Paragraph

Suppose \(f:\N \to \N\) satisfies the recurrence relation

\begin{equation*} f(n+1) = \begin{cases} \frac{f(n)}{2} \amp \text{ if } f(n) \text{ is even} \\ 3f(n) + 1 \amp \text{ if } f(n) \text{ is odd}\end{cases}\text{.} \end{equation*}

Note that with the initial condition \(f(0) = 1\text{,}\) the values of the function are: \(f(1) = 4\text{,}\) \(f(2) = 2\text{,}\) \(f(3) = 1\text{,}\) \(f(4) = 4\text{,}\) and so on, the images cycling through those three numbers. Thus \(f\) is NOT injective (and also certainly not surjective). Might it be under other initial conditions? 3 

  1. If \(f\) satisfies the initial condition \(f(0) = 5\text{,}\) is \(f\) injective? Explain why or give a specific example of two elements from the domain with the same image.

  2. If \(f\) satisfies the initial condition \(f(0) = 3\text{,}\) is \(f\) injective? Explain why or give a specific example of two elements from the domain with the same image.

  3. If \(f\) satisfies the initial condition \(f(0) = 27\text{,}\) then it turns out that \(f(105) = 10\) and no two numbers less than 105 have the same image. Could \(f\) be injective? Explain.

  4. Prove that no matter what initial condition you choose, the function cannot be surjective.

in-context