Sketch: Think of long division. Note that you can then set up inequalities. After calculating a bit we should get, that some integer is in some interval $I$. Now prove that there can't be an integer in that interval using $\pmod{10000}$.

Anyone can solve this?

My sketch would work, but this is a great proof given in the official solutions: Let the digits $7143$ show up at the $(k+1)$'th digit after the comma, so there exists a $r \in \mathbb{Z}_{\geq0}$ with \[ r+ 0.7143 \leq 10^k \cdot \frac{m}{n} < r + 0.7144. \]Multiplication by $7n$ yields \[ 7nr + 5.0001n \leq 7m \cdot 10^k < 7nr + 5.0008n, \]so \[ 1 \leq 7m \cdot 10^k - (7n+5)n < 0.0008n. \]But $1 < 0.0008n$ is equivalent to $1250 < n$.

What did you do after you multiplied by 7n?

Which step is not clear? I'm subtracting $7nr-5n$ from each term. (Perhaps the missing $n$ at the end of the inequality chain confused you - it's fixed now.)

muriloogps wrote: Anyone can solve this?

What do you mean? There is a full proof in Answer #4...