Chapter 9.5, Problem 17E

Solution

To prove : using mathematical induction to prove that statement $2^n\ge 2n$ holds true for all positive integers.

Given information :

The statement $2^n\ge 2n$ .

Formula used :

Follow the steps to prove that the statement is true for any positive integer.

Anchor step: Prove that any statement $P_n$ is true for $n=1$ .

Inductive hypothesis: Assume $P_n$ is true for $n=k$ .

Inductive Step: Prove that $P_n$ is true for $n=k+1$ .

Proof :

Conjecture:

Take $P_n$ to be statement $2^n\ge 2n$ .

Anchor step:

Prove that $P_n$ is true for $n=1$ .

For $n=1$ ,

  $\begin{array}{l}{2}^1\stackrel{?}{=}2\cdot 1\\ 2=2\end{array}$

Inductive hypothesis:

Assume $P_n$ is true for $n=k$ . Replace $n$ with $k$ in $P_n$ .

  $P_k:2^k\ge 2k$

Inductive Step:

Prove that $P_n$ is true for $n=k+1$ . This means it is needed to be proved that $P_{k+1}$ must be true.

Proceed with the inductive hypothesis.

  $P_k:2^k\ge 2k$

Multiply both sides with $2$ .

  $\begin{array}{c}2\cdot 2^k\ge 2\cdot 2k\\ 2^{k+1}\ge 4k\end{array}$

Since $k\ge 1$ ,

  $2^{k+1}\ge 2\left(k+1\right)$

The statement is same as the statement $P_{k+1}$ , so $P_{k+1}$ is true.

So, if $P_n$ is true for $n=k$ , then it is also true for $n=k+1$ .

Conclusion:

As $P_n$ is true for $n=1$ and $P_k$ further implies $P_{k+1}$ , this means $P_n$ is true for $n=2$ , $n=3$ and so on.

By mathematical induction, $P_n:2^n\ge 2n$ is true for all positive integers $n$ .