proving Fibonacci sequence for all $ n > 0, F(n) < 2^n$

Question

My base step: $ S(1)=F(1)< 2^1 $

Induction hypothesis: $S(k)=F(k)< 2^k$ for all $k>0$

Induction: $ S(k+1)=F(k+1)< 2^{(k+1)}$ for all $k>0$

So would this be a valid next step to prove that hypothesis holds? - $ S(k+1)=2^k + F(k+1)< 2^{(k+1)} $ for all $ k>0 $

Would that be correct?

In particular, note the difference between the phrases "for all $k>0$" and "for some $k>0$" — JMoravitz, Sep 24 '17 at 22:52
An alternative proof is as follows. Since $F_{n+2}$ is the number of binary strings (i.e., string formed by $0$ and $1$) of length $n$ such that there are no consecutive occurrences of $1$, it follows that $F_{n+2}$ is less than or equal to $2^n$. Thus, $$F_n=F_{n+2}-F_{n+1}<F_{n+2}\leq 2^n.$$ — Batominovski, Jan 19 '20 at 14:38

score 0 · Answer 1 · answered Sep 24 '17 at 22:46

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No, it is wrong. You stated that the hypothesis is $F(k)<2^k$ for all $k>0$. That is, you are assuming as a hypothesis the very thing that you want to prove.

answered Sep 24 '17 at 22:46

José Carlos Santos

427,504

Would it then be correct to say for induction hypothesis that theorem is true for all values of k > 0? – Mayer Sep 24 '17 at 23:04
@Mayer The inductive hypothesis is that it holds for $k$, and the inductive step is proving that, if it holds at $k$, it holds at $k+1$. – Carl Schildkraut Sep 24 '17 at 23:07
@Mayer No! It's the same problem. The induction hypothesis should be just: let $k\in\mathbb N$ and suppose that $F(k)<2^k$. Without a “for all”. – José Carlos Santos Sep 24 '17 at 23:08
that makes sense, thanks guys! – Mayer Sep 24 '17 at 23:12

proving Fibonacci sequence for all $ n > 0, F(n) < 2^n$

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