We want to extend the idea of function to include Vector Valued Function. The domain of such a function will contain real numbers and the range will contain vectors. We will use them to study objects moving in space. Our first question is, " What are vectors?"

We could answer the question so as to just solve the problem above or we can attempt to lay ground work for problems involving other kinds of vectors. We begin with a general definition from a Linear Algebra text:

A Vector Space is a set of objects with two operations on the objects in the set. We refer to the operations as addition and multiplication with the understanding that they are not the addition and multiplication we've studied before because the objects they act on are different. The set and the operation called addition form what is called an Abelian Group. The other operation, called multiplication, must satisfy four more properties (some might call these rules or laws or axioms) ; these include two distributive properties, an associative rule, and finally, the real number 1 must act as a multipicative identity. A formal definition follows:

V = {{ u,v,w,... } , + , ' } We will use juxtaposition ( that is, next to each other with no symbol between ) for multiplication instead of ' and V refers to the set and the two operations but we will also use V to refer to the set of objects { u,v,w,...} The objects in the set are called vectors and the set with the operations is called a Vector Space, that is, if the following ten properties(axioms) are present:

1. Addition on V is closed , that is, any two objects from V have a unique sum in V.

2. Scalar Multiplication on V is also closed. The vectors are multiplied by numbers(scalars) not vectors.

3. u + v = v + u Addition of vectors is commutative.

4. u + ( v + w ) = ( u + v ) + w Addition is associative

5. There is a object referred to by 0 and for any object from V we have u + 0 = u

(This is called the additive identity property)

6. For each object in V there is a unique element referred to by -u and -u + u = 0

( This is called the additive inverse property ) Note: At this point we have an Abelian Group.

7. c and d are arbitary real numbers while u is an arbitrary vector then c(du) = cd(u) (This is an associative property)

8. c ( u + v ) = cu + cv Multiplication distributes across addition.

9. ( c + d )u = cu + du This is a second distributive axiom where we have two reals and one vector.

10. For any object from V , say u , 1u = u

We will study vectors that are represented as ordered pairs or triplets or quadruples or quintuples etc. and they will obey these ten axioms under the addition and multiplication defined on them. There are many other objects that will be used for vectors in future studies. Thoughout the remainder of the Chapter we will be studying a particular kind of vector.  Our text refers to the vectors we'll be studying as vectors in a plane( ordered pairs ) or vectors in space ( ordered triplets ).

Many more rules follow from the ten axioms in the definition of the vector space. At first it may be difficult for you to "see" that we can prove subsequent rules for vectors whether or not we know the objects being used for the vectors or not. But if we do not know what the objects are, how can we know if they are vectors? The answer lies in the fact that we need only know that they obey the ten axioms listed in the definition. I'll try to illustrate this idea in the few examples that follow and will leave you a few to try on your own.

{ V = { u,v,w,...} , + , ' , } is a vector space.

Theorem 1.If u + v = u , then u = 0 .

Our first thought might be, " This is obvious. " . However, while it is fairly obvious if we are in a vector space, it may be false if we're not . [We'll have examples of that later in the course .]

To prove this theorem we begin by assuming that u + v = u [ Can you explain what "happens" if it is false.] Then we add -u to (u + v) because -u is available for our use. [ Explain 1 ] Then, using associativity and /or commutativity we get v as the result of the addition.[ Explain 2] Next we add -u to u and get 0 [ Explain 3 ] Since (u + v) and u were assumed to be the same object in V and addition produces unique objects, our results, namely, v and 0 , must be the same and the proof is finished. [ Did you "catch the part about addition producing unique results---Explain 4 ]

Here are some possible answers for the "[Explain]"

[Explain 1] Axiom 6 guarantees that -u is present for u, additive inverses

[Explain 2] -u + (u + v) = (-u + u) + v = 0 + v = v associative Use commutative if doing (u + v) + -u

[Explain 3] property of additive opposites, Axiom 6

[Explain 4] Addition is actually a binary function so that results are unique; also, see Axiom 1

You probably have guessed by now that " more is going on than meets the eye ". "How to Prove Things in Mathematics " is another study for another time. With limited attention to detail, we exhibit " Proofs " in our work so as to show relationships between different items in our discussion. In the case above, I'm showing you how some " Rules of Algebra " might follow from a definition. This example also illustrates that we do not need to know what the objects are like, only that they are from a vector space.

When we know more about the vectors, the proofs will have a different apppearance. For example:

V = { (x,y) where x and y are real numbers } also, () + () = ( )

and k() = ( ) define the operations on our set.

Note:We call the first operation vector addition and the second is scalar multiplication( scalar refers to the real number k ).

Can you prove that the set V with the two operations is a vector space ? We need to show that the ten axioms are satisfied. For example, Axiom 5: there is an additive identity.

This proof is different than the previous in that we have more information about the vectors, namely, that they can be represented as ordered pairs. Also, the operations are specific.

The proof would simply require that we show what the vector 0 is and that it works.

Proof: ( 0,0 ) is the 0 element and for any ( x,y ) in set V we have (0,0) + (x,y) = ( 0+x, 0+y ) = (x,y).

You might try proving some of the other axioms for homework.

Is it true in a vector space that -1u = -u  What do the proofs "look like" for the ordered pair vectors versus the general vector space in which we have no knowledge of the vectors other than the ten axioms?

Proof: u = (x,y) then -1(x,y) = (-1x,-1y) = (-x,-y) and (-x,-y) + (x,y) = (0,0) for any (x,y)

To prove the same theorem for the general vector space we need to first prove other theorems that are needed for the proof. The next two theorems can be used as homework problems.

Homework Thm. 1: If u + v = 0 then u = -v and v = -u

Homework Thm. 2: If u is any vector from V then 0u = 0 [ Obviously,the first zero is a real number and the second is a vector.]

Finally, we can prove that -1u = -u in any vector space.

Proof:For any vector u in V   -1u + u = -1u + 1u [explain 1] then -1u + 1u = (-1 + 1)u [explain 2]

we have (-1 + 1)u = 0u [explain 3] and 0u = 0 [explain 4] and this implies that -1u = -u [explain 5]

If you had trouble with any of the "[explain]" in the proof, then here are some possible answers.

[explain 1] : the property of the number 1 , axiom 10 in the definition of vector space

[explain 2]: distributive property, axiom 8 and commutative, axiom 3 for the reversing of order

[explain 3]: addition of opposites in real numbers is not in question and -1 + 1 is the real number 0

[explain 4]: by homework theorem 2

[explain 5]: -1u + u = 0 implies that -1u = -u by your homework theorem 1

This is the extent of work in this course on "theorem proving". It will serve those of you who go on to graduate work in Mathematics as an example of the "shift in thinking" and new "problem type" you will encounter in the Junior and Senior level courses you'll be required to study.