Definition: u − v = u + (−1)v.
Lemma: u − v = 0 iff u = v.
Proof: Assume u − v = 0
u + (−1)v = 0
u + (−1)v + 1v = 1v (add 1v to both sides)
u + ((−1) + 1)v = 1v
u + 0v = v
u + 0 = v
u = v
and conversely

This is not yet a proof!!

The set of n-tuples of the field is a vector space and any vector space with a basis with n elements is isomorphic to this space of tuples according to the obvious isomorphism: Σαiui ↔ (α0, ... αn−1) but to see this we must show that if Σαiui = Σβiui then for each i αi = βi.
Assume Σαiui = Σβiui
Σαiui − Σβiui = 0
Σ(αiui − βiui) = 0
Σ(αi − βi)ui = 0
but by the meaning of ‘independence of the {ui}’ we have:
for each i αi − βi = 0
for each i αi = βi

This establishes the one to one nature of the correspondence and thus the isomorphism. It also shows that two vector spaces with equinumerous bases are isomorphic to each other. A vector space with a basis of n elements is called an n-dimensional vector space and is unique up to isomorphism. Indeed this produces all isomorphisms between two such spaces.