logo Your Math Help is on the Way!

More Math Help

Home
Algebraic Symmetries
Radical Expressions and Equation
The Exponential Function
Math 1010-3 Exam #3 Review Guide
MATH 511 ASSIGNMENT SHEET
Rational Numbers Worksheet
Are You Ready for Math 65?
Solving Simultaneous Equations Using the TI-89
Number Theory: Fermat's Last Theorem
algorithms-in-everyday-mathematics
COLLEGE ALGEBRA
Course Syllabus for Intermediate Algebra
Solving Inequalities with Logarithms and Exponents
Introduction to Algebra Concepts and Skills
Other Miscellaneous Problems
Syllabus for Calculus
SYLLABUS FOR COLLEGE ALGEBRA
Elementary Linear Algebra
Adding and Subtracting Fractions without a Common Denominator
Pre-Algebra and Algebra Instruction and Assessments
Mathstar Research Lesson Plan
Least Common Multiple
Division of Polynomials
Counting Factors,Greatest Common Factor,and Least Common Multiple
Fractions
Real Numbers, Exponents and Radicals
Math 115 Final Exam Review
Root Finding and Nonlinear Sets of Equations
Math 201-1 Final Review Sheet
Powers of Ten and Calculations
Solving Radical Equations
INTERMEDIATE ALGEBRA WITH APPLICATIONS COURSE SYLLABUS
EASY PUTNAM PROBLEMS
INTRODUCTION TO MATLAB
Factoring Polynomials
Section 8
Declining Price, Profits and Graphing
Arithmetic and Algebraic Structures
Locally Adjusted Robust Regression
Topics in Mathematics
INTERMEDIATE ALGEBRA
Syllabus for Mathematics
The Quest To Learn The Universal Arithmetic
Solving Linear Equations in One Variable
Examples of direct proof and disproof
ELEMENTARY ALGEBRA
NUMBER THEORY
Algebra I
Quadratic Functions and Concavity
Algebra
More on Equivalence Relations
Solve Quadratic Equations by the Quadratic Formula
Solving Equations and Inequaliti
MATH 120 Exam 3 Information
Rational Number Ideas and Symbols
Math Review Sheet for Exam 3
Polynomials
Linear Algebra Notes
Factoring Trinomials
Math 097 Test 2
Intermediate Algebra Syllabus
How to Graphically Interpret the Complex Roots of a Quadratic Equation
The General, Linear Equation
Written Dialog for Problem Solving
Radian,Arc Length,and Area of a Sector
Internet Intermediate Algebra
End Behavior for linear and Quadratic Functions
Division of Mathematics
161 Practice Exam 2
Pre-calculus
General linear equations
Algebraic Symmetries
Math 20A Final Review Outline
Description of Mathematics
Math 150 Lecture Notes for Chapter 2 Equations and Inequalities
Course Syllabus for Prealgebra
Basic Operations with Decimals: Division
Mathematics Content Expectations
Academic Systems Algebra Scope and Sequence
Syllabus for Introduction to Algebra
Syllabus for Elementary Algebra
Environmental Algebra
Polynomials
More Math Practice Problems
INTERMEDIATE ALGEBRA COURSE SYLLABUS
Intermediate Algebra
Syllabus for Linear Algebra and Differential Equations
Intermediate Algebra
Rational Expressions and Their Simplification
Course Syllabus for Intermediate Algebra
GRE Review - Algebra
Foundations of Analysis
Finding Real Zeros of Polynomial Functions
Model Academic Standards for Mathematics
Visual-Fraction-Addition-Teaching-Method
Study Guide for Math 101 Chapter 3
Real Numbers
Math 9, Fall 2009, Calendar
Final Review Solutions
Exponential and Logarithmic Functions





Try the Free Math Solver or Scroll down to Tutorials!

 

 

 

 

 

 

 

 
 
 
 
 
 
 
 
 

 

 

 
 
 
 
 
 
 
 
 

Please use this form if you would like
to have this math solver on your website,
free of charge.


More on Equivalence Relations

1.1 quadratic polynomials

Let Q be the set of all real quadratic polynomials. That is .

Define a relation T on Q as follows:

with a ≠ 0 and f(ax + b) = g(x)} .

Exercise 1.1 Let f(x) = x2 and g(x) = x2 + x. Is (f(x), g(x)) ∈ T? That is, do there
exist real numbers a, b with a ≠ 0 such that (ax + b)2 = x2 + x? Show that the answer is
no.
Hint: If so (this is a proof by contradiction), then there exist real numbers a, b with a ≠ 0
such that

which implies (by comparing the coefficients) that a2 =?, 2ab =?, and b2 = 0. Why is this
impossible?

Exercise 1.2 Now let’s start with a different pair of quadratics, f(x) = x2 and g(x) =
4x2 + 4x + 1. Show that (f(x), g(x))∈ T: Again we seek real numbers a, b such that

Find a solution.

Exercise 1.3 Prove that T is an equivalence relation.

Hint: Here is a proof that the relation ~ is symmetric. (You should show that T is
reflexive and transitive.) Suppose that (f(x), g(x)) ∈ T, that is,

f(ax + b) = g(x), (1)

for some real numbers a ≠ 0 and b. To prove that (g(x), f(x)) ∈ T we need to show that

g(cx + d) = f(x), (2)

for some real numbers c ≠ 0 and d.

