Course notes (chap. 1 Number Theory, chap. 2 ... - McGill University

10011×01110 = 01001 since (x 4 +x+1)×(x 3 +x 2 +x) mod(x 5 +x 2 +1) =
x 3 +1.
× 000 001 010 011 100 101 110 111
000 000 000 000 000 000 000 000 000
001 000 001 010 011 100 101 110 111
010 000 010 100 110 011 001 111 101
011 000 011 110 101 111 100 001 010
100 000 100 011 111 110 010 101 001
101 000 101 001 100 010 111 011 110
110 000 110 111 001 101 011 010 100
111 000 111 101 010 001 110 100 011
multiplication of F23
Figure 3: operations of F 2 3