thiscomprehensive solution. Thank you Oliver.