Notes September132010

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					Class Notes for September 13, 2010

Do Now: The mapping from the set A to the set B is a function, if for every           there is
exactly one    such that the mapping takes a to b.

Review of Homework #1:

1)(a) An algebraic number over the field of rational numbers is the set of complex
numbers that are the roots to non-zero polynomial equations that have rational
coefficients. *Algebraic numbers over a field F have coefficients in that field.
**Algebraic numbers form a field.

(b) Is 5-3i an algebraic number? (5-3i)2 - 10(5-3i) + 34= 25 -15i –15i -9 -50 +30i +34= 0
Therefore 5-3i is a root of polynomial equation x2-10x+34=0.

2) (c) In the additive group xm-xn=x(m-n) is analogous to xm/xn=xm-n in the multiplicative
group since division is analogous to subtraction and in the additive relationship x is
being added (m-n) times and in the multiplicative relationship x is being multiplied (m-n)
times.

(d) In the multiplicative group    = xy ∙   is analogous to (x+y) – (w-z) = (x+y) + (z-w) in
the additive group since + and – are inverse operations in the additive group and / and ∙
are inverse operations in the multiplicative group. Also in the additive group (w-z) and
(z-w) are additive inverses and in the multiplicative group and are multiplicative
inverses.

(e) In the multiplicative group (bc)a =baca is analogous to a(b+c)=ab+ac in the additive
group since + is analogous to ∙ and in the additive relationship (b+c) is being added a
times and in the multiplicative relationship (bc) is being multiplied a times.
*Similarly ab+c = abac is also analogous to the additive relationship since a is being
multiplied (b+c) times.

(f) In the additive group 0∙b =0 is analogous to b0=1 in the multiplicative group since in
the additive relationship b is being added 0 times to get the additive identity and in the
multiplicative group b is being multiplied 0 times to get the multiplicative identity.
Key Points When Writing Mathematics:

Expression: vs. Equation:
 5x-9           5x-9=3

Roots vs. Factors
x2-1
(x+1)(x-1) [factors]
x=1 and x=-1 [roots]
 ═ >: “implies” or “If …then…” NOT “then”
**DO NOT write ‘if A ═ > B’ for ‘if A then B’

Equivalence Relations and Equivalence Classes

An equivalence relation, ~, on a set, S, is a relation that satisfies:
      1) Reflexivity: x~x,
      2) Symmetry: x~y ═ > y~x
      3) Transitivity: x~y and y~z ═ > x~z

An equivalence class is a partition on a set, S, such that       are in the same
equivalence class if x~y. Any two equivalence classes are either equal or disjoint.

Example 1:
 α~β iff sinα=sinβ and cosα=cosβ is an equivalence class on the set of all angles since:
        1) Reflexivity: sinα=sinα and cosα=cosα ═ > α~α
        2) Symmetry: α~β ═ > sinα=sinβ and cosα=cosβ. By symmetry of =, sinβ=sinα
            and cosβ=cosα ═ > β~α
        3) Transitivity: α~β and β~θ ═ > sinα=sinβ, cosα=cosβ, sinβ=sinθ and cosβ=cosθ
            ═ >by transitivity of =, sinα=sinθ and cosα=cosθ ═ > α~θ
There are infinitely many equivalence classes for this equivalence relation. They are
represented by 0 ≤ α < 360.

Example 2:
On the set of integers, a≡b mod 3 if a-b=3∙q,        is an equivalence relation since:
       1) Reflexivity: a-a=0=3∙0. Since        , a≡ a mod 3.
       2) Symmetry: a≡b mod 3 ═ > a-b=3∙q. –(a-b)=-(3∙q) ═> b-a=3∙(-q). Since
             , b≡a mod 3.
       3) Transitivity: a≡b mod 3 and b≡c mod 3 ═ > a-b=3∙q and b-c=3∙z;              .
            c=b-3∙z ═ > a-c=a-(b-3∙z)=a-b+3∙z=3∙q+3∙z=3(q+z). But                so a≡c
            mod 3.
The set of equivalence classes mod 3 are:
{…,-3n,…,-6, -3, 0, 3, 6,…,3n,…}
{…,-3n+1,…,-5, -2, 1, 4, 7,…,3n+1,…}
{…,-3n+2,…,-4, -1, 2, 5, 8,…,3n+2,…}
or we write Z3={0,1,2} for the set of equivalence classes.

Example 3:
On the set of integers, a~b if a-b is even is an equivalence relation proved in
Homework#1.
The equivalence classes for this relation is the set of even integers and the set of odd
integers or Z2={0,1}.


Extra Example:
On the set of all functions f: R->R, f~g if f is a constant non-zero multiple of g, f(x)= k∙g(x)
is an equivalence relation since:
        1) Reflexivity: f(x)=1∙f(x) ═ > f~f
        2) Symmetry: f~g ═> f(x)= k∙g(x) ═> g(x) = ∙ f(x) ═ > g~f
        3) Transitivity: f~g and g~t ═ > f(x)= k∙g(x) and g(x)= c∙t(x) ═ > f(x)= k∙(c∙t(x)) ═ >
           f(x) = (k∙c)∙t(x) ═ > f~t

How many equivalence classes are there for this relationship?



Reference Angles

A reference angle of an angle θ is an angle α with 0 ≤ α ≤ 90 such that    =
and |cosθ| = |cosα|. The reference angle is always the smallest angle between the
terminal side of an angle and the x-axis.

A reference angle is a representative of an equivalence class formed by the equivalence
relation on the set of all angles:
α~θ if|sinθ| = |sinα| and |cosθ| = |cosα| on the set of all angles
This relationship creates infinitely many equivalence classes represented by 0 ≤ α < 90.
 or example a 30 reference angle is a representa ve for its class {30 +k , 150 +k :
      }.

				
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