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I am working on exercises in an introductory Real Analysis textbook, but I am stumped on a question involving the Principle of Mathematical Induction. The question is the following:

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Here is what I have so far:

First, we must check the base case $P(n)$. For $n = 5$, we have $2(5) - 3 ≤ 2^{5-2}$, which is equivalent to $7 \leq 8$, which is true. Now, suppose $P(k)$ is true for some $k \in {\displaystyle \mathbb {N}}, k \geq 5$, so we have $2k - 3 \leq 2^{k-2}$. It follows that $2(k+1) - 3 = 2k + 2 -3 = 2k - 1 \leq 2^{k-2}$.

I'm not sure what to do after this point. I know the end result we're looking for should have the following form: $2^{(k+1) - 2}$, but I'm not exactly sure how to get there. Any advice would be much appreciated! For context, this is my first proof class.

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    $\begingroup$ You cannot conclude $2 k - 1 \le 2^{k - 2}$. $\endgroup$
    – player3236
    Sep 4, 2020 at 5:47

1 Answer 1

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If $2k-3\leq 2^{k-2}$ then $2^{(k+1)-2}=2^{k-1} =2(2^{k-2}) \geq 2(2k-3)=4k-6 \geq 2k-1=2(k+1)-3$.

[The second inequality above follows from the fact that $k \geq 5$ so we certainly have $2k \geq 5$].

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