Consider the sequence $\{a_n\}$, such that each item of the sequence is either $0$ or $1$. For any positive integer $i$, let $$f(i)=\dfrac{a_1+a_2+\dots+a_i}{i}$$
Suppose that for some positive integer $k$, $f(k)<4/5$, and for some positive integer $m>k$, $f(m)>4/5$.
Prove that there exist a positive integer $n$, such that $a_n=4/5$.
Suppose that there does not exist such an $n$, then there must be a point in the sequence where it increase from below $4/5$ to above $4/5$, so there is a value $j$ ($k<j<m$) suh that \begin{eqnarray*} \frac{a_1+\cdots+a_j}{j} < \frac{4}{5} < \frac{a_1+\cdots+a_{j}+1}{j+1} \end{eqnarray*} Rearrange this to obtain \begin{eqnarray*} 5(a_1+\cdots+a_j) < 4j < 5(a_1+\cdots+a_j)+1. \end{eqnarray*} Each of these values is a whole number, but the first and last differ by $1$, which is a contradiction.