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For every natural number n, consider P(n) a statement. Prove that if the following two statements are true:
1) for all n ∈ N, P(2^n);
2) for all n > 2, P(n) ⇒ P(n − 1);
then P(n) is true for all natural numbers n ≥ 2.

Other questions ask you to use the method to solve this problem, and I want to make sure I'm using the best method, even though I think the method I used works, I'm wondering the best solution.

My method:
For x=2, P(2^2) = P(4), and P(4)⇒P(3), thus P(3) (minimum case).
Then for i ∈ {1,...,∞}, given P(2^(n+1)), Because P(n)⇒P(n-1), P(x) for all X between P(2^n) and P(2^(n+i)) is true, and since 2^(n+i) goes to infinity as i does, P(x) is true for all x.

Matthew
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1 Answers1

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Yes, that works. You can note that $P(n)$ implies $P(n-1)$ means $P(n)$ implies $P(m)$ for all $m \leq n$. Then for any particular natural number $p$, choose $2^q \geq p$. Then we have $P(2^q)$ is true so $P(p)$ is true.

mtheorylord
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