Math 110A HW §1.4 – Solutions
1. Decide whether the given set B is closed w.r.t the binary operation defined.
(e) x ∗ y = |x| − |y|, B = Z+
B is not closed w.r.t ∗, since 1 ∗ 2 = −1 6∈ Z+
(f) x ∗ y = x + xy, B = Z+
If x and y are positive integers, then xy is a positive integer and hence x+xy is a positive
integer. Thus B is closed w.r.t ∗.
2. Determine whether or not ∗ is commutative and/or associative. Also whether or not there is
an identity element. If so, find all inverses.
(i) x ∗ y = x − y + 1
∗ is not commutative since 1 ∗ 2 = 1 − 2 + 1 = 0 6= 2 = 2 − 1 + 1 = 2 ∗ 1.
∗ is not associative since 1 ∗ (2 ∗ 3) = 1 ∗ (2 − 3 + 1) = 1 ∗ 0 = 1 − 0 + 1 = 1, but
(1 ∗ 2) ∗ 3 = (1 − 2 + 1) ∗ 3 = 0 ∗ 3 = 0 − 3 + 1 = −2.
We claim that Z does not contain an identity w.r.t ∗.
Proof. Suppose e ∈ Z is the identity for ∗. Thus 1 ∗ e must equal 1. Thus we have the
following.
1∗e = 1
1−e+1 = 1
e=1
Thus if e is the identity, then e must equal 1. (In other words, no element other than 1
could be the identity.) However, 1 ∗ 2 = 1 − 2 + 1 = 0. Since 1 ∗ 2 6= 2, 1 cannot be the
identity. Hence the integers do not contain an identity w.r.t. ∗.
9(b) Let E be the set of even integers. Prove E is closed w.r.t multiplication.
Proof. Let n, m ∈ E. Then there exists k, l ∈ Z such that n = 2k and m = 2l. Then we have
nm = (2k)(2l) = 2(2kl). Since k, l ∈ Z, we know that 2kl ∈ Z. Thus nm is an even integer.
Hence the set of even integers is closed w.r.t. multiplication.
13. Assume that ∗ is an associative binary operation on A with an identity element. Prove that
the inverse of an element is unique when it exists.
Proof. Let x ∈ A and let e ∈ A be the identity element. Assume x has an inverse with
respect to ∗. Suppose b and c are inverses of x. Then we know that x ∗ b = e = b ∗ x and
x ∗ c = e = c ∗ x. Therefore we have the following.
x∗b= x∗c
b ∗ (x ∗ b) = b ∗ (x ∗ c)
b ∗ (x ∗ b) = (b ∗ x) ∗ c,
b∗e = e∗c
b=c
Thus the inverse is unique.
since ∗ is associative