Understand the problem

Azambuja writes a rational number $q$ on a blackboard. One operation is to delete $q$ and replace it by $q+1$; or by $q-1$; or by $\frac{q-1}{2q-1}$ if $q \neq \frac{1}{2}$. The final goal of Azambuja is to write the number $\frac{1}{2018}$ after performing a finite number of operations. Show that if the initial number written is $0$, then Azambuja cannot reach his goal.

Source of the problem
Brazilian national mathematical olympiad 2018
Topic
Invariance
Difficulty Level
Easy
Suggested Book
Problem Solving Strategies by Arthur Engel

Start with hints

Do you really need a hint? Just try it yourself!

It is always a good idea to try using the invariance principle in such problems.
Make a change of variables to see patterns.
Note that the operation is restricted to rational numbers. Hence, writing q=\frac{r}{s}=(r,s) could help.
Let us denote by q_n the number on the board at the nth step. We shall use the new variable a_n=2q_n-1 (just to simplify the denominator). Clearly, a_{n+1} is either a_n\pm 2 or -\frac{1}{a_n}. Writing a_n=(r_n,s_n), this means that (r_{n+1},s_{n+1}) is either (r_n \pm 2s_n,s_n) or (-s_n,r_n). Thus, we need to find out if (-1008,1009) is reachable starting from (-1,1). However, (odd, odd) pairs can produce only other (odd,odd) pairs under this operation, and (-1008,1009) is an (even, even) pair. Hence \frac{1}{2018} cannot be reached starting from 0.

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