BASIC MATHEMATICS
ASSIGNMENT 5
5.2.3 
(8pt) Let A = {a,
 b, c, d, e, f, g, h, i, j, k, l, m, n, o, p} and M =
a b c d
 e f g h 


 i j k l  be a 4 × 4 matrix. Define R on the set A by
m n o p
xRy if and only if x and y are on the same row of the matrix
M.
(a) Prove that R is an equivalence relation on A.
(b) What are the equivalence classes under R?
Proof.
(a) For any x, y, z ∈ A, then x is in the same row of M as itself,
so xRx and hence R is reflexive. If xRy, then x and y are
in the same row of M , so we immediately have yRx, hence
R is symmetric. If xRy and yRZ, then x, y are in the same
row of M and y, z are in the same row of M . Thus, x, z
are in the same row of M , which implies that xRz. So R
is transitive. Therefore, R is an equivalence relation on A.
(b) By definition of R, there are four equivalence classes of R
which are
{a, b, c, d},
{e, f, g, h},
{i, j, k, l},
{m, n, o, p}
respectively.
5.2.6. (8pt) Let A be any nonempty set. Define S on A by S = A × A.
(a) Verify that S is an equivalence relation.
(b) Show that [a] = A for all a ∈ A.
Proof.
(a) Since S = A × A. Then
(x, x) ∈ A × A = S, ∀ x ∈ A ⇒ S is reflexive.
(x, y), (y, x) ∈ A × A = S, ∀ x, y, ∈ A ⇒ S is symmetric.
(x, y), (y, z), (x, z) ∈ A × A = S, ∀ x, y, z ∈ A ⇒ S is transitive.
Hence S is an equivalence relation.
(b) It is clear that [a] ⊆ A, ∀ a ∈ A. On the other hand,
for any a ∈ A. Choose any x ∈ A, we have that (a, x) ∈
A×A = S ⇒ aSx. So A ⊆ [a] and hence [a] = A, ∀ a ∈ A.
5.2.9. (10pt) Let D be the set of digits {0, 1, 2, · · · , 9}. Define a relation R on the power set P (D) by ARB if and only if the subsets
A and B have the same number of elements.
(a) Prove that R is an equivalence relation on P (D).
(b) List the elements in the equivalence class [∅] and in [D].
(c) List the elements in the equivalence class [{4}].
(d) What is in the equivalence class [{4, 7}]?
(e) How many equivalence classes are there under R?
Proof.
(a) For any A ∈ P (D), then A ⊆ D, which implies that A is
a finite set. Clearly, Card(A) = Card(A). So ARA ⇒ R
is reflexive. For A, B ∈ P (D) with ARB, then Card(A) =
Card(B), which immediately implies BRA. R is symmetric. Finally, for A, B, C ∈ P (D), with ARB and BRC,
then Card(A) = Card(B) and Card(B) = Card(C) ⇒
Card(A) = Card(C), this shows that ARC. Hence R
is transitive. Therefore, R is an equivalence relation on
P (D).
(b) For any X ⊆ A with X ∈ [∅] we must have that Card(X) =
Card(∅) = 0. The only possible is X = ∅. Thus [∅] =
{∅}. Next, for any Y ⊆ with Y ∈ [D], then Card(Y ) =
Card(D) = 10. The only possible is Y = D. Thus [D] =
{D}.
(c) For any X ⊆ D with X ∈ [{4}], then Card(X) = Card({4}) =
1. Thus
n
o
[{4}] = {0}, {{1}, {{2}, {{3}, {{4}, {{5}, {{6}, {{7}, {{8}, {{9}
(d) For any X ⊆ A with X ∈ [{4, 7}], we must have that
Card(X) = Card({4, 7}) = 2. Thus, we can conclude that
all subsets of D which consists exactly two elements are in
the equivalence class [{4, 7}].
(e) Since each equivalence class under R in P (D) is classified
by the number of elements, and Card(D) = 10. Therefore,
there are 11 equivalence classes under R.
5.2.12. (9pt) Let C be the set of all differentiable real-valued functions.
Define a relation R on C by f Rg if and only if f 0 = g 0 .
