João's scientific calculator has a special key that transforms any number $x$ displayed on the screen and that is different from 1 into the number $\frac{1}{1-x}$.

a) What happens if the number 2 is displayed on the screen and we press the special key three times?

b) What happens if the number 2 is displayed on the screen and we press the special key ten times?

c) Finally, what happens if the number 2 is displayed on the screen and we press the special key 2015 times?
Solution

a) After pressing the key three times, we obtain:

$$
2 \xrightarrow{1^{st}} \frac{1}{1-2}=-1 \xrightarrow{2^{nd}} \frac{1}{1-(-1)}=\frac{1}{2} \xrightarrow{3^{rd}} \frac{1}{1-1 / 2}=2
$$

b) Due to the previous item, every three presses of the special key, it is as if the number 2 had not been changed. Thus, after the sixth and ninth use of the special key, the number 2 will still be on the screen. Finally, with the tenth use of the special key, it will be transformed into $\frac{1}{1-2}=-1$. This repetition pattern is not specific to the number 2, as shown by the sequence:

$$
x \xrightarrow{1^{st}} \frac{1}{1-x} \xrightarrow{2^{nd}} \frac{1}{1-\frac{1}{1-x}}=-\frac{1-x}{x} \xrightarrow{3^{rd}} \frac{1}{1-\left(-\frac{1-x}{x}\right)}=x
$$

c) Since every three uses of the special key the number 2 will remain on the screen, after any number of uses that is a multiple of 3, the number 2 will still be there. Since 2013 is a multiple of 3, we only need to analyze the last two presses:

$$
\ldots \xrightarrow{2013^{rd}} 2 \xrightarrow{2014^{th}} \frac{1}{1-2}=-1 \xrightarrow{2015^{th}} \frac{1}{1-(-1)}=\frac{1}{2}
$$

Therefore, the number $\frac{1}{2}$ will remain on the screen.