1. Chapter 1- Preliminaries
1.1 History of Algebra
The word “algebra”-al jebr (in Arabic)
• was first used by Mohammed Al-Khwarizmi -muslim Math.
• ninth century, when taught mathematics in Baghdad.
• means “reunion”,decribes his method for collecting the terms
of an equations in order to solve it.
• Omar Khayyam, another Mathematician, defined it as
the science of solving equations.
Elementary Algebra (Classical age of Algebra)
• its central theme is clearly identified as the solving of eqs.
- method of solving linear, qudratic, cubic, quartic equations.
- 1824, Niels Abel – there does not exits any formula for
equations degree 5 or greater.
2. Modern Age
• new varieties of algebra arose
connection with the application in math to practical problems.
- Matrix Algebra
- Bolean Algebra
- Algebra of vectors and tensors
- ~200 different kinds of algebra.
• the awareness grew
- algebra can no longer be conceived merely as the
science of solving equations.
- It had to be viewed as much more broadly as a
branch of mathematics.
revealing general principles which apply equally to
all known and all possible algebras:
* What is it that all algebras have in common?
* What trait do they share which lets us refer to all of
them as algebras? Algebraic Structure
• Abstract Algebra (Modern Algebra) -more adv. course
The study algebraic structures.
3. 1.2 Logic and Proof
- Understand these terms and feel comfortable using them
to define new terms.
• Statement or Proposition
- Declarative sentence that is either true or false, but not both.
- Statements that are assumed to be true.
- A precise meaning to a mathematical term.
4. • Theorem
- A major landmark in the mathematical theory.
- Postulates and definitions are used to prove theorems.
- Once a theorem is proved to be true, it can be used.
- A result that is needed to prove a theorem.
- A result that follows immediately from a theorem.
- Is not a general result but is a particular case.
- Mathematical argument intended to convince us that
a result is correct.
5. Conjunction, Disjunction and Negation
Let P and Q be statements.
i) The statement P AND Q, P ∧ Q,
is called the conjunction of P and Q.
ii) The statement P OR Q, P ∨ Q,
is called the disjunction of P and Q.
iii) The negation of P is denoted by NOT P or ~ P
Conditional and Biconditional Statement
Conditional statement: “If P then Q”, P ⇒ Q.
Biconditional statement: “P if and only if Q”, P ⇔ Q.
Consider a statement P(x) :
- Statement P(x) is depending on the variable x.
- Adding quantifiers can convert statement P(x) into a
statement that is either true or false.
• Universal quantifier
P(x) is true for all values of x, denoted by
∀x, P ( x)
For all x, P(x).
For every x, P(x).
For each x, P(x).
P(x), for all x.
7. • Existential quantifier (∃)
There exist an x for which P(x) is true
For some x, P(x).
P(x), for some x.
: ∃x, P ( x)
∀x ∈ R, x −1 = ( x −1)( x + x + 1)
- True or false statement? Why?
∀x ∈ R, x + x − 6 = 0
- True or False statement? Why?
∃x ∈ R, x + x − 6 = 0
- True or False statement? Why?
- Many mathematical theorems can be expressed
symbolically in the form of
may consists of one or more
- The theorem says that if the assumption is true than the
conclusion is true.
- How do you go about thinking up ways to prove a
• Understand the definitions
• Try examples
• Try standard proof methods
9. Methods of Proof ( P ⇒ Q)
1. Direct Method
• find a series of statements P1,P2,…,Pn
• verify that each of the implications below is true
P →P , P →P2 , P2 →P3 .....Pn −1 →Pn and Pn → Q
An integer n is defined to be even if n = 2m for some integer m.
Show that the sum of two even integers is even.
10. 2. Contrapositive Method
• may prove ¬Q → ¬P
If x is a real number such that
x + 7 x < 9, then x < 1.1
11. 3. Proof by Contradiction
• assume that P is true and not Q is true (Q is false)
• will end up with a false statement S
• Conclude that not Q must be false, i.e., Q is true
If x is an integer and x2 is even then x is an even integer.
12. 4. Proof by Induction
• assume that for each positive integer n,
a statement P(n) is given. If
1. P(1) is a true statement; and
2. Whenever P(k) is a true statement, then P(k+1) is also true,
• then P(n) is a true statement for every n in positive integer.
+ ... +
1• 3 3 • 5 5 • 7
(2n − 1)(2n + 1) 2n + 1
13. 5. Proof by Counterexamples
• Sometimes a conjectured result in mathematics is not true.
• Would not be able to prove it.
• Could try to disprove it.
• The conjecture in the form of ∀x, P ( x )
• Take the negation: NOT (∀x, P ( x ))
∃x, NOT P ( x)
• Hence to disprove the statement
∀x, P ( x)
need only to find one value, say c, such that P(c) is false.
• The value c is called a counterexample to the conjecture.
Let x be a real number. Disprove the statement
If x2 >9 then x >3.
• To disprove the conjecture in the form of
∃x, P ( x)
cannot use counter example!!!
Its negation is equivalently in the form of ∀x, NOT
Need to show that P(x) is false for all values of x.
• To prove
P ( x)
P ⇔ Q : Prove P ⇒Q and Q ⇒P