Download SOLUTIONS FOR THE FIRST PROBLEM SET Page 4, problem 5(a

SOLUTIONS FOR THE FIRST PROBLEM SET
Page 4, problem 5(a): This f is not well-defined. For example, if we let x = 2/3, then
f (x) = 2. But we also have x = 4/6, and the stated definition would give f (x) = 4. Thus,
f is not well-defined.
Page 4, problem 7: The stated relation is an equivalence relation on the set A. We must
verify that the relation is reflexive, symmetric, and transitive.
Proof of reflexivity: Suppose a ∈ A. Then f (a) = f (a) and so we do indeed have a ∼ a.
Proof of symmetry: Suppose a, b ∈ A and that a ∼ b. Then f (a) = f (b). Therefore,
f (b) = f (a). Therefore, we indeed have b ∼ a.
Proof of transitivity: Suppose that a, b, c ∈ A, that a ∼ b, and that b ∼ c. Since a ∼ b, we
have f (a) = f (b). Since b ∼ c, we have f (b) = f (c). Hence, we have f (a) = f (c). Therefore,
we indeed have a ∼ c.
Note that the above proof does not use the assumption that f is surjective.
Suppose that a ∈ A. The equivalence class of a under the relation ∼ is:
{ b ∈ A | b ∼ a } = { b ∈ A | f (b) = f (a) } .
This set is precisely the fiber of f over f (a). (See page 2 in the text for the definition of the
term “fiber.”) Thus, every equivalence class for ∼ is indeed a fiber of the map f .
Now consider any fiber F of the map f . By definition, for some c ∈ B, we have
F = { b ∈ A | f (b) = c } .
Under the assumption that the map f is surjective, we have f (a) = c for some a ∈ A. Then,
F = { b ∈ A | f (b) = f (a) } ,
which is precisely the equivalence class for a under the relation ∼. Thus, under the assumption that f is surjective, every fiber of f is an equivalence class for ∼.
Under the assumption that f is surjective, we have shown that the fibers of f are precisely
the equivalence classes for ∼. However, if f is not surjective, then there exists an element
c ∈ B such that c is not in the image of f . The fiber of f over c will then be empty. This
fiber will not be an equivalence class for the relation ∼.
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Page 11, problem 9: Suppose that a is an odd integer. This means that we can write a
as a = 2k + 1 for some k ∈ Z. Then we obtain
a2 = (2k + 1)2 = 4k 2 + 4k + 1 = 4(k 2 + k) + 1 .
Notice that k 2 + k = k(k + 1), a product of two consecutive integers. One of those integers
must be even. Therefore, we can write k 2 + k = 2m for some integer m. It follows that
a2 = 4(k 2 + k) + 1 = 4(2m) + 1 = 8m + 1
and so a2 indeed leaves a remainder of 1 when divided by 8.
Page 11, problem 11: We use the definition on page 10: Since a and b are in (Z/nZ)× ,
there exist elements c and d in (Z/nZ) such that
a·c=1
and
b·d=1 .
Multiplication in Z/nZ is a commutative and associative operation. Using associativity and
commutativity many times, we have
(a · b) · (c · d) = (a · c) · (b · d) = 1 · 1 = 1 .
Furthermore, we have c · d = cd by definition. Hence (a · b) · (cd) = 1. Since cd is in (Z/nZ),
we have indeed verified that a · b is in (Z/nZ)× .
Page 21, problem 1: We do parts (a), (b), and (d).
(a):
Suppose a, b, c ∈ Z. Notice that
(a ⋆ b) ⋆ c = (a − b) − c = a − b − c ,
and
a ⋆ (b ⋆ c) = a − (b − c) = a − b + c .
The binary operation ⋆ is not associative. As a counterexample, take a = b = 0 and c = 1.
We then have (a ⋆ b) ⋆ c = −1 and a ⋆ (b ⋆ c) = 1, and so (a ⋆ b) ⋆ c 6= a ⋆ (b ⋆ c).
(b):
(1)
This binary operation ⋆ is associative. The fact that
(a ⋆ b) ⋆ c = a ⋆ (b ⋆ c)
for all a, b, c ∈ R can be verified by using elementary algebra. But we will proceed in
a different way, reducing the equality (1) to the associative law for multiplication of real
numbers. Consider the real numbers 1 + a, 1 + b, and 1 + c, Then
(2)
(1 + a)(1 + b) (1 + c) = (1 + a) (1 + b)(1 + c) .
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This is true because multiplication of real numbers is an associative operation.
We will use the follow equation. Let x, y ∈ R. Then
(3)
(1 + x)(1 + y) = 1 + x + y + xy = 1 + x ⋆ y
Using (3) several times, we find that
(1 + a)(1 + b) (1 + c) =
and
.
1 + a ⋆ b (1 + c) = 1 + (a ⋆ b) ⋆ c
(1 + a) (1 + b)(1 + c) = (1 + a) 1 + b ⋆ c = 1 + a ⋆ (b ⋆ c) .
