(Please explain every step) Multiplication in GF(22) is polynomial multiplication reduced using the irreducible polynomial (x2 + x + 1) . Complete the multiplication table for GF(22). Do the multiplication as polynomials, reduce the answer mod (x2 + x + 1), write the result as a 2-bit pattern, then enter it in the table as hex. The row headed by 0 and the column headed by 0 are easy, as are the row headed by 1 and the column headed by 1. That leaves four entries. 2*2 : 2*3 : 3*2 : 3*3 : * 0 1 2 3 0 1 2 3 Consider the set {1, 2, 3} and the operation * as shown in your table. (Notice that 0 has been removed from the set.) Show that the Group Axioms are followed: Closed: Associative Identity: Inverse: Commutative: Is this a multiplicative Abelian group? Now consider the distributive property: a*(b+c) = a*b + a*c Using your tables, demonstrate that this holds for several representative a, b, and c. A proof would require that you show that it works for all choices. a = 0 b = 1 c = 2 a = 1 b = 1 c = 2 a = 2 b = 1 c = 1 a = 2 b = 1 c = 2 GRAND CONCLUSION: Is the set {0, 1, 2, 3} with the operations + and * as defined here, a Finite Field?
(Please explain every step) Multiplication in GF(22) is polynomial multiplication reduced using the irreducible polynomial (x2 + x + 1) . Complete the multiplication table for GF(22). Do the multiplication as polynomials, reduce the answer mod (x2 + x + 1), write the result as a 2-bit pattern, then enter it in the table as hex. The row headed by 0 and the column headed by 0 are easy, as are the row headed by 1 and the column headed by 1. That leaves four entries.
2*2 :
2*3 :
3*2 :
3*3 :
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Consider the set {1, 2, 3} and the operation * as shown in your table. (Notice that 0 has been removed from the set.) Show that the Group Axioms are followed:
Closed:
Associative
Identity:
Inverse:
Commutative:
Is this a multiplicative Abelian group?
Now consider the distributive property: a*(b+c) = a*b + a*c
Using your tables, demonstrate that this holds for several representative a, b, and c. A proof would require that you show that it works for all choices.
a = 0 b = 1 c = 2
a = 1 b = 1 c = 2
a = 2 b = 1 c = 1
a = 2 b = 1 c = 2
GRAND CONCLUSION: Is the set {0, 1, 2, 3} with the operations + and * as defined here, a Finite Field?
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