Explanation Given information: f : A → A be defined by f ( x ) = a x + b mod n . and ( a , n ) = 1 Formula used: 1) The function is bijective if and only if its inverse exist. 2) If ( g f ) ( x ) = ( f g ) ( x ) = x , f and are inverse of each other. Calculation: Let f : A → A be defined by f ( x ) = a x + b mod n . To show inverse of f exist, it is sufficient to show that, f is bijective. One-one: Let f ( x ) = f ( y ) ⇒ a x + b ( mod n ) = a y + b ( mod n ) ⇒ a x − a y = b − b ( mod n ) ⇒ a ( x − y ) = 0 ( mod n ) ⇒ n | a ( x − y ) as ( a , n ) = 1 ⇒ n | ( x − y ) ⇒ x = y mod n Thus, f is one-one. Onto: For every element in the codomain, there exist an element in the domain. Thus, f is onto. Therefore, f inverse exist. Now, to show g is inverse of f . Let g : A → A be defined by g ( x ) = a ' x + b ' mod n . Where 1 = a ' a mod n ⇒ a ' a = 1 mod n and b ' = − a ' b mod n

8th Edition

ISBN: 9781285463230

Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 2.8, Problem 13E

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To prove: Let f: A → A be defined by f(x)= ax + b mod n, then f−1: A → A exist if (a, n) = 1, and is given by f−1(x)= a'x + b' mod n, where a' is defined by 1 = a'a mod n, with 0 < a' < n and b' = −a'b mod n.

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Chapter 2.8, Problem 12E

Chapter 2.8, Problem 14E

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To prove: Let f : A → A be defined by f ( x ) = a x + b mod n , then f − 1 : A → A exist if ( a , n ) = 1 , and is given by f − 1 ( x ) = a ' x + b ' mod n , where a ' is defined by 1 = a ' a mod n , with 0 < a ' < n and b ' = − a ' b mod n .

Solution Summary: The author explains that the function is bijective if and only when its inverse exist.

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To prove: Let f: A → A be defined by f(x)= ax + b mod n, then f−1: A → A exist if (a, n) = 1, and is given by f−1(x)= a'x + b' mod n, where a' is defined by 1 = a'a mod n, with 0 < a' < n and b' = −a'b mod n.

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Chapter 2 Solutions

To prove: Let f : A → A be defined by f ( x ) = a x + b mod n , then f − 1 : A → A exist if ( a , n ) = 1 , and is given by f − 1 ( x ) = a ' x + b ' mod n , where a ' is defined by 1 = a ' a mod n , with 0 &lt; a ' &lt; n and b ' = − a ' b mod n .

Solution Summary: The author explains that the function is bijective if and only when its inverse exist.

Concept explainers

Videos

To prove: Let f: A → A be defined by f(x)= ax + b mod n, then f−1: A → A exist if (a, n) = 1, and is given by f−1(x)= a'x + b' mod n, where a' is defined by 1 = a'a mod n, with 0 < a' < n and b' = −a'b mod n.

Want to see the full answer?

Chapter 2 Solutions

To prove: Let f : A → A be defined by f ( x ) = a x + b mod n , then f − 1 : A → A exist if ( a , n ) = 1 , and is given by f − 1 ( x ) = a ' x + b ' mod n , where a ' is defined by 1 = a ' a mod n , with 0 < a ' < n and b ' = − a ' b mod n .