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Abstract Algebra: An Introduction

Found in: Page 30

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

Show that $a^{p - 1} = 1 (m o d p)$ for the given p and a:

$a = 2, p = 5$ .
$a = 4, p = 7$ .
$a = 3, p = 11$ .

It is proved that $2^{5 - 1} \equiv 1 (m o d 5)$ .
It is proved that $4^{7 - 1} \equiv 1 (m o d 7)$ .
It is proved that $3^{11 - 1} \equiv 1 (m o d 11)$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the definition of Congruence

Consider that a, b, n as an integer with $n > 0$ . If n divides $a - b$ , then a is congruent tob modulo n (expressed as $a \equiv b (m o d n)$ ).

Step 2: Show that ap-1=1modp for the given p and a

a)

Put $a = 2$ and $p = 5$ in the expression $a^{p - 1}$ as shown below:

$\begin{array}{rcl} 2^{5 - 1} & = & 2^{4} \\ = & 18 \end{array}$

Divide 16 by 5 as follows:

$16 = 5 (3) + 1$

Then, $2^{5 - 1}$ can be expressed as follows:

$2^{5 - 1} = 1 (m o d 5)$

Thus, it is proved that $2^{5 - 1} = 1 (m o d 5)$ .

Step 3: Show that ap-1=1modp for the given p and a

b)

Put $a = 4$ and $b = 7$ in the expression $a^{p - 1}$ as shown below:

$\begin{array}{rcl} 4^{7 - 1} & = & 4^{6} \\ = & 4096 \end{array}$

Divide 4096 by 7 as follows:

$4096 = 7 (585) + 1$

Then, $4^{7 - 1}$ can be expressed as follows:

$4^{7 - 1} = 1 (m o d 7)$

Thus, it is proved that $4^{7 - 1} = 1 (m o d 7)$ .

Step 4: Show that ap-1=1modp for the given p and a

c)

Put $a = 3$ and $p = 11$ in the expression $a^{p - 1}$ as shown below:

$\begin{array}{rcl} 3^{11 - 1} & = & 3^{10} \\ = & 59049 \end{array}$

Divide 59049 by 11 as follows:

$59049 = 5368 (11) + 1$

Then, $3^{11 - 1}$ can be expressed as follows:

$3^{11 - 1} \equiv 1 (m o d 11)$

Thus, it is proved that $3^{11 - 1} \equiv 1 (m o d 11)$ .

Most popular questions for Math Textbooks

Prove or disprove: If $[a] ⊝ [b] = [0]$ in $ℤ_{n}$ , then $[a] = [0]$ or $[b] = [0]$ .

Use Exercise 13 to solve the following equations.

(a) $15 x = 9$ in $ℤ_{18}$ (b) $25 x = 10$ in $ℤ_{65}$

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if $a < b$ and $b < c$ , then $a < c$ ;
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