Download Math 325 - Dr. Miller - HW #3: Identities, Inverses, and a bit of

Math 325 - Dr. Miller - HW #3: Identities, Inverses, and a bit of Groups - SOLUTIONS, 9/14/12
1. Prove whether each set contains an identity for the given operation.
(a) Z+ , a ∗ b = max(a, b), the maximum of the numbers
Given a, e ∈ Z+ , a ∗ e = a means that max(a, e) = a. Consider e = 1, which is an
element of Z+ . For all a ∈ Z, a ∗ 1 = max(a, 1) = a and 1 ∗ a = max(1, a) = a
because 1 is the smallest positive integer. Therefore, Z+ contains an identity for ∗.
OR YOU COULD TRY THE “IF” STYLE OF PROOF SET-UP:
If a ∗ e = a, that means max(a, e) = a. Consider e = 1 ∈ Z+ . For all a ∈ Z,
a ∗ 1 = max(a, 1) = a and 1 ∗ a = max(1, a) = a because 1 is the smallest positive
integer. Therefore, Z+ contains an identity e = 1 for ∗.
(b) Z+ , a ∗ b = min(a, b), the minimum of the numbers
Given a, e ∈ Z+ , a ∗ e = a means that min(a, e) = a. This requires that every
positive integer a be less than or equal to some fixed positive integer e, which is not
possible. Thus, Z+ doesn’t contain an identity for ∗.
OR YOU COULD TRY THE “IF” STYLE OF PROOF SET-UP:
If a ∗ e = a, that means min(a, e) = a. This requires that every positive integer a
be less than or equal to some fixed positive integer e, which is not possible. Thus,
Z+ doesn’t contain an identity for ∗.
OR YOU COULD USE PROOF BY CONTRADICTION:
Suppose there exists an identity e ∈ Z+ for ∗. Then a ∗ e = e for all a ∈ Z+ means
min(a, e) = a for such elements a, so that a ≤ e for all a ∈ Z+ . Yet e + 1 ∈ Z+ ,
and e ∗ (e + 1) = e by definition of minimum yet also equals e + 1 by definition of
identity. The equality e + 1 = e leads to 1 = 0, a contradiction. Therefore, no such
e exists; Z+ does not contain an identity for ∗.
(c) Z, a ∗ b = ab − a − b + 2
Given a, e ∈ Z, a ∗ e = a means that ae − a − e + 2 = a, or e(a − 1) − 2(a − 1) = 0,
whence a = 1 or e = 2. Consider e = 2 ∈ Z. For all a ∈ Z, a∗2 = 2a−a−2+2 = a
and 2 ∗ a = 2a − 2 − a + 2 = a, as desired. Thus, Z contains an identity, 2, for ∗.
OR YOU COULD TRY THE “IF” STYLE OF PROOF SET-UP:
If a ∗ e = a, that means ae − a − e + 2 = a, or e(a − 1) − 2(a − 1) = 0, whence
a = 1 or e = 2. Consider e = 2 ∈ Z. For all a ∈ Z, a ∗ 2 = 2a − a − 2 + 2 = a and
2 ∗ a = 2a − 2 − a + 2 = a, as desired. Thus, Z contains an identity, 2, for ∗.
(d) T = {a + bi ∈ C | ab 6= 0}, (a + bi) ∗ (c + di) =
a b
+ i
c d
Given a + bi, e + e0 i ∈ T , where ab 6= 0 and ee0 6= 0, (a + bi) ∗ (e + e0i) = a + bi
means that ae + eb0 i = a + bi, or ae = a, whence a = 0 (not allowed anyway) or
e = 1, and eb0 = b, whence b = 0 (again, irrelevant because not allowed) or e0 = 1.
Consider 1 + i, which belongs to T because 1, 1 ∈ R and 1 · 1 6= 0. It is not true
that (1 + i) ∗ (a + bi) = a + bi for all a + bi ∈ T , for consider 2 + 2i ∈ T (it belongs
to T because 2 · 2 6= 0). We have (1 + i) ∗ (2 + 2i) = 12 + 12 i 6= 2 + 2i. Because 1 + i
was the only candidate for an identity, no such element exists in T .