Actually we can compute and express c and d in terms of a and b. Here’s how it works.
Let c, d be any real numbers and substitute cx + d for x in Equation (1):



By choosing c = 1/a and d = -b/a, we have acx + ad + b = x so that g(cx + d) = f(x).
Thus c = 1/a and d = -b/a are the real numbers that make Equation (2) true.

Since this is an equivalence relation we will drop the use of T and the ordered pairs and
simply write f(x) ~ g(x) to mean (f(x), g(x)) ∈ T.

There are two natural questions that arise in this context. Given two quadratics, how can
we tell if they are equivalent? Is there a set of “simple” quadratics such that every other
quadratic is congruent to one of these?

We try to tackle the first question first. A fairly simple observation will help here. Suppose
f(x) = Ax2 + Bx + C and g(x) = Dx2 + Ex + F are quadratics and that f(x) ~ g(x).
Then the leading coefficients A,D have the same sign. That is, A and D are either both
positive or both negative. To see this, suppose that a, b are real numbers with a ≠ 0 and
f(ax + b) = g(x). Then compute

Thus D = Aa2and so A and D have the same sign. To state this observation in a more
symbolic way, let sgn(f(x)) denote the sign of the leading coefficient of f(x). That is,
sgn(f(x)) = +1 if the leading coefficient is positive and sgn(f(x)) = -1 if the leading
coefficient is negative. The observation proved above can be stated in the following way:

Theorem 1.4 Let f(x), g(x) be quadratic polynomials. If f(x) ~ g(x) then sgn(f(x)) =
sgn(g(x)).

The function sgn is called an invariant of the equivalence relation. (We will give a more
general definition later.) This invariant is useful because it allows us to prove that certain
quadratics are not equivalent. For example, f(x) = 2x2 + 3x + 4 is not equivalent to
g(x) = -5x2 + 7x - 9 because sgn(f(x)) = +1 and sgn(g(x)) = -1. Thus using the
contrapositive form of Theorem 1.4, we know that

Unfortunately, the converse of Theorem 1.4 does not hold. The quadratics f(x) =
x2, g(x) = x2+x discussed above provide a counterexample. sgn(f(x)) = sgn(g(x)) = +1,
but f(x) is not equivalent to g(x). However there is another invariant which when combined
with the sign invariant will provide necessary and sufficient conditions for equivalent.
More on this later.

Exercise 1.5 Find a counterexample to this statement: If f(x) = Ax2 + Bx + C and
g(x) = Dx2 + Ex + F and f(x) ~ g(x) then C and F have the same sign.

1.2 invariants and reduced forms

Let’s pause now to explain what an invariant is and what a reduced form is for a general
equivalence relation.

Definition 1.6 Let ~ be an equivalence relation on a set A. A function
from A onto a set S is an invariant for the equivalence relation if x~ y implies that
I(x) = I(y) for all x, y ∈ A.

Now suppose that I is an invariant. Sometimes we can use I to determine that two
elements x, y ∈ A are not equivalent. That is, if I(x) ≠ I(y) then On the other
hand, it may happen that I(x) = I(y) but for some x, y ∈ A.

Definition 1.7 Let ~ be an equivalence relation on a set A and let   be invariants
for the equivalence relation. The set of invariants is sufficient
if the following property holds:

  for all k = 1, . . . , r implies x ~y.

A sufficient set of invariants gives us a way to determine if two given elements x, y ∈ A
are equivalent. If for any one of the invariants, then But if
  for all k = 1, . . . , r then x ~ y. Unfortunately we do not have a sufficient
set of invariants our equivalence relation. The function sgn(f(x)) is an invariant for the
equivalence relation on the set of quadratics Q But sgn alone is not a sufficient set of
invariants. Later we will add one other invariant, I, which will make {sgn, I} a sufficient
set of invariants.

Now we turn to the idea of a reduced form for an equivalence relation.

Definition 1.8 Let ~be an equivalence relation on a set A. A subset is a reduced
subset if every x ∈ A is equivalent to exactly one element of R.

In addition, and this is where the word “reduced” comes in, we would like the elements
in the reduced set R is be simple in some sense.

1.3 A reduced form for quadratic polynomials

Let’s start with an example.

Exercise 1.9 Let g(x) = -x2 + 6x + 1. Is g(x) equivalent to a quadratic of the form
-x2 + D, where D is a real number? (We are hoping that this will be our reduced form.)
To restate the question in terms of the definition of the equivalence relation, we ask: do
there exist real numbers a ≠ 0, b and D such that -(ax + b)2 + D = -x2 + 6x + 1?
Hint: By expanding (ax+b)2, the question becomes this: do there exist real numbers a ≠ 0,
b, and D such that

Solve for a, b, and D by comparing coefficients and fill
Here’s what we are getting at: The quadratics of the form ±x2 + D are fairly simple.
Could it be that every quadratic is equivalent to one of this form? The answer is yes.

Theorem 1.10 Let f(x) be a quadratic polynomial in Q. Then there exists a real number
D such that either f(x) ~ x2 + D or f(x) ~ -x2 + D.

Exercise 1.11 Prove Theorem 1.10. Hints: Let f(x) = Ax2 + Bx + C. Divide the proof
into two cases, A > 0,A < 0. Then complete the square.

Exercise 1.12 Write an explicit expression for the D in terms of A, B, C. The graph of
the quadratic f(x) is a parabola. What is the geometric meaning of D. Is D an invariant?