# Math 3121 Abstract Algebra I

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Math 3121 Abstract Algebra I
Lecture 2 Sections 0-1: Sets and Complex Numbers

Questions on HW (not to be handed in)
HW: pages 8-10: 12, 16, 19, 25, 29, 30

Finish Section 0: Sets and Relations
Correction to slide on functions Equivalence Relations and Partitions

Corrected Slide: Functions
Definition: A function f mapping a set X into a set Y is a relation between X and Y with the properties: 1) For each x in X, there is a y in Y such that (x, y) is in f 2) (x, y1) ∊ f and (x, y2) ∊ f implies that y1 = y2. When f is a function from X to Y, we write f: X Y, and we write “(x, y) in f” as “f(x) = y”.

Functions (Corrected Version)
Definition: A function f mapping a set X into a set Y is a relation between X and Y with the property that each x in X appears exactly once as the first element of an ordered pair (x, y) in f. In that case we write f: X Y. This means that 1) For each x in X, there is a y in Y such that (x, y) is in f 2) (x, y1) ∊ f and (x, y2) ∊ f implies that y1 = y2. When f is a function, we write “(x, y) in f” as “f(x) = y”.

Recall: Equivalence Relation
Definition: An equivalence relation R on a set S is a relation on S that satisfies the following properties for all x, y, z in S. Reflexive: x R x Symmetric: If x R y, then y R x. Transitive: If x R y and y R z, then x R z.

Equivalence Classes Definition: Suppose ~ is an equivalence relation on a nonempty set S. For each a in S, let a̅ = {x∊S | x~a}. This is called the equivalence class of a ∊ S with respect to ~.

1) By definition of a̅ = {x ∊ S | x ~ a}: x ∊ a̅ ⇔ x ~ a. 2) By symmetry, a ~ a. Thus: a ∊ a̅

Functions and Equivalence Relations
Theorem: Suppose f: X  Y is a function from a set X to a set Y. Define a relation ~ by (x ~ y)⇔ (f(x) = f(y)). Then ~ is an equivalence relation on X. Proof: Left to the reader.

Theorem Theorem: Let ~ be an equivalence relation on a set S, and let a̅ denote the equivalence class of a with respect to ~. Then x ~ y ⇔ x̅ = y̅.

Proof Proof: We show each direction of the implication separately
x ~ y ⇒ x̅ = y̅: We will show that x ~ y implies that x̅ and y̅ have the same elements. 1) Start with transitivity of ~: x ~ y and y ~ z ⇒ x ~ z 2) Rewrite 1) as: x ~ y ⇒ (y ~ z ⇒ x ~ z) (Note: P and Q ⇒ R is equivalent to P ⇒ (Q ⇒ R )) 3) By symmetry of ~, replace y ~ z by z ~ y and x ~ z by z ~ x in 2): x ~ y ⇒ (z ~ y ⇒ z ~ x) 4) Reversing x and y in 3) gives: y ~ x ⇒ (z ~ x ⇒ z ~ y). 5) By symmetry of ~, replace y ~ x by x ~ y in 4): x ~ y ⇒ (z ~ x ⇒ z ~ y). 6) Combining 5) and 3): x ~ y ⇒ (z ~ x ⇔ z ~ y). 7) By definition of equivalence class x ~ y ⇒ (z ∊ x̅ ⇔ z ∊ y̅). Thus x ~ y ⇒ x̅ = y̅ x̅ = y̅ ⇒ x ~ y : Suppose x̅ = y̅. Since x ∊ x̅, equality of sets implies that x ∊ y̅. Thus x ~ y. Thus x ~ y ⇔x̅ = y̅. QED

Partitions Definition: A partition of a set S is a set P of nonempty subsets of S such that every element of S is in exactly one of the subsets of P. The subsets (elements of P) are called cells. Note that a subset P of the power set of S is a partition whenever 1) ∀ x ∊ S, x is in some member of P 2) ∀ X, Y ∊ P, (X⋂Y ≠ Ø) ⇒ X=Y. Note that 1) is equivalent to: 1’) The union of all members of P is equal to S: ∪(P) = S and 2) is equivalent to: 2’) The intersection of any two different members of P is empty. ∀ X, Y ∊ P, X ≠ Y ⇒ X ÇY =Ø

Theorem Theorem (Equivalence Relations and Partitions): Let S be a nonempty set and let ~ be an equivalence relation on S. Then ~ corresponds to a partition of S whose members are the equivalence classes a̅ = {x ∊ S | x ~ a}.

Proof Proof: Let P = {a̅ | a ∊ S} We show that P is a partition of S. We must show that P is a collection of nonempty subsets of S such that each element of S is in exactly one member of P. We will show that 1) for each x in S, x ∊ x̅, 2) for any x, y in S, (x̅⋂y̅ ≠ Ø) ⇒ (x̅=y̅). From 1) we conclude that each member of P is nonempty, and that each member of S is in at least one member of P. From 2) we conclude that each element of S is in at most one member of P.

Proof of 1) Proof of 1) For each a ∊ S, a ∊ a̅: Let a ∊ S.
Because ~ is reflexive, a ~ a Thus a ∊ a̅ = {x ∊ S |x~a} .

Proof of 2) Proof of 2) for any x, y in S, (x̅⋂y̅ ≠ Ø)⇒ (x̅=y̅): Assume x̅⋂y̅ ≠ Ø. Then there is an element z in x̅ ⋂ y̅. Then z ~ x and z ~ y. Applying symmetry to the first and then transitivity to the pair, we get x ~ y. By the previous theorem x̅ = y̅. Thus (x̅⋂y̅ ≠ Ø)⇒ (x̅=y̅).

Section 1: Complex Numbers
This section covers complex numbers. It summarizes: Definition: ℂ = { a + b i | a, b ∊ ℝ}, where i2 = -1 Addition and multiplication of complex numbers: (a + b i)+(c + d i) = (a + c) + (b + d) i (a + b i)(c + d i) = a c + a d i + b i c + b i d i = (a c – bd) + (a d + b c) i Note: follows from distributive and commutative laws. Absolute value: |a + b i | = sqrt (a2 + b2) Euler’s formula: e i ϑ = cos ϑ + i sin ϑ. Polar coordinates in the complex plane. r e i ϑ =r cos ϑ + i r sin ϑ. Solving for roots using polar coordinates. The unit circle in the complex plane. Roots of unity: e i 2π/n = cos (2π/n) + i sin (2π/n)

More on the Complex Unit Circle
Let U is the unit circle on the complex plane. U = {z ∊ ℂ | |z| = 1} U is closed under multiplication of complex numbers. The function f(ϑ) = e i ϑ maps the real numbers into the complex circle. It wraps the real line around the circle. Note the addition formula: f(a+b) = f(a) f(b). Expand this in terms of Euler and get the addition formulas for sine and cosine (in class). Note that a ~ b ⇔ f(a) = f(b) defines an equivalence relation on ℝ.

HW – not to hand in Pages 19-20: 1, 3, 5, 13, 17, 23, 38, 41