Now we can use equation (2). We obtain
1 + (a ⋆ b) ⋆ c = 1 + a ⋆ (b ⋆ c) .
It follows that (a ⋆ b) ⋆ c = a ⋆ (b ⋆ c) for all a, b, c ∈ R. Therefore, (1) does hold for all
a, b, c ∈ R, as we wanted to show.
(d): Suppose that (a, b), (c, d), and (e, f ) are elements of Z × Z. We compute
(a, b)⋆(c, d) ⋆(e, f ) = (ad+bc, bd)⋆(e, f ) = (ad+bc)f +(bd)e, (bd)f = (adf +bcf +bde, bdf )
and
(a, b)⋆ (c, d)⋆(e, f ) = (a, b)⋆(cf +de, df ) = a(df )+b(cf +de), b(df ) = (adf +bcf +bde, bdf ) .
We have used the fact that multiplication in Z is associative many times, as well as the
distributive law. It is clear from the above equations that ⋆ is associative.
An alternative approach is to use the fact that matrix multiplication is an associative
operation. Note that
bd ad + bc
b a d c
.
=
0
bd
0 b 0 d
Page 21, problem 2: We do parts (a), (b), and (d).
(a): This operation ⋆ is not commutative. As a counterexample, take a = 1 and b = 2.
Then
a ⋆ b = 1 ⋆ 2 = 1 − 2 = −1 ,
but
b⋆a=2⋆1=2−1=1
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and so a ⋆ b 6= b ⋆ a.
(b): This operation ⋆ is commutative. For if a, b ∈ R, then
a ⋆ b = a + b + ab = b + a + ba = b ⋆ a .
We have used the facts that addition and multiplication of real numbers are commutative
operations.
(d): This operation ⋆ is commutative. Suppose that (a, b) and (c, d) are in Z × Z. Then,
by definition,
(a, b) ⋆ (c, d) = (ad + bc, bd) ,
and
(c, d) ⋆ (a, b) = (cb + da, db) .
Since ad + bc = cb + da and bd = db for all a, b, c, d ∈ Z, it follows that
(a, b) ⋆ (c, d) = (c, d) ⋆ (a, b)
for all (a, b) and (c, d) in Z × Z.
Page 21, problem 5: Suppose that n > 1. We want to show that Z/nZ is not a group
under multiplication (as defined on page 9 in the text). Assume to the contrary that Z/nZ
is a group under multiplication. Suppose that e is the identity element in that group. In
particular, we would then have e · 1 = 1.
Note that since n > 1, we have 0 6= 1. Also, note that 0 · 1 = 0 and so 0 · 1 6= 1. Therefore,
e cannot be 0. That is, we have e 6= 0.
Since we are assuming that Z/nZ is a group under multiplication, requirement (d) would
imply that there exists an element b in Z/nZ such that
0·b = e .
However, by definition, we have 0 · b = 0. It follows that e = 0. This is a contradiction
since we have previously pointed out that e 6= 0.
Page 21, problem 6: We discuss parts (a),(c), (e), and (f). In each part, we will let S
denote the described set. We will refer to the four requirements (a) - (d) occurring in the
definition of the term “group” on our first handout.
(a): This set S is a group under addition. First of all, S is closed under addition. To see
this, assume that r and s are in S. We can write r and s in lowest terms. Thus, we write
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r = a/b and s = c/d, where a, b, c, d are assumed to be integers, a and b are relatively prime,
and c and d are relatively prime. Also, since r, s ∈ S, their denominators b and d are both
odd. Then
r + s = a/b + c/d = (ac + bd)/(bd)
Note that ac + bd and bd are integers. Since b and d are odd, so is bd. The denominator of
the rational number r + s will be a divisor of bd and hence will also be odd. Therefore, r + s
is in S.
The associative law is certainly satisfied for addition on S. Note that 0 = 0/1 has
denominator 1. Since 1 is odd, we have 0 ∈ S. We can take 0 as the identity element
e in requirement (c). Finally, if r is any rational number, then r and −r have the same
denominator. Thus, if r ∈ S, then so is −r. Hence it is clear that requirement (d) is
satisfied.
(c):
This set S is not a group under addition. It is not closed under addition. As a
counterexample, take a = 2/3 and b = 1/3, both of which are in the set S. But their sum
a + b = 1 is not in the set S.
(e):
This set S is a group under addition. One can describe S as follows:
S = { m/2 | m ∈ Z }.
The set S is closed under addition. For if r, s ∈ S, we can write r and s in the form
r = a/2, s = b/2, where a, b ∈ Z. Then
r + s = (a + b)/2
which is indeed in the set S. We use the fact that a + b ∈ Z.
The associative law is obviously satisfied. Furthermore, it is clear that 0 ∈ S because
0 = 0/2. We can take e to be 0 for requirement (c). As for requirement (d), note that if
r ∈ S, then −r has the same denominator as r and is therefore also in S.
