Chapter 6.6, Problem 42WE

(a)

To determine

To Prove:

Each repeating decimal represents a rational number N = the number , p = the number of digits in the repeating block of N.

Prove $N\cdot {10}^p$ has the same repeating block as N.

Explanation of Solution

Calculation:

Let $N=a_0.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}$

To Prove:

  $N\cdot {10}^p$ has p digits in the repeating block .

Proof:

  $\begin{array}{l}N=a_0.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\\ \Rightarrow N\cdot {10}^p=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\\ \text{So, }N\cdot {10}^p\text{ has p digits in the repeating block}\text{.}\end{array}$

Hence proved.

(b)

To determine

To Prove:

Each repeating decimal represents a rational number N = the number , p = the number of digits in the repeating block of N.

Prove

  $N\cdot {10}^p-N$ is a terminating decimal.

Explanation of Solution

Given :

From part (a),

  $\begin{array}{l}N=a_0.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\\ N\cdot {10}^p=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\end{array}$

Calculation:

To Prove:

  $N\cdot {10}^p-N$ is a terminating decimal.

Proof:

  $\begin{array}{l}N=a_0.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\\ N\cdot {10}^p=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\\ \Rightarrow N\cdot {10}^p-N\\ =a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}-a_0.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\\ =a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p+0.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}-a_0-0.\overline{a_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p}\\ =a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p-a_0\\ \text{Since integers are closed under subtraction, and}\\ a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p\text{ and }{a}_0\text{ are integers , so,}\\ N\cdot {10}^p-N=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p-a_0\text{ is an integer}\text{.}\end{array}$

Hence, $N\cdot {10}^p-N$ is a terminating decimal.

Hence proved.

(c)

To determine

To Prove:

Each repeating decimal represents a rational number N = the number , p = the number of digits in the repeating block of N.

Prove $N\left({10}^p-1\right)$ is rational .

Explanation of Solution

Concept Used:

Rational number is of the form $\frac{p}{q}\text{ where p and q are integers and q}\ne \text{0}$ .

They have decimal form that is either terminating or repeating.

Calculation:

To Prove:

  $N\left({10}^p-1\right)$ is rational .

Proof:

From part (b),

  $N\cdot {10}^p-N=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p-a_0\text{ is an integer}\text{.}$

and $N\cdot {10}^p-N$ is a terminating decimal.

Hence , $N\left({10}^p-1\right)$ is a terminating decimal.

So, $N\left({10}^p-1\right)$ is rational .

Hence proved.

(d)

To determine

To Prove:

Each repeating decimal represents a rational number N = the number , p = the number of digits in the repeating block of N.

Prove $\frac{N\left({10}^p-1\right)}{{10}^p-1}$ is rational .

Explanation of Solution

Concept Used:

Rational number is of the form $\frac{p}{q}\text{ where p and q are integers and q}\ne \text{0}$ .

Calculation:

To Prove:

  $\frac{N\left({10}^p-1\right)}{{10}^p-1}$ is rational .

Proof:

From part (b),

  $N\cdot {10}^p-N=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p-a_0\text{ is an integer}\text{.}$

From part (c),

  $N\left({10}^p-1\right)$ is rational and is integer .

So,

  $\begin{array}{l}N\cdot {10}^p-N=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p-a_0\\ \Rightarrow N\left({10}^p-1\right)=a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p-a_0\\ \Rightarrow N=\frac{a_0{a}_1{a}_2{a}_3{a}_4\mathrm{...}{a}_p-a_0}{{10}^p-1}=\frac{N\left({10}^p-1\right)}{{10}^p-1}\\ and{10}^p-1\ne {\text{0 is an integer since 10}}^p\text{ and 1 are}\\ \text{integers and integers are closed under subtraction}\text{.}\end{array}$

So, by the definition of a rational number, N is a rational number.

Hence, each repeating decimal represents a rational number.

Hence proved.