(a) Prove that R is an equivalence relation.
(b) Define the real-valued function g by g(x) = x2 . Use a
theorem of the calculus to characterized the elements in
[g].
(c) Let f ∈ C be given. Specify the elements in [f ] without
reference to a derivative.
Proof.
(a) For f, g, h ∈ C, then f 0 = f 0 ⇒ f Rf . So R is reflexive. If
f Rg, then f 0 = g 0 , this immediately implies that gRf . So
R is symmetric. If f Rg and gRh, then f 0 = g 0 , g 0 = h0 ⇒
f 0 = h0 which shows that f Rh, then R is transitive. Hence
R is an equivalence relation.
(b) For any f ∈ C with f ∈ [g], then f Rg ⇒ f 0 = g 0 = 2x.
Thus
Z
f (x) = 2xdx = x2 + α
for some constant α ∈ R. Hence
[g] = {x2 + α : α ∈ R}
(c) According to (b), we can conclude that for any g ∈ [f ],
g(x) = f (x) + α for some constant α ∈ R
5.2.15. (8pt) Let R be a symmetric and transitive binary relation on a
set A. Assume that dom(R) = A. Prove that R is an equivalence relation on A.
Proof. For any x ∈ A. Since A = dom(R), so xRy for some y ∈
A. Since R is symmetric, we have yRx. Since R is transitive,
we have xRx. This shows that R is reflexive, and thus R is an
equivalence relation on A.
5.2.18. (8pt) Let A be a given set and Π be a partition of A. Prove
that A/(A/Π) = Π.
Proof. For any x ∈ A. Since Π is a partition of A, so x ∈ W for
some W ∈ Π. Let [x] denote the equivalence of class x under
A/Π. For any y ∈ A, we have that
y ∈ [x] ⇔ x(A/Π)y ⇔ y ∈ W
This shows that [x] = W . Hence A/(A/Π) ⊆ Π. Similarly, for
any W ∈ Π. Since W 6= ∅, so there exists x ∈ A such that
x ∈ W . Thus
W = A ∩ W = {y ∈ A : x(A/Π)y} = {y ∈ A : y ∈ W } = [x]
This shows that Π ⊆ A/(A/Π). Therefore, A/(A/Π) = Π.
5.3.3. (10pt) Let partial order is defined on A = {a, b, c, d, e} by
the Hasse diagram
(a)
(b)
(c)
(d)
(e)
Find
Find
Find
Find
Find
any maximal or minimal elements.
any greatest or least elements.
all lower bounds and upper bounds of {c, e}.
the glb({c, e}), if it exists.
the lub({c, e}), if it exists.
Sol.
(a) e is a maximal element in A. a, b are minimal elements in
A.
(b) e is the greatest element in A. There is no least element in
A.
(c) c is the only lower bound of {c, e} and e is the only upper
bound of {c, e}.
(d) glb({c, e}) = c.
(e) lub({c, e}) = e.
5.3.6. (6pt) Let A = {1, 2, 4, 5, 10}, and define a partial order by x y
if and only if x divides y.
(a) Draw the Hasse diagram for the poset (A, ).
(b) Find any maximal or minimal elements.
(c) Find any greatest or least elements.
(d) Find all lower bounds and upper bounds of {2, 4, 5}.
(e) Find the glb({2, 4}), if it exists.
(f ) Find the lub({2, 4, 5}), if it exists.
Sol.
(a)
4
2
(b)
(c)
(d)
(e)
(f)
10
ooo
o
o
oo
ooo
o
o
oo
5
==
~~
==
~
==
~~
= ~~~
1
4 and 10 are maximal elements in A. And 1 is a minimal
element in A.
There is no greatest element in A. 1 is the least element
in A.
1 is the only lower bound of {2, 4, 5} and there is also no
upper bound for {2, 4, 5}.
glb({2, 4, 5}) = 1.
lub({2, 4, 5}) does not exist.
5.3.9. (8pt) Let B = {r : r ∈ Q and r2 ≤ 2}.
(a) For the poset (R, ≤), find glb(B) and lub(B), if they exist.
