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10E

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Abstract Algebra: An Introduction

Found in: Page 520

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

Do exercise 9 when $A = ℤ$ .
Exercise 9: Let $A = {1, 2, 3, 4}$ . Exhibit functions f and g from A to A such that $f o g \neq g o f$ .

It can be concluded that the exhibit functions f and g from A to A, then $fog \neq gof$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Consider the given statement

It is given that the exhibit functions f and g from A to A, then $fog \neq gof$ .

As it is known that $ℤ$ is the set of integers, therefore, $ℤ = \{- 1, 2, - 3, 4\}$ .

Now, functions f and g are defined as;

$f (- 1) = - 3 f (2) = 2 f (- 3) = 4 f (4) = - 1$

And,

$g (- 1) = 2 g (2) = - 3 g (- 3) = - 1 g (4) = 4$

Step 2: Determine the values of functions fog and gof at A={-1,2}

The value of fog and gof at $A = \{- 1, 2\}$ are;

$\begin{array}{rcl} fog (- 1) & = & f (2) \\ = & 2 \\ gof (- 1) & = & f (- 3) \\ = & - 1 \\ fog (2) & = & f (3) \\ = & 4 \\ gof (2) & = & f (2) \\ = & 3 \end{array}$

Step 3: Determine the values of functions fog and gof at A={-3,4}

The value of fog and gof at $A = \{- 3, 4\}$ are;

$\begin{array}{rcl} fog (- 3) & = & f (- 1) \\ = & - 3 \\ gof (- 3) & = & f (4) \\ = & 4 \\ fog (4) & = & f (4) \\ = & - 1 \\ gof (4) & = & f (- 1) \\ = & 2 \end{array}$

Therefore, it can be observed from all the cases that $fog \neq gof$ . So, the provided statement has been proved.

Most popular questions for Math Textbooks

Let $S = {(a, b) a, b \in □ □ a n d b \neq 0}$ and define $(a, b) □ (c, d)$ if and only if $a d = b c$ . Prove that $□$ is an equivalence relation on S.

Let G be a group and define $a ~ b$ if and only if there exists $c \in G$ such that $b = c^{1} a c$ . Prove that $~$ is an equivalence relation on G.

Assume A and B are matrices over $ℤ$ . Find $A + B$ .

$A = [\begin{matrix} 1 & 2 & - 2 & 0 \\ 3 & 5 & 7 & 11 \end{matrix}]$ and $B = [\begin{matrix} 0 & - 8 & 2 & 4 \\ 6 & 0 & 4 & 1 \end{matrix}]$
$A = [\begin{matrix} 3 & 0 & 2 \\ 4 & 1 & 6 \\ 0 & 1 & 0 \\ 2 & - 5 & 7 \end{matrix}]$ and $B = [\begin{matrix} 1 & - 2 & 0 \\ 3 & 0 & 4 \\ 0 & 7 & - 6 \\ 1 & 6 & 0 \end{matrix}]$

Let B be a finite set and $f : B \to B$ is a function. Prove that f is injective if and only if f is surjective.

Describe each set by listing:

(a) The integers strictly between -3 and 9.

(b) The negative integers greater than -10.

(c) The positive integers whose square roots are less than or equal to 4.

NOTE: Z is the set of integers, Q is the set of rational numbers, and Z the set of real numbers.

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