(f ): This set S is not a group under addition. The set S is not closed under addition. As
a counterexample, take a = 1/3 and b = 1/2, both of which are in the set S. However,
a + b = 5/6
has denominator 6 and is therefore not in the set S
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Page 22, problem 8: In this problem, we consider the set
G = { z ∈ C | z n = 1 for some n ∈ Z+ } .
G is a subset of C. There are two operations on C, addition and multiplication, which we
write as + and ·.
Part (a): To show that G is a group under multiplication, there are four requirements in
the definition to verify. First of all, we verify that G is closed under multiplication. This
will show that requirement (a) in the definition is satisfied. Suppose that z1 and z2 are in
G. Then there exist positive integers n1 and n2 such that
z1n1 = 1,
and
z2n2 = 1 .
Let n = n1 n2 , which is a positive integer. Then
z1n = (z1n1 )n2 = 1n2 = 1,
and
z2n = (z2n2 )n1 = 1n1 = 1 ,
and so we have
(z1 · z2 )n = z1n · z2n = 1 · 1 = 1 .
Therefore, we have shown that z1 · z2 is in G.
Requirement (b) is satisfied. Multiplication of complex numbers is an associative operation. Restricting that operation to G will still be an associative operation.
For requirement (c), we can take e = 1. It obviously is in G because 11 = 1. Clearly, we
have 1 · z = z · 1 = z for all z ∈ G.
Finally, we verify requirement (d). Suppose that z ∈ G. Then z n = 1 for some positive
integer n. It follows that z 6= 0. Hence z has an inverse under multiplication in C, which we
will denote by 1/z. We have
(1/z)n = 1/z n = 1/1 = 1
and so we see that 1/z ∈ G. Note also that z · (1/z) = (1/z) · z = 1. Thus, requirement (d)
is satisfied.
Part (b): Now we consider G under the operation of addition. Requirement (a) fails to be
true. As a counterexample, take z1 = z2 = 1. Then z1 ∈ G and z2 ∈ G, but z1 + z2 = 2 and
it is clear that 2 6∈ G. For if n is any positive integer, then 2n ≥ 2 > 1 and therefore 2n 6= 1.
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A: The group operation in SΩ is composition. We find for part (a):
1 2 3
1 2 3
1 2 3
2
,
=
σ = σσ =
3 1 2
2 3 1
2 3 1
1 2 3
1 2 3
1 2 3
3
2
=ε ,
=
σ = σσ =
1 2 3
3 1 2
2 3 1
1 2 3
1 2 3
1 2 3
2
= ε,
=
τ = ττ =
1 2 3
2 1 3
2 1 3
1 2 3
1 2 3
1 2 3
=
στ =
3 2 1
2 1 3
2 3 1
1 2 3
1 2 3
1 2 3
,
=
τσ =
1 3 2
2 3 1
2 1 3
1 2 3
1 2 3
−1
−1
.
,
τ =
σ =
2 1 3
3 1 2
Part (b):
We already computed τ σ in part (a). We find that
1 2 3
1 2 3
1 2 3
2
,
=
σ τ=
1 3 2
2 1 3
3 1 2
which is indeed equal to τ σ.
Here is the deduction that τ στ −1 = σ 2 .
τ σ = σ2τ
=⇒ (τ σ)τ −1 = (σ 2 τ )τ −1 =⇒ (τ σ)τ −1 = σ 2 (τ τ −1 )
=⇒ τ στ −1 = σ 2 ε =⇒ τ στ −1 = σ 2 .
We have used the associative law.
Part (c): There are only six elements in SΩ . The identity element ε has order 1 and
certainly satisfies the equation ε6 = ε. The other five elements in SΩ have the following cycle
decompositions:
(1 2 3),
(1 3 2),
(1 2),
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(1 3),
(2 3) .
If ϕ is either one of the first two elements listed above, then ϕ has order 3 and satisfies the
equation ϕ3 = ε. Hence we have ϕ6 = (ϕ3 )2 = ε2 = ε
If ϕ is one of the last three elements listed above, then ϕ has order 2 and satisfies the equation
ϕ2 = ε. Hence we have ϕ6 = (ϕ2 )3 = ε3 = ε.
Therefore, if ϕ is any one of the six elements in SΩ , then we have verified that ϕ6 = ε.
Problem B: The function f : R → R+ defined by f (x) = ex has all x ∈ R has the required
properties. We explain this briefly.
The function f (x) is a strictly increasing function on R because its derivative is positive
for all x ∈ R. Therefore, f is injective.
As x → −∞, f (x) → 0. As x → +∞, f (x) → +∞. Since f (x) is continuous, the
intermediate value theorem implies that f is surjective. (Note that R+ denotes the set of
positive real numbers. )
It follows that f is injective and surjective. Hence f is a bijection from R to R+ . Also,
we have
f (x + y) = ex+y = ex ey = f (x)f (y)
as required.
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