Chapter 2.4, Problem 6E

a.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false.

  $n^3<n$ for all $n\ge 6$

Explanation of Solution

Given :

It is given in the question that $n^3<n$ for all $n\ge 6$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basic step:

  $\begin{array}{l}n=6\Rightarrow 6^3=216<720=6!\\ 6^3<6!\end{array}$

  $2$ .Inductive step:

Assume there is a natural number $n\ge 6$ such that $n^3<n!$

Then,

  $\begin{array}{}$ {(n+1)}^3=n^3+3n^2+3n+1<n^3+3n^2<n^3+n^4<n!+nn!=(n+1)!$$ If{n}^3<n!,then{(n+1)}^3<(n+1)!$\end{array}$

By the PMI, $(\forall n\ge 6)(n^3<n!)$ .

Example for when the inequality y is false:

For $n=1$ ; $1^3=1=1!$

b.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

  $2^n>n^2$ for all $n>4$ .

Explanation of Solution

Given :

It is given in the question that $2^n>n^2$ for all $n>4$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basis step:

  $\begin{array}{l}n=5\Rightarrow 2^5=32>25=5^2\\ 2^5>5^2\end{array}$

  $2$ .Inductive step:

Assume there is a natural number $n>4$ such that $2^n>n^2$ .

Then,

  $2^{n+1}=2\cdot 2^n>2n^2=n^2+n^2>n^2+2n+1={(n+1)}^2$

  $If{2}^n>n^2,then{2}^{n+1}>{(n+1)}^2$

By the PMI, $(\forall n>4)(2^n>n^2)$

Example for when the inequality is false:

For $n=1$ : $2^1=2>1^2$

c.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

  $n+1!>2^{n+3}$ for all $n\ge 5$

Explanation of Solution

Given :

It is given in the question that $n+1!>2^{n+3}$ for all $n\ge 5$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basic step:

  $\begin{array}{l}n=5\Rightarrow (1+5)!=720>256=2^8\\ (1+5)!>2^{5+3}\end{array}$

  $2$ .Inductive step:

Assume there is a natural number $n\ge 5$ such that $(1+n)!>2^{n+3}$

Then,

  $\begin{array}{l}(1+n+1)!=(1+n+1)(1+n)!>(2+n)2^{n+3}>2^{n+1+3}\\ If(1+n)!>2^{n+3},then(1+n+1)!>2^{n+3}\end{array}$

By the PMI, $(\forall n\ge 5)(1+n)!>2^{n+3}$ .

Example for when the inequality is false:

For $n=1$ : $2!=2<16=2^{1+3}$

d.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false.

  $2n-8\le n^2-8n+16$ for all $n\ge 6$

Explanation of Solution

Given :

It is given in the question that $2n-8\le n^2-8n+16$ for all $n\ge 6$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basic Step :

  $\begin{array}{l}n=6\Rightarrow 2\cdot 6-8=4=6^2-8\cdot 6+16\\ 2\cdot 6-8\le 6^2-86+16\end{array}$

  $2$ .Inductive Step:

Assume there is a natural number $n\ge 6$ such that $2n-8\le n^2-8n+16$

Then,

  $\begin{array}{}$ 2(n+1)-8=2+2n-8\le 2+n^2-8n+16\le 2+n^2+2n-8-8n+16={(n+1)}^2-8(n+1)+16$$ If2n-8\le n^2-8n+16,then2(n+1)-8\le {(n+1)}^2-8(n+1)+16.$\end{array}$

By the PMI, $(\forall n\ge 6)(2\cdot n-8\le n^2-8n+16)$

Example for when the inequality is false:

Foe $n=5$ : $10-8=2>1={(5-4)}^2$

e.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

  $n!>3n$ for all $n\ge 4$ .

