Is R a Finite-Dimensional Vector Space Over Q?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
33 replies · 27K views
something much neater!

If a vector space V is finite dimensional , of dim n say, over a field F
then, (assume, {v1,v2,...,vn} is a basis for V)
V is isomorphic to F^n
through the correspondence
a1.v1 +a2v2 + a3v3 + ... + anvn -----> (a1,a2,a3,...,an) {the n component tuple of F^n) (verify the bijective linear transformation)
It follows that R (reals) would be isomorphic to Q^n (Q denotes field of rationals)
But, Q^n is countable (Q being countable)
implying that R is countable (absurd)
Q.E.D.
 
Physics news on Phys.org
sihag said:
something much neater!

If a vector space V is finite dimensional , of dim n say, over a field F
then, (assume, {v1,v2,...,vn} is a basis for V)
V is isomorphic to F^n
through the correspondence
a1.v1 +a2v2 + a3v3 + ... + anvn -----> (a1,a2,a3,...,an) {the n component tuple of F^n) (verify the bijective linear transformation)
It follows that R (reals) would be isomorphic to Q^n (Q denotes field of rationals)
But, Q^n is countable (Q being countable)
implying that R is countable (absurd)
Q.E.D.
How is this neater? It's exactly what HallsofIvy had in post #2.
 
oops, i missed that ! : )