Tom Ferretti
MAE 501
10/3/07
In relation to Homework #4, there were problems within the class in expressing a proper
definition for a function, injective function, and surjective function. The first priority in
class was to get the students to properly represent these definitions in their own words.
The following were stated by classmates with encouragement and guidance from
Professor Berger and the rest of the class:
Function: A function is a rule or correspondence which associates to each element X in
a set A, a unique element f(x) in a set B.
Injective Function: An injective function is:
1) A function where whenever f(x) = f(y) in the range, then x = y in the domain.
OR
2) A function where f(x) = f(y) implies x = y.
Surjective Function: A surjective function is:
1) A function where range = codomain or image = codomain.
OR
2) A function f: A→B is surjective if: for every y in B, there exists an x in A such that
f(x) = y.
The following is a non-example:
Now suppose we rewrote the definition for a surjective function as:
There exists an x in A such that f(x) = y, for every y in B.
We can find counterexamples to this argument to prove this is not an adequate definition
of a surjective function.
Suppose we have two sets: A and B, such that (a1, a2, a3) are the elements of A,
and (b1, b2, b3) are the elements of B. Suppose f is a surjective function from A to B by
satisfying the definition: There exists an x in A such that f(x) = y, for every y in B.
Suppose f(a1) = b1, f(a1) = b2, and f(a1) = b3. This satisfies our definition, and it hits every
element of B, but it is not a function.
Using our previous sets, the function where f(a1) = b1, f(a2) = b2, and f(a3) = b3
would be a surjective function, since for every bi in B, there exists an ai in A such that f(ai)
= bi.
A common misconception in the homework for the definition of injective functions
was as follows:
Each x in set A maps to precisely one y in set B.
We can obtain a counter-example to disprove this as a definition for an injective function.
Suppose f(a1) = b1 and f(a2) = b1. This is a contradiction in an injective function. We
conclude that this is closer to the definition of a function than an injective function.
We now question the following definition of a function:
All elements of A map to exactly one element of B.
This means that every element ai in A maps to the same element bn in B.
This definition would be of a special function, but not a general definition of a function.
The English in this definition is poor, so we would want to more clearly define how
exactly we want a function to behave.
Suppose we have the following function:
f: A→B from the set of natural numbers to the set of all integers such that:
f(n) = {n/2 if n even, (1-n)/2 if n odd.
To prove this is a surjection, we want to show the following: given any integer w in the
set Z of all integers, find a natural number n such that f(n) = w.
Proof: Let w in Z denote an arbitrary integer. Suppose w is positive. Then, if x = 2w, x
is even, so we have f(x) = 2w/2 = w. Now suppose w is negative. Then, if x = -2w+1, x
is odd, then f(x) = [1- (-2w+1)]/2 = (1+2w-1)/2 = w. If w = 0, then x = 2w, so x is even,
and f(x) = 0/2 = 0. So, no matter what integer w is, we can find an x so that f(x) = w.
One student claimed that when f(n) = n/2 (if n is even), we can plug in any even natural
number for n, we obtain all positive integers as values for f(n). This is an unproven
statement because it only assumes that we obtain only positive integers for f(n), but we
must prove this result. We should stick to the first proof for correctness.
Equivalence Relations
For a relation (~) between sets A and B, to be classified as an equivalence relation, it
must satisfy the following properties:
- Reflexivity
(A~A)
- Symmetry
(A~B implies B~A)
- Transitivity
(A~B and B~C implies A~C)
Suppose we have the following relation:
A, B non-empty sets with A~B ↔ A∩B is non-empty.
First, we check the reflexive property:
A∩A = A which is non-empty. Hence, A~A.
Secondly, we check the symmetric property:
We know A∩B is non-empty. Since A∩B = B∩A, we can conclude B∩A is nonempty as well, therefore A∩B is non-empty → B∩A is non-empty, and this
relation is symmetric.
Third, we check the transitive property:
We know A∩B and B∩C are non-empty, does this imply A∩C is non-empty?
Suppose A∩B and B∩C each have one element, such that the element of A∩B is
not in C, and the element of B∩C is not in A. Then this is a counterexample to
the assumption that the transitive property holds. Therefore, this relation is not
transitive and is not an equivalence relation.
Now suppose we have the following relation between people:
A~B ↔ B is a brother of A.
First, we notice this is not a reflexive relation since one cannot be a brother of
themselves.
Second, we check the symmetry of the relation. Suppose B is the brother of A,
but A is the sister of B. This means A is a girl and B is a boy. In this situation,
A~B does not imply B~A. Therefore this is not a symmetric relation.
Third, we check the transitive property:
If B is a brother of A, and C is a brother of B, then C is also a brother of A.
Therefore, the transitive property holds, but it is still not an equivalence relation.
Some examples of equivalence relations:
1) A~B ↔ A=B
2) A~B ↔ A and B are same color.
3) A~B ↔ A and B are parallel
Def: A and B are parallel wherever slope A (m(A)) = slope B (m(B)).
Def: A and B are parallel if A and B never intersect transversally (intersect
at one point).
* These definitions were created to comply with parallel being an
equivalence relation.
4) (Triangles) A~B ↔ A is similar to B.
5) (Polygons) A~B ↔ A is similar to B.
We often see equivalence classes without seeing them formally.
If we have an equivalence relation on a set, then we can partition the set into disjoint
sets called equivalence classes.
Suppose: 2 lines are parallel if they don’t intersect transversally.
This means y = x, y = x+5, y = x-9 all lie in the same equivalence class, and we can name
the class m=1. The equivalence class m=0 would mean all constant functions. There is
also an equivalence class where m is undefined.
Now suppose our set consists of all the days. We can say A~B ↔ A differs from B by
168 hours. The classes are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday,
and Sunday.
Now take the set of all angles: A~B ↔ sin A = sin B and cosine A = cosine B.
What are the equivalence classes?
One student asserts that each equivalence class {A} would consist of all angles (A + 2nπ)
where 0 < A ≤ 2π. This means that each point on the unit circle corresponds to one class.
Other examples of equivalence relations are rotations, and modular theory.