Explanation of Solution

Given :

It is given in the question that $n!>3n$ for all $n\ge 4$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basis step :

  $\begin{array}{l}n=4\Rightarrow 4!=24>12=3\cdot 4\\ 4!>3\cdot 4\end{array}$

  $2$ .Inductive step:

Assume there is a natural number $n\ge 4$ such that $n!>3n$

Then,

  $\begin{array}{l}(n+1)!=(1+n)n!>(n+1)3n>3(n+1)\\ Ifn!>3n,then(n+1)!>3(n+1)\end{array}$

By the PMI, $(\forall n\ge 4)(n!>3n)$

Example for when the inequality y is false:

For $n=1$ : $1<3=3\cdot 1$

f.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

If x > 1 is a real number,then ${(x+1)}^n>n{x}^2+1$ for all $n\ge 3$ .

Explanation of Solution

Given :

It is given in the question that If x > 1 is a real number,then ${(x+1)}^n>n{x}^2+1$ for all $n\ge 3$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

Let x be an arbitrary real number greater than $1$ .

  $1$ .Basic step:

  $\begin{array}{}$ n=3\Rightarrow {(x+1)}^3=x^3+3x^2+3x+1>3x^2+1$$ {(x+1)}^3>3x^2+1$\end{array}$

  $2$ .Inductive Step:

Assume there is a natural number $n\ge 3$ such that ${(x+1)}^n>n{x}^2+1$

Then,

  $\begin{array}{}$ {(x+1)}^{n+1}>(x+1)(n{x}^2+1)=n{x}^3+n{x}^2+x+1>(n+1)x^2+1$$ If{(x+1)}^n>n{x}^2+1,then(n+1)!>(n+1)n^2+1$\end{array}$

By the PMI, $(\forall n\ge 3)({(x+1)}^n>n{x}^2+1)$

Since x was arbitrary $(\forall n>1)(\forall n\ge 3)({(x+1)}^n>n{x}^2+1)$

Example for when the inequality is false:

For $n=1$ : $x+1<x^2+1$

g.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

  ${\displaystyle \prod _{i=2}^n\frac{i^2-1}{i^2}=\frac{n+1}{2n}}$ for all $n\ge 2$

Explanation of Solution

Given :

It is given in the question that ${\displaystyle \prod _{i=2}^n\frac{i^2-1}{i^2}=\frac{n+1}{2n}}$ for all $n\ge 2$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basis step:

  $\begin{array}{l}n=2\Rightarrow \frac{2^2-1}{2^2}=\frac{3}{4}=\frac{2+1}{2\cdot 2}\\ {\displaystyle \prod _{i=2}^2\frac{i^2-1}{i^2}=\frac{n+1}{2n}}\end{array}$

  $2$ .Inductive step:

Assume there is a natural number $n\ge 2$ such that ${\displaystyle \prod _{i=2}^n\frac{i^2-1}{i^2}=\frac{n+1}{2n}}$

Then,

  $\begin{array}{l}{\displaystyle \prod _{i=2}^{n+1}\frac{i^2-1}{i^2}=}{\displaystyle \prod _{i=2}^n\frac{i^2-1}{i^2}\frac{{(n+1)}^2-1}{{(n+1)}^2}=\frac{n+1}{2n}\frac{{(n+1)}^2-1}{{(n+1)}^2}=\frac{(n+1)+1}{2(n+1)}}\\ If{\displaystyle \prod _{i=2}^n\frac{i^2-1}{i^2}=}\frac{n+1}{2n},then{\displaystyle \prod _{i=2}^{n+1}\frac{i^2-1}{i^2}=}\frac{n+1+1}{2(n+1)}\end{array}$

By the PMI, $(\forall n\ge 2)(If{\displaystyle \prod _{i=2}^n\frac{i^2-1}{i^2}=}\frac{n+1}{2n}$ )

Example for when the inequality y is false:

For $n=1$ : $0\ne 1$

h.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

  ${\displaystyle \prod _{i=1}^n\frac{1}{i}\le 2^{-n}}$ for all $n\ge 4$

Explanation of Solution

Given :

