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Proof: Let $a_0, a_1, a_2, \ldots$ be defined as $x_0 = 2, a_k = \frac{a_{k-1}}{2a_{k-1} - 1}$ for all $k \ge 1 \in \mathbb Z. $ Define $P(n)$ by $$a_n = \begin{cases} 2, & \text{if $n$ is even} \\ \frac23, & \text{if $n$ is odd} \end{cases}$$

$P(0) , P(1)$ obviously true. Let $k \ge 0 \in \mathbb Z$ and suppose for all $i \in \mathbb Z$ with $0 \le i \le k,$

$$a_i = \begin{cases} 2, & \text{if $i$ is even} \\ \frac23, & \text{if $i$ is odd} \end{cases}$$

But
$$a_{k+1} = \frac{a_k}{2a_k - 1}$$

$$= \begin{cases} \frac{2}{2 \cdot 2 - 1}, & \text{if $k$ is even} \\ \frac{\frac23}{2 \cdot \frac23 - 1}, & \text{if $k$ is odd} \end{cases}$$

$$= \begin{cases} \frac23, & \text{if $k$ is even} \\ \frac{\frac23}{\frac13}, & \text{if $k$ is odd} \end{cases}$$

$$= \begin{cases} \frac23, & \text{if $k + 1$ is odd} \\ 2, & \text{if $k + 1$ is even} \end{cases}$$

Do we need the hypothesis hold for all $i \le k$ above? To me it looks like we could do without the strong induction. Why do we need $P(n)$ hold for all $i \le k$?

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It appears that Strong Induction is in fact not required, nor is $P(1)$ required as a base case.

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