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This is a repost of https://math.stackexchange.com/questions/732343/period-3dynamical-systems. I posted an answer to that question. Someone voted that answer down so that the Community bot would delete the question from the site. I feel that my answer contributed something worthwhile to the site, so I'm reposting the question, and posting my answer.

Let $f:{\bf R}\to{\bf R}$ be strictly decreasing. How can we prove that there is no point of (minimal) period 3?

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Gerry Myerson
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You can prove it by first establishing

  1. if $f$ is strictly decreasing, then so is its third iterate,

  2. If $f$ is strictly decreasing, it has exactly one fixed point, and

  3. if $x$ is a fixed point of $f$, then it's a fixed point of the third iterate of $f$.

Gerry Myerson
  • 179,216