It is given in the question that ${\displaystyle \prod _{i=1}^n\frac{1}{i}\le 2^{-n}}$ for all $n\ge 4$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$.Basis step:

  $\begin{array}{l}n=4\Rightarrow \frac{1}{1}\frac{1}{2}\frac{1}{3}\frac{1}{4}=\frac{1}{24}\le \frac{1}{16}=2^{-4}\\ {\displaystyle \prod _{i=1}^4\frac{1}{i}\le }{2}^{-4}\end{array}$

  $2$ .Inductive Step:

Assume there is a natural number $n\ge 4$ such that ${\displaystyle \prod _{i=1}^n\frac{1}{i}\le }{2}^{-n}$

Then,

  ${\displaystyle \prod _{i=1}^{n+1}\frac{1}{i}=}{\displaystyle \prod _{i=1}^n\frac{1}{i}\frac{1}{(n+1)}\le 2^{-n}\frac{1}{(n+1)}\le }{2}^{-(n+1)}$

  $If{\displaystyle \prod _{i=2}^n\frac{1}{i}\le 2^{-n}},then{\displaystyle \prod _{i=2}^{n+1}\frac{1}{i}\le }{2}^{-(n+1)}$

By the PMI, $(\forall n\ge 4)({\displaystyle \prod _{i=1}^n\frac{1}{i}\le 2^{-n})}$

Example for when the inequality is false:

For $n=1$ : $1>\frac{1}{2}$

i.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

For all $n>2$ ,the sum of the angle measures of the interior angles of a convex polygon of n sides is

  $(n-2)\cdot 180°$ .

Explanation of Solution

Given:

It is given in the question that ifall $n>2$ ,the sum of the angle measures of the interior angles of a convex polygon of n sides is $(n-2)\cdot 180°$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basis step:

  $n=3\Rightarrow (3-2)180°$

The sum of the interior angles of a triangle is 180°.

  $2$ .Inductive step:

Assume there is a natural number $n\ge 3$ such that the sum of the interior angles of a polygon of n sides is

  $(n-2)180°$ .

The sum of the interior angles of a polygon of $n+1$ sides is:

  $(n-2)180°+180°=(n+1-2)180°$

By the PMI,the prove is done.

Example for when the inequality is false:

For $n=1$ : A polygon of $1$ side is a line.

  

j.

To determine

Prove this by using generalized PMI and show that the equation or inequality is false .

  $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\mathrm{......}+\frac{1}{\sqrt{n}}$ for all $n\ge 2$

Explanation of Solution

Given :

It is given in the question that $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\mathrm{......}+\frac{1}{\sqrt{n}}$ for all $n\ge 2$ .

Concept Used:

In this we have to use the concept of sets,induction and PMI.

Proof:

  $1$ .Basis step:

  $\begin{array}{l}n=2\Rightarrow \sqrt{2}<1.5<1.7=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}\\ \sqrt{n}<{\displaystyle \sum _{i=1}^2\frac{1}{\sqrt{i}}}\end{array}$

  $2$ .Inductive step:

Assume there is a natural number $n\ge 2$ such that $\sqrt{n}<{\displaystyle \sum _{i=1}^n\frac{1}{\sqrt{i}}}$

Then,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n+1}\frac{1}{\sqrt{i}}}={\displaystyle \sum _{i=1}^n\frac{1}{\sqrt{i}}}+\frac{1}{\sqrt{n+1}}>\sqrt{n}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1}\\ If\sqrt{n}<{\displaystyle \sum _{i=1}^n\frac{1}{\sqrt{i}}},then\sqrt{n+1}<{\displaystyle \sum _{i=1}^{n+1}\frac{1}{\sqrt{i}}}\end{array}$

By the PMI, $(\forall n\ge 2)(\sqrt{n}<{\displaystyle \sum _{i=1}^n\frac{1}{\sqrt{i}}})$

Example for when the inequality is false:

For $n=1$ : $\sqrt{1}\ge \frac{1}{\sqrt{1}}$