OR YOU COULD TRY THE “IF” STYLE OF PROOF SET-UP:
If (a + bi) ∗ (e + e0i) = a + bi, that means ae + eb0 i = a + bi, or ae = a, whence
a = 0 (not allowed anyway) or e = 1, and eb0 = b, whence b = 0 (again, irrelevant
because not allowed) or e0 = 1. Consider e + e0 i = 1 + i, which belongs to T because 1, 1 ∈ R and 1 · 1 6= 0. It is not true that (1 + i) ∗ (a + bi) = a + bi for all
a + bi ∈ T , for consider 2 + 2i ∈ T (it belongs to T because 2 · 2 6= 0). We have
(1 + i) ∗ (2 + 2i) = 12 + 12 i 6= 2 + 2i. Because 1 + i was the only candidate for an
identity, no such element exists in T .
=
"
ax 2by
3cz 4dw
Given A, E ∈ M2×2 (Q ), with A =
"
a b
c d
+
(e) M2×2 (Q ),
"
a b
c d
#
∗
"
x∗y
z∗w
+
that
"
#
ae 2be0
3ce00 4de000
#
#
#
and E =
=
"
a b
c d
"
#
e e0
e00 e000
#
A ∗ E = A means
,
yielding
a = 0 or e = 1,
1
b = 0 or e0 = ,
2
1
00
c = 0 or e = , and
3
1
000
d = 0 or e = .
4
The matrix E =
A∗E =
"
a b
c d
"
#
1 1/2
1/3 1/4
#
∈ M2×2 (Q+ ), and for all A =
"
a b
c d
#
∈ M2×2 (Q+ ),
= E ∗ A. Thus, M2×2 (Q+ ) contains an identity for ∗.
OR YOU COULD TRY THE “IF” STYLE OF PROOF SET-UP, which I won’t
produce this time to save space.
2. For each set in Problem #1 that DOES contain an identity, determine whether the set
contains inverses for each of its elements. Prove your claim.
(a) Z+ , a ∗ b = max(a, b), e = 1
If a ∗ b = e, that means max(a, b) = 1. Consider a = 2 ∈ Z+ . For every positive
integer b, max(a, b) ≥ a = 2, so that a ∗ b can never equal 1. Therefore, there does
not exist an element 2−1 ∈ Z+ .
(b) Z+ , a ∗ b = min(a, b), no identity - Then inverses cannot exist at all.
(c) Z, a ∗ b = ab − a − b + 2, e = 2
If a ∗ b = e, that means ab − a − b + 2 = 2, so that b = a/(a − 1). Consider
a = 3 ∈ Z (you thought I’d pick a = 1, didn’t you?). Assume 3 ∗ b = 2, whence
3b − 3 − b + 2 = 2, so that b must equal 3/2. However, b = 3/2 is not an integer,
so there does not exist an element 3−1 in Z.
(d) T = {a + bi ∈ C | ab 6= 0}, (a + bi) ∗ (c + di) =
+
(e) M2×2 (Q ),
"
a b
c d
#
∗
"
x y
z w
If A ∗ B = E, that means
ax = 1,
"
"
#
=
a b
c d
"
#
a b
+ i, no identity - So no inverses.
c d
#
∗
ax 2by
,E=
3cz 4dw
"
1 1/2
1/3 1/4
"
1
2
1
4
#
2by = 1/2,
x y
z w
#
=
"
1
1
3
3cz = 1/3,
#
"
#
, so that
and 4dw = 1/4.
#
1
1
a b
Let
∈ M2×2 (Q+ ) and consider a1 4b
. Because a, b, c, d ∈ Q+ , which
1
c d
9b
16d
is closed under multiplication and division, the entries of this new matrix are also
in Q+ . Furthermore,
"
a b
c d
#
∗
"
1
a
1
9b
1
4b
1
16d
#
=
"
1
1
3
1
2
1
4
#
=
"
1
a
1
9b
1
4b
1
16d
#
∗
"
a b
c d
#
,
so that every A ∈ M2×2 (Q+ ) has an inverse A−1 ∈ M2×2 (Q+ ) as well.
3. Use the definition of inverse to DERIVE the formula for the multiplicative inverse of the
non-zero complex number a + bi; you may NOT use the formula for complex division.
