H
2 Binary Operations
1-4, 7-20, 26-28
1. b ∗ d = e, c ∗ c = b, [(a ∗ c) ∗ e] ∗ a = [c ∗ e] ∗ a = a ∗ a = a
2. (a ∗ b) ∗ c = b ∗ c = a
a ∗ (b ∗ c) = a ∗ a = a
This is not enough to say ∗ is associative - only one example.
3. (b ∗ d) ∗ c = e ∗ c = a
b ∗ (d ∗ c) = b ∗ b = c
No, ∗ is not associative.
4. No, it is not commutative. b ∗ e 6= e ∗ b
7. ∗ defined on Z by letting a ∗ b = b − a.
Not associative
(a ∗ b) ∗ c = (a − b) ∗ c = a − b − c
a ∗ (b ∗ c) = a ∗ (b − c) = a0b + c
Not commutative
a ∗ b = a − b but b ∗ a = b − a
8. ∗ is defined on Q by letting a ∗ b = ab + 1.
Commutative
a ∗ b = ab + 1 and b ∗ a = ba + 1
Not associative
(a ∗ b) ∗ c = (ab + 1) ∗ c = abc + c + 1
a ∗ (b ∗ c) = a ∗ (bc + 1) = abc + a + 1
9. ∗ is defined on Q by letting a ∗ b =
Commutative
ba
a ∗ b = ab
2 and b ∗ a = 2
Associative
abc
(a ∗ b) ∗ c = ab
2 ∗c= 4
bc
a ∗ (b ∗ c) = a ∗ 2 = abc
4
ab
2
10. ∗ is defined on Z + by letting a ∗ b = 2ab
Commutative
a ∗ b = 2ab and b ∗ a = 2ba
Not associative
ab
(a ∗ b) ∗ c = 2ab ∗ c = 22 c
bc
a ∗ (b ∗ c) = a ∗ 2bc = 2a2
11. ∗ is defined on Z + by letting a ∗ b = ab
Not commutative
a ∗ b = ab but b ∗ a = ba
Not associative
c
a ∗ (b ∗ c) = a ∗ bc = ab but (a ∗ b) ∗ c = ab ∗ c = (ab )c = abc
12.
1 element: (element must be identity) 1
3 elements: 39
2
n elements: nn
Think of a multiplication table, n rows and n columns with n choices for each value.
n(n+1)
13. n 2
For each pair of distinct elements, where order doesn’t matter, there are n choices. So think
of it as for each case, it is the number of elements that lie above or on the main diagonal
if we write all the relationships in a matrix. There are n main diagonal entries and there
are n(n−1)
entries above the main diagonal. So, n + n(n−1)
= n(n+1)
. And since there are n
2
2
2
elements, we have the desired result.
14. Need to add ∀ a, b ∈ S
15. Not iff
16. ∀ a, b ∈ H
17. Not closed (condition 2)
18. Closed
19. Closed
20. Closed
26. Let a, b, c, d ∈ S and let ∗ be an associative and commutative binary operation. Consider
(a ∗ b) ∗ (c ∗ d) = (a ∗ b) ∗ (d ∗ c) = (d ∗ c) ∗ (a ∗ b) = [(d ∗ c) ∗ a] ∗ b
27. Proof: Let s ∈ S. Since ∗ is a binary operation, it must be that S is closed under ∗, so s ∗ s = s
and so ∗ is commutative. Also, s ∗ (s ∗ s) = s ∗ s = (s ∗ s) ∗ s, so ∗ is associative.
28. Consider
∗
a
b
a
b
b
b
b
a
Then, a ∗ (a ∗ b) = a ∗ b = b but (a ∗ a) ∗ b = b ∗ b = a, so this ∗ is not associative.