Abstract Algebra Beamer Lesson1

Abstract Algebra
Abstract Algebra
Gemma P. Salasalan,Ph.D
Institute of Arts and Sciences
Davao del Sur State College
Matti, Digos City
January 2022
Gemma P. Salasalan,Ph.D Abstract Algebra

Abstract Algebra
Textbook: Algebra by Hungerford
References:
1. Fundamental Abstract Algebra by Malik
2. Abstract Algebra by Fraleigh

Abstract Algebra
Sets
The composition of Real Numbers

Abstract Algebra
Sets
Definition 1.1.
A set is a collection of objects. A set S with only a finite number of
elements is called a finite set; otherwise S is called an infinite set. We
let |S| denote the number of elements of S. We denote a finite set by a
listing of its elements within braces {}.
Given a set S, we use the notation x ∈ S and x < S to mean x is a member
{element} of S and x is not a member {element} of S, respectively.
Definition 1.2.
A set A is said to be a subset of a set S if every element of A is an
element of S. In this case, we write A ⊆ S and say that A is contained in
S. If A ⊆ S, but A , S, then we write A ⊂ S and say that A is a proper
subset of S.

Abstract Algebra
Sets
Theorem 1.3.
Let A and B be sets. Then A = B if and only if A ⊆ B and B ⊆ A.

Abstract Algebra
Sets
Definition 1.4.
The null set or empty set is the set with no elements. We usually
denote the empty set by ∅. For any set A, we have ∅ ⊆ A.
Given a set S, the notation
A = {x ∈ S|P(x)}
Definition 1.5.
Definition 1.6.

Abstract Algebra
Sets
Definition 1.7.

Abstract Algebra
Sets
Example 1.8.

Abstract Algebra
Sets

Abstract Algebra
Sets
Definition 1.9.

Abstract Algebra
Sets
Example 1.10.
Let A = {1, 2, 3, 4} and B = {3, 4, 5, 6}. Then AB = {1, 2}.
Definition 1.11.

Abstract Algebra
Sets
Example 1.12.
Let (x, y), (z, w) ∈ A × B. Show that (x, y) = (z, w) if and only if x = z
and y = w.

Abstract Algebra
Sets
Definition 1.13.
For a subset A of a set S, let A′
denote the subset SA. A′
is called the
complement of A in S.

Abstract Algebra
Arithmetic in Z
Definition 2.1 (Well-ordering Principle).
Every nonempty subset of the set of nonnegative integers contains a
smallest element.
Example 2.2.
1. The set of Natural Numbers N is a well-ordered set.
2. The open interval (0, 2) is a non-empty subset of R but it has no
smallest element.
Consider the following grade-school division problem:

Abstract Algebra
Arithmetic in Z
Denote a− dividend, b− divisor, q− quotient and r− remainder
Theorem 2.3 (The Division Algorithm).
Let a, b be integers with b > 0. Then there exist unique integers q and r
such that
a = bq + r and 0 ≤ r < b.
Example 2.4.

Abstract Algebra
Arithmetic in Z
Proof :
Suppose a, b ∈ Z and b > 0. Consider
S = {a − bq|q ∈ Z and a − d ≥ 0}.
We need to show that:
1. S , ∅.
2. By WOP, we can find a least element r ∋ r < b.
2. Uniqueness, that is, r and q are unique.
1. Consider the following cases:
Case 1. a ≥ 0.
We take q = 0, then a − bq = a − b(0) = a . This implies that a ∈ S.
Case 2. a < 0 .
We take q = a, then a − b(a) = a(1 − b). Note that a < 0 , b > 0 (from
assumption), and a(1 − b) ≥ 0 . This implies that a − b(a) ∈ S.
Thus, S , ∅. By WOP, S has a least element r = a − bq for some integer
q. Hence, a = bq + r and r ≥ 0.

