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Solutions
Math 310, Homework #5
Problem HW5.1. Write out the addition and multiplication tables for Z/5Z and Z/6Z.
Solution. For Z/5Z, we have:
+
[0]5
[1]5
[2]5
[3]5
[4]5
[0]5
[0]5
[1]5
[2]5
[3]5
[4]5
[1]5
[1]5
[2]5
[3]5
[4]5
[0]5
[2]5
[2]5
[3]5
[4]5
[0]5
[1]5
[3]5
[3]5
[4]5
[0]5
[1]5
[2]5
[4]5
[4]5
[0]5
[1]5
[2]5
[3]5
·
[0]5
[1]5
[2]5
[3]5
[4]5
[0]5
[0]5
[0]5
[0]5
[0]5
[0]5
[1]5
[0]5
[1]5
[2]5
[3]5
[4]5
[2]5
[0]5
[2]5
[4]5
[1]5
[3]5
[3]5
[0]5
[3]5
[1]5
[4]5
[2]5
[4]5
[0]5
[4]5
[3]5
[2]5
[1]5
and
and for Z/6Z, we have:
+
[0]6
[1]6
[2]6
[3]6
[4]6
[5]6
[0]6
[0]6
[1]6
[2]6
[3]6
[4]6
[5]6
[1]6
[1]6
[2]6
[3]6
[4]6
[5]6
[0]6
[2]6
[2]6
[3]6
[4]6
[5]6
[0]6
[1]6
[3]6
[3]6
[4]6
[5]6
[0]6
[1]6
[2]6
[4]6
[4]6
[5]6
[0]6
[1]6
[2]6
[3]6
[5]6
[5]6
[0]6
[1]6
[2]6
[3]6
[4]6
·
[0]6
[1]6
[2]6
[3]6
[4]6
[5]6
[0]6
[0]6
[0]6
[0]6
[0]6
[0]6
[0]6
[1]6
[0]6
[1]6
[2]6
[3]6
[4]6
[5]6
[2]6
[0]6
[2]6
[4]6
[0]6
[2]6
[4]6
[3]6
[0]6
[3]6
[0]6
[3]6
[0]6
[3]6
[4]6
[0]6
[4]6
[2]6
[0]6
[4]6
[2]6
[5]6
[0]6
[5]6
[4]6
[3]6
[2]6
[1]6
and
Theorem HW5.2. Let m ≥ 2 be a composite integer (that is, not prime). Prove that there are
elements [x]m and [y]m of Z/mZ with [x]m , [y]m 6= [0]m such that [x]m · [y]m = [0]m .
Proof. Let m ≥ 2 be a composite integer. Then by definition there are integers x, y with 0 < x, y < m
such that xy = m. As 0 < x, y < m, then m - x, y so that [x]m , [y]m 6= 0 in Z/mZ. However,
[x]m · [y]m = [xy]m = [m]m = [0]m .
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