Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 1 Exercise 1.8

Solution:

### Part a

The $n$-th proposition is

$$

P_n: \quad n^2>n+1.

$$ Clearly, $P_2$ is true because $2^2=4$ which is greater than $2+1=3$. We have the induction basis.

Now we assume $P_n$ is true (here we assume that $n\ge 2$), that is $n^2>n+1$. We would like to show $P_{n+1}$ is true based on $P_n$. Note that $n>0$, we have

\begin{align*}

(n+1)^2=n^2+2n+1>n+1+2n+1>(n+1)+1.

\end{align*} Therefore $P_{n+1}$ is true if $P_n$ is true. By the principle of mathematical induction, we make the conclusion that $P_n$ is true for all positive integers $n$.

### Part b

The $n$-th proposition is

$$

P_n: \quad n!> n^2.

$$ Clearly, $P_4$ is true because $4!=24$ which is greater than $4^2=16$. We have the induction basis.

Now we assume $P_n$ is true (here we assume that $n\ge 4$), that is $n!> n^2$. We would like to show $P_{n+1}$ is true based on $P_n$. Note that $n\ge 4$, we have

\begin{equation}\label{eq:1-8-1}

n^2\ge 4n>n,\quad n^2 >1.

\end{equation} Therefore

\begin{align*}

(n+1)!=&\ (n+1)\cdot n!>4\cdot n^2\\

=&\ n^2+2n^2+n^2\\

\text{use \eqref{eq:1-8-1}}\quad >&\ n^2+2n+1=(n+1)^2.

\end{align*} Therefore $P_{n+1}$ is true if $P_n$ is true. By the principle of mathematical induction, we make the conclusion that $P_n$ is true for all positive integers $n$.