Proving Mathematical Induction

Question

I have the following exercise which I am trying to understand:

For which nonnegative integers $n$ is $2n + 3 ≤ 2^n$? Prove your answer.

I know that we first have to inspect for the basis case and so we find that $n > 3$, since $n=1,2,3$ do not hold but $4$ plus does, so $n > 3.$ But for the Induction step I don't understand how to start it and the textbook solution is unclear, could anyone explain to me what this means?

Textbook solution: For the inductive step assume that $P(k)$ is true. Then, by the inductive hypothesis, $2(k + 1) + 3 = (2k + 3) + 2 < 2^k + 2.$ But because $k ≥ 1 , 2^k + 2 ≤ 2^k + 2^k = 2^k+1$. This shows that $P(k+1)$ is true.

Answer 1

$$2^1<2^2<\cdots<2^k$$ Hence $$2^k +2<2^k+2^k=2\cdot 2^k=2^{k+1}$$

Answer 2

Normally, when you have two functions $f(n)$ and $g(n)$,
where for all $n \in \Bbb{Z^+}$, you have that $0 < f(n), g(n),$
and you establish a base case $n_0 \in \Bbb{Z^+}$ such that $f(n_0) < g(n_0),$
you then have two separate strategies that you might follow
to inductively prove that for all $n \in \Bbb{Z^+}$ such that $n \geq n_0,$
it is true that $f(n) < g(n)$.

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The first strategy would be to examine the general ratios:

$\displaystyle R_n = \frac{f(n+1)}{f(n)}, ~S_n = \frac{g(n+1)}{g(n)}.$

If you can show that for all $n \in \Bbb{Z^+},$ such that $n \geq n_0$,
that $0 < R_n < S_n$, then you can reason as follows:

Suppose that $f(n) < g(n)$.
Then, you also have that $f(n+1) = R_n \times f(n)$
and $g(n+1) = S_n \times g(n)$.

Therefore, since $0 < f(n) < g(n)$ and $0 < R_n < S_n,$
then $R_n \times f(n) < S_n \times g(n).$

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The second strategy is very similar to the first strategy, except that instead of focusing on ratios, you focus on differences.

That is,
let $F_n$ denote $f(n+1) - f(n)$ and
let $G_n$ denote $g(n+1) - g(n)$.

Suppose that you can show that for all $n \in \Bbb{Z^+}$
such that $n \geq n_0$, you have that
$F_n < G_n$.

Then, you can reason as follows:

Assume that $f(n) < g(n)$.
Then, $f(n+1) = F_n + f(n)$
and $g(n+1) = G_n + g(n)$.
Then, because $f(n) < g(n)$ and $F_n < G_n$
you have that $F_n + f(n) < G_n + g(n).$

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The second strategy above is what the offered solution focused on, differences.

To prove:
For all $n \in \Bbb{Z^+}$ let $f(n) = 2n + 3$ and let $g(n) = 2^n.$
Then for all $n \in \Bbb{Z^+}$ such that $n \geq 4$
$f(n) < g(n)$.

First, the assertion was manually verified for $n = 4$.

Then, (in effect), the offered solution noted that
$F_n = f(n+1) - f(n) = [2(n + 1) + 3] - [2(n) + 3] = 2.$ $G_n = g(n+1) - g(n) = 2^{n+1} - 2^n = 2^n.$

Then, the offered solution noted that for $n \geq 4$, you have that:
$F_n < G_n$.

This conclusion allowed the second strategy above, that focuses on differences, to be invoked.