Error Control Coding (2E, Lin and Costello)- Solutions Manual

# Error Control Coding (2E, Lin and Costello)- Solutions Manual

This preview shows pages 1–4. Sign up to view the full content.

Chapter 2 2.3 Since m is not a prime, it can be factored as the product of two integers a and b , m = a · b with 1 < a, b < m . It is clear that both a and b are in the set { 1 , 2 , · · · , m - 1 } . It follows from the definition of modulo- m multiplication that a ¡ b = 0 . Since 0 is not an element in the set { 1 , 2 , · · · , m - 1 } , the set is not closed under the modulo- m multiplication and hence can not be a group. 2.5 It follows from Problem 2.3 that, if m is not a prime, the set { 1 , 2 , · · · , m - 1 } can not be a group under the modulo- m multiplication. Consequently, the set { 0 , 1 , 2 , · · · , m - 1 } can not be a field under the modulo- m addition and multiplication. 2.7 First we note that the set of sums of unit element contains the zero element 0 . For any 1 ‘ < λ , X i =1 1 + λ - X i =1 1 = λ X i =1 1 = 0 . Hence every sum has an inverse with respect to the addition operation of the field GF ( q ) . Since the sums are elements in GF ( q ) , they must satisfy the associative and commutative laws with respect to the addition operation of GF ( q ) . Therefore, the sums form a commutative group under the addition of GF ( q ) . Next we note that the sums contain the unit element 1 of GF ( q ) . For each nonzero sum X i =1 1 with 1 ‘ < λ , we want to show it has a multiplicative inverse with respect to the multipli- cation operation of GF ( q ) . Since λ is prime, and λ are relatively prime and there exist two 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
integers a and b such that a · + b · λ = 1 , (1) where a and λ are also relatively prime. Dividing a by λ , we obtain a = + r with 0 r < λ. (2) Since a and λ are relatively prime, r 6 = 0 . Hence 1 r < λ Combining (1) and (2), we have · r = - ( b + k‘ ) · λ + 1 Consider X i =1 1 · r X i =1 1 = · r X i =1 1 = - ( b + k‘ ) · λ X i =1 +1 = ( λ X i =1 1)( - ( b + k‘ ) X i =1 1) + 1 = 0 + 1 = 1 . Hence, every nonzero sum has an inverse with respect to the multiplication operation of GF ( q ) . Since the nonzero sums are elements of GF ( q ) , they obey the associative and commutative laws with respect to the multiplication of GF ( q ) . Also the sums satisfy the distributive law. As a result, the sums form a field, a subfield of GF ( q ) . 2.8 Consider the finite field GF ( q ) . Let n be the maximum order of the nonzero elements of GF ( q ) and let α be an element of order n . It follows from Theorem 2.9 that n divides q - 1 , i.e. q - 1 = k · n. Thus n q - 1 . Let β be any other nonzero element in GF ( q ) and let e be the order of β . 2
Suppose that e does not divide n . Let ( n, e ) be the greatest common factor of n and e . Then e/ ( n, e ) and n are relatively prime. Consider the element β ( n,e ) This element has order e/ ( n, e ) . The element αβ ( n,e ) has order ne/ ( n, e ) which is greater than n . This contradicts the fact that n is the maximum order of nonzero elements in GF ( q ) . Hence e must divide n . Therefore, the order of each nonzero element of GF ( q ) is a factor of n . This implies that each nonzero element of GF ( q ) is a root of the polynomial X n - 1 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern