Chapter 3.4, Problem 1E

(a)

To determine

To find the value $6+6\text{ in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $6+6\text{ =5 in }{\mathbb{Z}}_7$ .

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is ${\mathbb{Z}}_7$ .

Formula used: Definition congruence modulo m.

For $x,y\in \mathbb{Z}$ , x is congruentto y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}6+6\text{ =12}\\ \equiv \text{5}\left(\mathrm{mod}7\right)\end{array}$

  $\therefore \text{6+6=5 in }{\mathbb{Z}}_7$

(b)

To determine

To find the value $5\cdot 4\text{in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $5\cdot 4\text{in }{\mathbb{Z}}_7\text{ is 6}$ .

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulo m.

For $x,y\in \mathbb{Z}$ , x is congruent to y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}5\cdot 4=20\\ \equiv 6\left(\text{mod 7}\right)\end{array}$

  $5\cdot 4=6\text{in }{\mathbb{Z}}_7$

c.

To determine

To find $3\cdot 3+5\cdot 2\text{ in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $3\cdot 3+5\cdot 2\text{ =5 in }{\mathbb{Z}}_7$

Explanation of Solution

Given information: Equivalence class under congruence modulo 7is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulo m.

For $x,y\in \mathbb{Z}$ , x is congruentto y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}3\cdot 3+5\cdot 2\text{ =9+10}\\ \equiv \text{2(mod 7)+3(mod7)}\\ \equiv \text{5(mod}\text{7)}\end{array}$

  $\therefore 3\cdot 3+5\cdot 2=5\text{in }{\mathbb{Z}}_7$

d.

To determine

To find $2\cdot 4+3\cdot 5\text{ in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $2\cdot 4+3\cdot 5\text{ in }{\mathbb{Z}}_7\text{ is 2}$

Explanation of Solution

Given information: Equivalence class under congruence modulo 7is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulom.

For $x,y\in \mathbb{Z}$ , x is congruentto y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}2\cdot 4+3\cdot 5\text{ =8+15}\\ \equiv 1\text{(mod 7)+1(mod7)}\\ \equiv 2\text{(mod}\text{7)}\end{array}$

  $\therefore \text{ 2}\cdot 4+3\cdot 5=2\text{in }{\mathbb{Z}}_7$

e.

To determine

To find the value $\text{5}\cdot 1+3\cdot 2\text{in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $\text{5}\cdot 1+3\cdot 2\text{ =4}\text{in }{\mathbb{Z}}_7$

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulo m.

For $x,y\in \mathbb{Z}$ , x is congruent to y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}5\cdot 1+3\cdot 2\text{ =5+6}\\ =11\\ \equiv 4\text{(mod}\text{7)}\end{array}$

  $\therefore \text{ 5}\cdot 1+3\cdot 2\text{ =4}\text{in }{\mathbb{Z}}_7$

f.

To determine

To find $0\cdot 3+2\cdot 4\text{ in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $0\cdot 3+2\cdot 4\text{ in }{\mathbb{Z}}_7\text{ is 1}$ .

Explanation of Solution

Given information: Equivalence class under congruence modulo 7is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulo m.

For $x,y\in \mathbb{Z}$ , x is congruent to y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}0\cdot 3+2\cdot 4\text{ =0+8}\\ =8\\ \equiv 1\text{(mod}\text{7)}\end{array}$

  $\therefore \text{ 0}\cdot 3+2\cdot 4\text{ =1}\text{in }{\mathbb{Z}}_7$

h.

To determine

To find $(4+5)\cdot (5+6)$

  $\text{in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $(4+5)\cdot (5+6)$

  $\text{in }{\mathbb{Z}}_7$ is 1

Explanation of Solution

Given information: Equivalence class under congruence modulo7 is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulom.

For $x,y\in \mathbb{Z}$ , x is congruent to y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}(4+5)\cdot (5+6)\text{=9}\cdot \text{11}\\ \equiv 2(\text{mod 7})\cdot 4(\mathrm{mod}7)\\ =8\\ \equiv 1(\mathrm{mod}7)\end{array}$

∴ $(4+5)\cdot (5+6)$

  $\text{in }{\mathbb{Z}}_7$ is 1

i.

To determine

To find $2^{25}\text{ in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $2^{25}\text{=2 in }{\mathbb{Z}}_7$

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulom.

For $x,y\in \mathbb{Z}$ , x is congruent to y ifm divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}{2}^{25}\text{=}{\left(2^5\right)}^5\\ \text{=}{\left(2^3\right)}^8\cdot \text{2}\\ \text{=}{\left(8\right)}^8\cdot \text{2}\\ \equiv {\left\{1(\text{mod 7})\right\}}^8\cdot \text{2}\\ {\text{=(1)}}^8\cdot \text{2}\\ \text{=2 }\end{array}$

∴ $2^{25}\text{=2 in }{\mathbb{Z}}_7$

j.

To determine

To find $5^{23}$

  $\text{in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $5^{23}$

  $\text{in }{\mathbb{Z}}_7$ is 3.

Explanation of Solution

Given information: Equivalence class under congruence modulo7 is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulo m.

For $x,y\in \mathbb{Z}$ , x is congruentto yif m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}{5}^{23}=5^{18}\cdot 5^5\\ ={\left(5^6\right)}^3\cdot 5^5\\ \equiv 1^3\left(\mathrm{mod}7\right)\cdot 5^5\\ \equiv 1\cdot 5^4\cdot 5\left(\mathrm{mod}7\right)\\ \equiv {\left(25\right)}^2\cdot 5\left(\mathrm{mod}7\right)\\ \equiv 4^2\cdot 5\left(\mathrm{mod}7\right)\\ \equiv 2\left(\mathrm{mod}7\right)\cdot 5\\ =10\left(\mathrm{mod}7\right)\\ \equiv 3\left(\mathrm{mod}7\right)\end{array}$

∴ $5^{23}$

  $\text{in }{\mathbb{Z}}_7$ is 3

k.

To determine

To find $4^{44}$

  $\text{in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $4^{44}$

  $\text{in }{\mathbb{Z}}_7$ is 2

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is ${\mathbb{Z}}_7$ .

Formula used: Definitionof congruence modulom.

For $x,y\in \mathbb{Z}$ ,x is congruent to y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}{4}^{44}\text{=}{\left(4^3\right)}^{14}\cdot 4^2\\ ={(}^{64}\cdot 16\\ \equiv {\left\{1\left(\mathrm{mod}\text{ 7}\right)\right\}}^8\cdot \left\{2\left(\mathrm{mod}\text{ 7}\right)\right\}\\ {\text{=(1)}}^8\cdot \text{2}\\ \text{=2 }\end{array}$

∴ $4^{44}$

  $\text{in }{\mathbb{Z}}_7$ is 2

l.

To determine

To find $2^{26}$

  $\text{in }{\mathbb{Z}}_7$ .

Answer to Problem 1E

  $2^{26}$

  $\text{in }{\mathbb{Z}}_7$ is 4.

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is ${\mathbb{Z}}_7$ .

Formula used: Definition of congruence modulo m.

For $x,y\in \mathbb{Z}$ , x is congruent to y if m divides $x-y$ and it is written as

  $x\equiv y\left(\text{mod }m\right)$

Calculation:

  $\begin{array}{c}{2}^{26}=2^{25}\cdot 2\\ ={\left(2^5\right)}^5\cdot 2\\ \equiv {\left\{4(\mathrm{mod}7)\right\}}^5\cdot 2\\ =2\mathrm{mod}\left(7\right)\end{array}$

∴ $2^{26}$

  $\text{in }{\mathbb{Z}}_7$ is 4.