# 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) = x^{2} and g(x) = x^{2} + x. Is (f(x), g(x)) ∈ T? That
is, do there

exist real numbers a, b with a ≠ 0 such that (ax + b)^{2} = x^{2} + 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 a^{2} =?,
2ab =?, and b^{2} = 0. Why is this

impossible?

**Exercise 1.2 **Now let’s start with a different pair of quadratics, f(x) = x^{2} and
g(x) =

4x^{2} + 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) = Ax^{2} + Bx + C and g(x) = Dx^{2} + 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 = Aa^{2}and 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) = 2x^{2} + 3x + 4 is not
equivalent to

g(x) = -5x^{2} + 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) =

x^{2}, g(x) = x^{2}+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) = Ax^{2} + Bx + C and

g(x) = Dx^{2} + 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) = -x^{2} + 6x + 1. Is g(x) equivalent to a quadratic of the
form

-x^{2} + 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 =
-x^{2} + 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 ±x^{2} + 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) ~ x^{2} + D or f(x) ~ -x^{2} + D.

**Exercise 1.11** Prove Theorem 1.10. Hints: Let f(x) = Ax^{2} + 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?