Is R a Finite-Dimensional Vector Space Over Q?

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The set of all real numbers, R, is not a finite-dimensional vector space over the field of rational numbers, Q. If R were finite-dimensional, it would imply that every real number could be expressed as a finite linear combination of a finite basis of real numbers with rational coefficients, leading to a contradiction regarding the countability of R. The discussion highlights that R has uncountable dimension over Q, as any basis for R must also be uncountable. The argument presented involves cardinality and the properties of finite and countable sets. Ultimately, the complexity of the problem lies in establishing the uncountability of the dimension rather than just its infinitude.
  • #31
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.
 
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  • #32
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.
 
  • #33
oops, i missed that ! : )
 
  • #34
And what could be neater than to quote me?
 

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