No Division Implies Chaos

Let $I$ be a closed interval in ${R^1}$ and $f:I \to I$ be continuous. Let ${x_0} \in I$ and \[ {x_{i + 1}} = f({x_i})\quad {\text {for}}\;i > 0.\] We say there is no division for $({x_1},{x_2}, \ldots ,{x_n})$ if there is no $a \in I$ such that ${x_j} < a$ for all $j$ even and ${x_j} < a$ for all $j$ odd. The main result of this paper proves the simple statement: no division implies chaos. Also given here are some converse theorems, detailed estimates of the existing periods, and examples which show that, under our conditions, one cannot do any better.