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# 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.

Invariance
Easy
##### Suggested Book
Problem Solving Strategies by Arthur Engel

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 $n$th 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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