(b) For the poset (Q, ≤), find glb(B) and lub(B), if they exist.
Sol.
√
√
√
(a) Since in R, r2 √
≤ 2 ⇔ − 2 ≤ r ≤ 2. So glb(B) = − 2
and lub(B) = 2.
(b) Since Q is dense in R. So both glb(B) and lub(B) does
not exist in (Q, ≤).
1−3n
5.3.12. (8pt) Consider the poset (R, ≤). Let B = 1+n : n ∈ N .
Find glb(B) and lub(B), if they exist.
Sol. Observe that
1 − 3n
1 + n − 4n
4n
=
=1−
1+n
1+n
1+n
1 − 3n
4n
Since n ∈ N , so n ≥ 0 and thus
=1−
≤ 1. This
1+n
1+n
shows that lub(B) = 1 in (R, ≤). On the other hand, since
1 − 3n
lim
= lim
n→∞ 1 + n
n→∞
1
n
1
n
−3
= −3
+1
1 − 3n
4
+3 =
> 0 for n ∈ N . So we can conclude
1+n
1+n
that glb(B) = −3 in (R, ≤).
And
5.3.15. (8pt) Let R be a relation on A. Then R is irreflexive on A
if ¬ xRx for all x ∈ A. Let R be a transitive and irreflexive
relation on A, and define a relation on A as follows: x y if
and only if xRy or x = y. Show that is a partial ordering on
A.
Proof. For any x ∈ A. Clearly we have that x = x, and hence
we have x x. So is reflexive. Next, for any x, y ∈ A
such that x y and y x. Suppose that x 6= y, then by the
definition of the relation , we have that xRy and yRx. Since
R is transitive, so we must have that xRx, which contradict
to that R is irreflexive. Hence we have that x = y. So is
antisymmetric Finally, if x, y, z ∈ A with x y and y z,
then one of the following cases hold:
xRy
x=y
xRy
x=y
and
and
and
and
y=z
yRz
yRz
y=z
⇒xRz
⇒xRz
⇒xRz
⇒xRz
⇒x z
⇒x z
⇒x z, since R is transitive
⇒x z
In either case, we all have that x z. Hence is transitive,
which shows that is a partial ordering on A.
5.3.18. (9pt) Let R be a transitive and reflexive relation on A. Define
a relation ≈ on A by x ≈ y if and only if xRy and yRx.
(a) Show that ≈ is an equivalence relation on A.
(b) Define a relation on A/ ≈ by [x] [y] if and only if xRy.
Show that is well defined.
(c) Show that (A/ ≈, ) is a poset.
Proof.
(a) Since R is reflexive, so xRx, ∀ x ∈ A ⇒ x ≈ x, ∀ x ∈
A, which implies that ≈ is reflexive. If x, y ∈ A with
x ≈ y, then xRy and yRx. This immediately implies that
y ≈ x. Hence ≈ is symmetric. Finally, if x ≈ y and
y ≈ z, x, y, z ∈ A,. Then we have that xRy and yRx
and yRz and zRy, then by transitivity of R we have that
xRz and zRx ⇒ x ≈ z. Thus ≈ is transitive. So ≈ is an
equivalence relation.
(b) For any x0 ∈ [x] and any y 0 ∈ [y], we have x ≈ x0 , y ≈ y 0 .
By definition of ≈, we have that xRx0 , x0 Rx, yRy 0 , y 0 Ry.
Since [x] [y], so xRy. And by transitivity of R, we have
that xRy 0 . Since x0 Rx, by transitivity of R again, we get
x0 Ry 0 . This shows that is well defined.
(c) By reflexivity of R, we have that xRx, ∀ x ∈ A ⇒ [x] [x], ∀ x ∈ A. Hence is reflexive. For [x], [y] ∈ A/R
such that [x] [y] and [y] [x], then xRy and yRx which
immediately implies that x ≈ y, that is, [x] = [y]. This
shows that is antisymmetric. If [x] [y] and [y] [z], z ∈ A, then xRy and yRz. Since R is transitive,
we xRz which implies that [x] [z]. And hence is
transitive. Therefore, (A/ ≈, ) is a poset.