(In fact, this problem is asking you to demonstrate how that formula actually comes
about in the first place.) A little high school algebra-style system solving – albeit ramped
up slightly – should be necessary in this problem.
If (a + bi) ∗ (c + di) = e + e0i, that means (a + bi)(c + di) = 1, so (ac − bd) + (ad+ bc)i = 1,
whence
ac − bd = 1
ad + bc = 0
and
(1)
(2)
(We need to solve for c and d.) Because a + bi is nonzero, a and b aren’t both 0, so
assume without loss of generality that a 6= 0. Then we can solve Eqn #2 for d to get
d = −(bc)/a. Substituting this expression into Eqn #1 produces
bc
ac − b −
a
!
= 1, so that a2 c + b2c = a, whence c = a/(a2 + b2).
Back-substituting this result for c into the previous one for d creates
d=
a
1
−b
−bc
= −b · 2
·
=
.
a
a + b2 a
a 2 + b2
(I know at one stage we assumed that a 6= 0, but because that condition does not feature
in our expressions for c and d any more, it’s possible that it may be ignored. We’re
about to prove that it truly doesn’t matter in the end which of a or b was nonzero.)
a
−b
Let a + bi be a nonzero complex number, and consider c + di = a2 +b
2 + a2 +b2 i. Because
a + bi is nonzero, a2 + b2 6= 0. Also a and b are real, so the quotients a/(a2 + b2) and
−b/(a2 + b2) are also real, making c + di a complex number. (Note that the problem’s
instructions don’t require us to prove that c + di is nonzero as well.)
2
2
a
−b
a
b
−ab
ab
Finally, (a + bi)( a2+b
2 + a2 +b2 i) = a2 +b2 + a2 +b2 + ( a2 +b2 + a2 +b2 )i = 1 and similarly
a
−b
−1
( a2 +b
exists in C.
2 + a2 +b2 i)(a + bi) = 1, so that the element (a + bi)
4. Prove that the subset P of M2×2 (R) defined by P =
forms a multiplicative group.
("
a b
c d
# )
ad > 0 and c = 0
You may use without proof facts about matrix multiplication in M2×2 (R) – that it is
associative, and what the identity and formulas for inverses are – but be sure to address these claims for the actual set
" P . For
# example, you are allowed to know that
1 0
in M2×2 (R), the identity matrix is
, but how does that help you prove that P
0 1
contains an identity?
"
#
"
#
a b
x y
Proof of closure: Let A =
,B =
∈ P , where ad > 0, xw > 0, and
c d
z w
"
#
ax + bz ay + bw
c = z = 0. Then A ∗ B =
. Because c = z = 0, the lower left
cx + dz cy + dw
entry cx + dz = 0 also. The product (ax + bz)(cy + dw) = axcy + axdw + bzcy + bzdw =
0 + axdw + 0 + 0 = (ad)(xw) and is greater than 0 because it is the product of positive
real numbers. Thus, A ∗ B ∈ P , proving that P is closed under multiplication.
Proof of associativity. We know that matrix multiplication is associative for all matrices
in M2×2 (R); therefore, it is associative for the subset P .
Proof of existence
of
"
# an identity. We know that the multiplicative identity matrix in
1 0
M2×2 (R) is
. This matrix belongs to P as well, for its (2,1)-entry is 0, and the
0 1
product of its diagonal entries is 1, a positive number.
Proof of existence of
" inverses
# for all elements. We know that if we have an
" INVERT#
a b
d −b
1
−1
IBLE matrix A =
in M2×2 (R), then its inverse is A = ad−bc
.
c d
−c a
Every A ∈ P is invertible, for the determinant ad − bc is nonzero because ad > 0
−1
and
c = 0. (We
actually belongs to P .) Rewrite A−1 =
"
# must still prove that A
d
ad−bc
−c
ad−bc
−b
ad−bc
a
ad−bc
. Because c = 0 and the determinant is nonzero, the (2,1)-entry of this
a
ad
d
· ad−bc
= ad−0
= 1, clearly a positive
matrix is 0, as desired. Also, the product ad−bc
−1
number. Thus, every element A ∈ P has an inverse A that is also an element of P .