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I was doing some homework problems and was proving several facts about vector spaces (and was quite proud at how easily I was doing it) and suddenly noticedmy proofs were circular. So I picked one to prove from scratch. I did it, but I want someone to verify that it is, in fact, a proof. Here it goes.
In the following, V is an Abelian group and F is a field. We have a defined function:defined as scalar multiplication. Under these assumptions, we may call V an F-vector space.
Proposition: Givenand v any vector in V, show that
.
Proof: I will assume that there exists ansuch that
for all
. Existence of such an f will be demonstrated.
Now, there exists an element 1 of F such that 1 is the multiplicative identity (F is a field). We have that 1f = f (for all f) for multiplication in F. Thus:
(The last step is a part of the definition of the scalar product.)
We defined above that
By comparing this result towe see that the required f = 1. Therefore such an f exists and the theorem is proved.
(End of proof.)
It just seems too...convenient I guess...to assume an f, then basically cancel the darned thing out, then state what it is. I can't find an argument against the above being a valid proof, but I'm uncomfortable with it.
Any thoughts? Thanks!
-Dan
) You're right, that's one of the axioms.
)
