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My question here is whether the point I make is sufficient to make the proof work, and in general whether relying on this kind of argument is ok.

The property $P(n)$ we're doing induction on is that $2n+1 < 2^n$ for all integers $n \geq 3$.

Base case $k=3$, $7 < 8$ works.

Now, we say assume $P(k)$ holds for some arbitrary integer $k \geq 3$, we'll show that it also holds for $P(k+1)$. In other words, we want to prove the implication $P(k) \rightarrow P(k+1)$.

$$2(k+1)+1 < 2^{k+1}$$ $$2(k+1) + 1 < 2 \cdot 2^k$$ $$k+1 + \frac{1}{2} < 2^k$$

But now, since $k$ is necessarily $\geq3$, we know that (this is the part I'm not sure about) $$k+1+\frac{1}{2} < 2k+1$$ and by induction hypothesis $$2k+1 < 2^k$$ therefore $$k+1+\frac{1}{2} < 2^k$$ and $$2(k+1)+1 < 2^{k+1}$$

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    Instead of starting with $2(k+1)+1<2^{k+1}$ and working down, it should instead be written $2(k+1)+1\leq \dots \leq \dots < 2^{k+1}$ – 2017-01-19
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    @JMoravitz, thank you, but is the way I did it wrong (in the sense that the proof is invalid regardless of considerations of form)? Or is it just not as clear? – 2017-01-19

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From the inductive step:

Assume that $P(n)$ is true: $2n+1 < 2^n$, for $n \geq 3$. Then: (by the induction hypothesis) $$2n + 1 + 2 < 2^n + 2 $$ (Since $n\geq 3 >1$) $$2^n + 2 < 2^n + 2^n=2 (2^n) =2^{n+1}$$

Therefore by transitivity of $<$: $$2(n+1) + 1 < 2^{n+1}$$

Since we have derived $P(n+1)$ from $P(n)$, we conclude $P(n) $ is true for all $n \geq 3$. Hope it helps.

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    Thank you, I see how this is clearer than what I have, but is what I have incorrect (in the sense of not sufficient to prove what I"m trying to prove), or is it just not as good? – 2017-01-19
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    @jeremyradcliff I think it's alright but keep in mind JMoravitz's comment. – 2017-01-19