Partition of the Plane into Parabolas, Assignments of Mathematics

A proof that the family of parabolas of the form ax2 + bx + c, where a is a real number, forms a partition of the plane. The document also introduces an equivalence relation that partitions the plane into parabolas and proves that this relation is reflexive, symmetric, and transitive.

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MAT 300 Spring 06
Dr. Zandieh
Homework #8
Solutions
1. Consider the family of sets consisting of all parabolas of the form where a
is a real number.
axy += 2
a. Prove that this set of parabolas forms a partition of the plane,
2.
Let and let
a
()
{
}
axyyxAa+== 22 |, . Define the family . We
wish to prove that F is a partition of
{}
=a
a
AF
2.
(i) First, note that (0, a) since
2
0 and
a. Also, , so (0,
a) . Thus, aa += 2
0
a
A
φ
aa AFA , . So, F satisfies the first property of a
partition.
(ii) Next, suppose FAA ba
, and
φ
ba AA . Then there exists
by the negation of the definition of the empty set. Thus,
and
ba AAyx ),(
a
Ayx ),(b
Ayx
),( by definition of intersection. By definition of ,
, or . Similarly, or by definition
of . So, by the symmetric and transitive properties of equality,
, so
a
A
axy += 2axy = 2bxy += 2bxy = 2
b
A
bxya == 2ba
. Thus, ba AA
=
, and F satisfies the second property
of a partition.
(iii) Finally, we wish to show that . So, first let . Then
or some
2
=
U
aa
AU
aa
Ayx ),(
a
Ayx ),( f
a by definition of the union of a family of sets.
Thus, by definition of , . Now, let . Consider the
set : note that , so
a
A2
),( yx 2
),( yx
2
xy
A)(22 xyxy += 2
),( xy
Ayx
. Since is closed
under subtraction and multiplication, , so . Thus,
by the definition of the union of a family of sets. So,
if and only if , and we have that by
definition of set equality. Thus, F satisfies the third property of a partition.
2
xy FA xy
2
U
aa
Ayx ),(
U
aa
Ayx ),(2
),( yx 2
=
U
aa
A
Since F satisfies all three properties, F is a partition of
2.
b) Write an equivalence relation that will partition the plane into parabolas of the form
where a is a real number.
axy += 2
We want two points and in
),( yx ),( vu
2 to be equivalent if and only if they are on the
same parabola, i.e. and . In other words, we want and
axy += 2auv += 2axy = 2
pf2

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MAT 300 Spring 06 Dr. Zandieh Homework # Solutions

  1. Consider the family of sets consisting of all parabolas of the form where a is a real number.

y = x^2 + a

a. Prove that this set of parabolas forms a partition of the plane, ℜ 2.

Let a ∈ℜand let Aa = {( x , y )∈ℜ^2 | y = x^2 + a }. Define the family. We

wish to prove that F is a partition of

F = { Aa } a ∈ℜ

(i) First, note that (0, a ) ∈ ℜ^2 since 0 ∈ℜand a ∈ℜ. Also, , so (0, a ). Thus,

a = 02 + a

∈ A a ∀ Aa ∈ F , Aa ≠ φ. So, F satisfies the first property of a

partition.

(ii) Next, suppose A a , Ab ∈ F and Aa ∩ Ab ≠ φ. Then there exists

by the negation of the definition of the empty set. Thus, and

( x , y )∈ AaA b ( x , y )∈ A a ( x , y )∈ Ab by definition of intersection. By definition of , , or. Similarly, or by definition of. So, by the symmetric and transitive properties of equality, , so

A a y = x^2 + a yx^2 = a y = x^2 + b yx^2 = b A b a = yx^2 = b a = b. Thus, Aa = Ab , and F satisfies the second property of a partition. (iii) Finally, we wish to show that. So, first let. Then or some

= ℜ^2

∈ℜ

U

a

Aa U

∈ℜ

a

( x , y ) Aa ( x , y )∈ A a f a ∈ℜby definition of the union of a family of sets. Thus, by definition of ,. Now, let. Consider the set : note that , so

A a ( x , y )∈ℜ^2 ( x , y )∈ℜ^2 A (^) yx 2 y = x^2 +( yx^2 ) ( x , y ) ∈ Ayx 2. Since is closed under subtraction and multiplication, , so. Thus, by the definition of the union of a family of sets. So, if and only if , and we have that by definition of set equality. Thus, F satisfies the third property of a partition.

yx^2 ∈ ℜ A (^) yx 2 ∈ F

U

∈ℜ

a

( x , y ) Aa

U

∈ℜ

a

( x , y ) Aa ( x , y )∈ℜ^2 = ℜ^2 ∈ℜ

U

a

Aa

Since F satisfies all three properties, F is a partition of ℜ 2.

b) Write an equivalence relation that will partition the plane into parabolas of the form y = x^2 + a where a is a real number.

We want two points (^) ( x , y )and (^) ( u , v )inℜ 2 to be equivalent if and only if they are on the same parabola, i.e. y = x^2 + a and v = u^2 + a. In other words, we want yx^2 = a and

vu^2 = a , or. So, define an equivalence relation if and only if.

yx^2 = vu^2 ( x , y )≡( u , v ) yx^2 = vu^2

c) Prove that your answer in part b. is an equivalence relation. We need to show that ≡ is reflexive, symmetric, and transitive. (i) Note first that, by the reflexive property of equality, for any , so

yx^2 = yx^2 ( x , y )∈ℜ^2 ( x , y )≡ ( x , y ) and ≡ is reflexive. (ii) Next, let ( x , y ), ( u , v )∈ ℜ^2 , and suppose that( x , y )≡ ( u , v ). Then , and by the symmetric property of equality,. Thus, , and

yx^2 = vu^2 vu^2 = yx^2 ( u , v )≡ ( x , y ) ≡ is symmetric. (iii) Finally, let , , and , and suppose and

. Then and by definition of

( x , y ) ( u , v ) ( r , s )∈ ℜ^2 ( x , y )≡( u , v ) ( u , v )≡ ( r , s ) yx^2 = vu^2 vu^2 = sr^2 ≡. So, by the transitive property of equality, , and thus by definition of

yx^2 = sr^2 ( x , y )≡ ( r , s ) ≡. So, ≡ is transitive. Since ≡is reflexive, symmetric, and transitive, it is an equivalence relation.