Abstract Algebra
Arithmetic in Z
Proof : cont...
2. Suppose r ≥ b. Then r = b + r’ , where 0 ≤ r′
< r. So,
a = bq + r = bq + b + r′
= (q + 1)b + r′
, so that r′
= a − (q + 1)b is an element of S smaller than r. This
contradicts the fact that r is the least element of S. Thus, r < d.
3. Suppose that there are integers q1 and r1 such that a = bq1 + r1 and
0 ≤ r1 < b. We need to show that, q1 = q and r1 = r.
Since a = bq + r and a = bq1 + r1, we have
bq + r = bq1 + r1
so,
b(q − q1) = r1 − r (1)
Note that,
0 ≤ r < b (2)
0 ≤ r1 < b (3)

Abstract Algebra
Arithmetic in Z
Multiplying −1 to inequality (2), we obtain
−b ≤ −r < 0 (4)
0 ≤ r1 < b (2)
Adding these two inequalities
−b < r1 − r ≤ b
−b < b(q − q1) < b from Equation (1)
−1 < q − q1 < 1 Divide each term by b
But q − q1 is an integer (because q and q1 are integers) and the only
integer strictly between -1 and 1 is 0. Therefore q − q1 = 0 and q = q1.
Substituting q − q1 = 0 in Equation (1),
b(q − q1) = r1 − r (1)
b(0) = r1 − r
0 = r1 − r
r1 = r.
Thus the quotient and remainder are unique.

Abstract Algebra
Arithmetic in Z
Definition 2.5 (Divisibility).
Let a and b be integers with b , 0. We say that b divides a (or that b is a
divisor of a, or that b is a factor of a) if a = bc for some integer c. In
symbols,”b divides a” is written b|a and ”b does not divide a” is written
b ∤ a.
Example 2.6.
3|24 because 24 = 3 • 8, but 3 ∤ 17. Negative divisors are allowed: −6|54
because 54 = (−6)(−9), but −6 ∤ −13.

Abstract Algebra
Arithmetic in Z
Remark 2.7.
1. If b divides a, then a = bc for some c. Hence −a = b(−c), so that
b|(−a). An analogous argument shows that every divisor of −a is also
a divisor of a. Therefore, a and −a have the same divisors.
2. Suppose a , 0 and b|a. Then a = bc, so that |a| = |b||c|.
Consequently, 0 ≤ |b| ≤ |a|. This last inequality is equivalent to
−|a| ≤ |b| ≤ |a| . Therefore
i. every divisor of the nonzero integer a is less than or equal to |a|;
ii. nonzero integer has only finitely many divisors.

Abstract Algebra
Arithmetic in Z
Definition 2.8.
Let a and b be integers, not both 0. The greatest common divisor
(gcd) of a and b is the largest integer d that divides both a and b. In other
words, d is the gcd of a and b provided that
1. d|a and d|b;
2. if c|a and c|b, then c ≤ d.
The greatest common divisor of a and b is usually denoted GCD(a, b).
Definition 2.9.
If p is an integer greater than 1, then p is a prime number if the only
divisors of p are 1 and p. A positive integer greater than 1 that is not a
prime number is called composite.
Example 2.10.
The greatest common divisor of 12 and 30 is 6, that is, GCD(12, 30) = 6.
The only common divisors of 10 and 21 are 1 and −1. Hence
GCD(10, 21) = 1. Two integers whose greatest common divisor is 1, such
as 10 and 21, are said to be relatively prime.

Abstract Algebra
Arithmetic in Z
Example 2.11.
Here are some examples to illustrate the definitions above.
1. GCD(45, 60) = 15, since 45 = 15 • 3 and 60 = 15 • 4 and 15 is the
largest number that divides both 45 and 60.
2. 45 and 60 are not relatively prime.
3. 45 and 16 are relatively prime since GCD(45, 16) = 1.

Abstract Algebra
Arithmetic in Z
Theorem 2.12.
Let a and b be integers, not both 0, and let d be their greatest common
divisor. Then there exist (not necessarily unique) integers u and v such
that d = au + bv.
Proof :
Let S be the set of all linear combinations of a and b, that is
S = {am + bn|m, n ∈ Z}.
Step 1. Find the smallest positive element of S.
Note that a2
+ b2
= aa + bb is in S and a2
+ b2
≥ 0. Since a and b are not
both 0, a2
+ b2
must be positive. By WOP, S contains positive integer
and hence, contains a least element. Let t be the smallest positive integer
in S. Hence, we can write t = au + bv for some integers u and v.

Abstract Algebra
Arithmetic in Z
Proof : cont... Step 2. Prove that t is the gcd of a and b, that is, t = d
We must prove that t satisfies the two conditions in the definition of the
gcd:
1. t|a and t|b;
2. if c|a and c|b, then c ≤ t.
Proof of (1): By the Division Algorithm, there are integers q and r such
that
a = tq + r , with 0 ≤ r t.
Consequently,
r = a − tq, (1)
r = a − (au + bv)q = a − aqu − bvq, (2)
r = a(1 − qu) + b(−vq) (3)
Thus r is a linear combination of a and b, and hence r ∈ S.

