Math 110A HW §2.6 – Solutions
15. Prove theorem 2.29.
Proof. (a) First note that clearly [ab] ∈ Zn . So Zn is closed w.r.t multiplication. We now
check well-defined. Assume [a] = [c] and [b] = [d]. We need to show [ab] = [cd].
Since [a] = [c] and [b] = [d], then a ≡ c(mod n) and b ≡ d(mod n). By theorem 2.23,
ab ≡ cd(mod n). Thus [ab] = [cd]. Therefore multiplication in Zn is a binary operation.
(b) Let [a], [b], [c] ∈ Zn .
[a]([b][c]) = [a][bc]
= [a(bc)]
= [(ab)c], since multiplication of integers is associative
= [ab][c]
= ([a][b])[c]
Hence multiplication is associative in Zn .
(c) Let [a] ∈ Zn . Then [a][1] = [a · 1] = [a] = [1 · a] = [1][a]. Hence [1] is an identity.
(d) Let [a], [b] ∈ Zn . Then [a][b] = [ab] = [ba] = [b][a], where the middle equality is due to
the fact that multiplication of integers is commutative. Hence multiplication in Zn is
commutative.
16. Prove the distributive property of multiplication over addition holds in Zn .
Proof. Let [a], [b], [c] ∈ Zn .
[a]([b] + [c]) = [a][b + c]
= [a(b + c)]
= [ab + ac], since the distributive property of integer multiplication over
integer addition holds.
= [ab] + [ac]
= [a][b] + [a][c]
Thus the distributive property holds in Zn .
18. Let p be a prime integer. Prove that if [a][b] = [0] in Zp , then either [a] = [0] or [b] = [0].
Proof. Let p be prime. Assume [a][b] = [0] in Zp . Thus [ab] = [0]. Therefore p|ab and thus
by Euclid’s lemma p|a or p|b. Therefore [a] = [0] or [b] = [0].
22. Let p be a prime integer. Prove the following cancellation law in Zp : If [a][x] = [a][y] and
[a] 6= [0], then [x] = [y].
Proof. Let p be prime. Assume [a][x] = [a][y] in Zp and [a] 6= [0]. Thus ax ≡ ay(mod n) and
so p|ax − ay. Consequently p|a(x − y), so by Euclid’s lemma p|a or p|x − y. However p 6 |a
since [a] 6= [0]. Thus p|x − y and hence [x] = [y].
23. Show that if n is not a prime, the cancellation law stated in Excercise 22 does not hold in Zp .
Let n = 4. Then [2][1] = [2] = [6] = [2][3], but [1] 6= [3]