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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 # Solutions
y = x^2 + a
a. Prove that this set of parabolas forms a partition of the plane, ℜ 2.
wish to prove that F is a partition of
(i) First, note that (0, a ) ∈ ℜ^2 since 0 ∈ℜand a ∈ℜ. Also, , so (0, a ). Thus,
a = 02 + a
partition.
by the negation of the definition of the empty set. Thus, and
( x , y )∈ Aa ∩ A 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 y − x^2 = a y = x^2 + b y − x^2 = b A b a = y − x^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
a
∈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 (^) y − x 2 y = x^2 +( y − x^2 ) ( x , y ) ∈ Ay − x 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.
y − x^2 ∈ ℜ A (^) y − x 2 ∈ F
∈ℜ
a
( x , y ) Aa
∈ℜ
a
( x , y ) Aa ( x , y )∈ℜ^2 = ℜ^2 ∈ℜ
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 y − x^2 = a and
v − u^2 = a , or. So, define an equivalence relation if and only if.
y − x^2 = v − u^2 ( x , y )≡( u , v ) y − x^2 = v − u^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
y − x^2 = y − x^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
y − x^2 = v − u^2 v − u^2 = y − x^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 ) y − x^2 = v − u^2 v − u^2 = s − r^2 ≡. So, by the transitive property of equality, , and thus by definition of
y − x^2 = s − r^2 ( x , y )≡ ( r , s ) ≡. So, ≡ is transitive. Since ≡is reflexive, symmetric, and transitive, it is an equivalence relation.