Abstract Algebra
Arithmetic in Z
Since r t (the smallest positive element of S), we know that r is not
positive. Since r ≥ 0, the only possibility is that r = 0. Therefore,
a = tq + r = tq + 0 = tq so that t|a.
Similarly, t|b. Hence, t is a common divisor of a and b.
Proof of (2): Let c be any other common divisor of a and b, so that c|a
and c|b. Then a = ck and b = cs for some integers k and s. Consequently,
t = au + bv = (ck)u + (cs)v (4)
= c(ku + sv). (5)
This implies that c|t. Hence, c ≤ |t| by Remark 2.7 . But t is positive, so
|t| = t. Thus, c ≤ t. This implies that t is the greatest common divisor d.


Abstract Algebra
Arithmetic in Z
Corollary 2.13.
Let a and b be integers, not both 0, and let d be a positive integer. Then d
is the greatest common divisor of a and b if and only if d satisfies these
conditions:
i. d|a and d|b;
i. if c|a and c|b, then c|d.

Abstract Algebra
Arithmetic in Z
Theorem 2.14.
If a|bc and (a, b) = 1, then a|c.
Proof :
Since (a, b) = 1, Theorem 2.12 shows that au + bv = 1 for some integers u
and v. Multiplying this equation by c shows that acu + bcv = c. But a|bc,
so that bc = ar for some r. Therefore,
c = acu + bcv = acu + (ar)v = a(cu + ro).
This shows that a|c.

Abstract Algebra
Arithmetic in Z
Congruence and Congruence Classes
Congruence and Congruence
Classes

Abstract Algebra
Arithmetic in Z
Congruence and Congruence Classes
Definition 2.15.
Let a,b, n be integers with n 0. Then a is congruent to b modulo n
[written ”a = b(modn)”], provided that n divides a − b.
Example 2.16.
1. 17 = 5(mod6) because 6 divides 17 − 5 = 12.
2. 4 = 25(mod7) because 7 divides 4 − 25 = −21.

Abstract Algebra
Arithmetic in Z
Relation and Operations

Abstract Algebra
Arithmetic in Z
Definition 2.17.
A relation R on a set S (more precisely, a binary relation on S, since it
will be a relation between pairs of elements of S) is a subset of S x S.
Example 2.18.
Let S = {2, 3, 5, 6} and let R mean ”divides.” Since
2R2, 2R6, 3R3, 3R6, 5R5, 6R6, we have
R = {(2, 2), (6, 2), (3, 3), (6, 3), (5, 5), (6, 6)}.
PROPERTIES OF BINARY RELATIONS
Let R be a relation on a set S and a, b, c ∈ S. Then
1. R is called reflexive if aRa for every a ∈ S.
2. R is called symmetric if whenever aRb then bRa.
2. R is called transitive if whenever aRb and bRc then aRc.

Abstract Algebra
Arithmetic in Z
Example 2.19.
1. Let R be the set of real numbers and R mean “is less than or equal
to.” Thus, any number is less than or equal to itself so R is reflexive.
2. Let R be the set of real numbers and R mean “is less than or equal
to.” Now 3 is less than or equal to 5 but 5 is not less than or equal to
3. Hence R is not symmetric.
• Let R be the set of real numbers and R mean “is less than or equal
to.” If x ≤ y and y ≤ z, then x ≤ z. Hence, R is transitive.
Definition 2.20.
A relation R on a set S is called an equivalence relation on S when R is
reflexive, symmetric and transitive.

Abstract Algebra
Arithmetic in Z
EQUIVALENCE SETS

Abstract Algebra
Arithmetic in Z
Definition 2.21 (Equivalence set or Equivalence class).
Let S be a set and R be an equivalence relation on S. If a ∈ S, the
elements y ∈ S satisfying yRa constitute a subset, [a], of S, called an
equivalence set or equivalence class.
Thus, formally,
[a] = {y : y ∈ S, yRa}

Abstract Algebra
Arithmetic in Z
Theorem 2.22.

Abstract Algebra
Arithmetic in Z
Definition 2.23 (Partition).
A set {A, B, C, ...} of non empty subsets of a set S will be called a
partition of S provided
1. A ∪ B ∪ C∪, ... = S
2. The intersection of every pair of distinct subsets is the empty set.
Theorem 2.24.

Abstract Algebra
Arithmetic in Z
Example 2.25.
Let a relation ∼ be defined on the set R of real numbers by xRy if and
only if |x| = |y|. Show that ∼ is an equivalence relation.
Solution:
1. Let a ∈ R. Then |a| = |a|. So a ∼ a and R is reflexive.
2. Let a, b ∈ R and suppose that a ∼ b. Then |a| = |b|. So |b| = |a|.
Thus, b ∼ a and R is symmetric.
3. Let a, b, c ∈ R and suppose that a ∼ b and b ∼ c. Then |a| = |b| and
|b| = |c|. So |a| = |c|. Thus, a ∼ c and R is transitive.
Therefore, R is an equivalence relation on R.

Abstract Algebra
Arithmetic in Z
Example 2.26.
Consider the set S = R where x ∼ y if and only if x2
= y2
. Prove that ∼
is an equivalence relation on S.
Proof :
Let x, y, z ∈ S.
i. Since x2
= x2
∀ x ∈ S. So, x ∼ x and ∼ is reflexive.
ii. Suppose x ∼ y. Then x2
= y2
.
Note that y2
= x2
. So, y ∼ x and ∼ is symmetric.
iii. Suppose x ∼ y and y ∼ z. Then x2
= y2
and y2
= z2
.
x2
= y2
⇒ = z2
Thus, x ∼ z and ∼ is an equivalence relation on S.

Abstract Algebra
Problem Set I
Submit your output @google classroom and handwritten @ DSSC
guardhouse addressed to Dr. Gemma P. Salasalan .
1. Prove or disprove. Let a, b, c, d ∈ Z.
a. If a|b and a|c, then a|(b + c).
b. If ab|cd, then a|c and a|d.
2. Given a set S and a relation ∼ on S. Check if ∼ is an equivalence
relation on S. If ∼ is not an equivalence relation on S, provide
counterexample.
a. S = R where a ∼ b iff a ≤ b.
b. S = Z where a ∼ b iff a|b.
c. S is a set of subsets of N where A ∼ B iff A ⊆ B.
3. Prove for sets A and B that A ⊆ B if and only if A ∪ B = B.
4. Prove that (A ∩ B)′
= A′
∪ B′
.

Abstract Algebra
Lesson 2
Operations
Definition 4.1.
A binary operation “ * ” on a non-empty set S is a mapping which
associates with each ordered pair (a, b) of elements of S a uniquely defined
element a ∗ b of S.
A binary operation on a set S is a mapping of S × S into S.
∗ : S × S → S such that a ∗ b ∈ S, ∀ a, b ∈ S
Example 4.2.
Let S = 0, 1, 2, 3, 4. Neither addition or multiplication are binary operation
on set S. Since 2 + 3 = 5 S and 2 · 3 = 6 S.

Abstract Algebra
Lesson 2
Operations
Properties of Binary Operation
i. Closure
A binary operation ∗ on a non-empty set S has closure property, if
a ∈ S, b ∈ S ⇒ a ∗ b ∈ S.
ii. Associative
The associative property of binary operations holds if, for a
non-empty set S, we can write (a ∗ b) ∗ c = a ∗ (b ∗ c), where a, b, c ∈ S.
iii. Commutative
A binary operation * on a non-empty set S is commutative, if
a ∗ b = b ∗ a, for all (a, b) ∈ S.
iv. Identity
A set S is said to have an identity element with respect to a binary
operation * on S if there exists an element e ∈ S with the property
e ∗ x = x ∗ e = x for every x ∈ S:

Abstract Algebra
Lesson 2
Operations
iv. Inverse
Consider a set S having the identity element e with respect to a
binary operation *. An element y ∈ S is called an inverse of x ∈ S
provided that, x ∗ y = y ∗ x = e. We can write, y = a−1
.
Definition 4.3.

Abstract Algebra
Lesson 2
Operations
Example 4.4.

Abstract Algebra Beamer Lesson1

Recommended

Recommended

More Related Content

Similar to Abstract Algebra Beamer Lesson1

Similar to Abstract Algebra Beamer Lesson1 (20)

More from Monica Waters

More from Monica Waters (20)

Recently uploaded

Recently uploaded (20)

Abstract Algebra Beamer Lesson1