Is R a Finite-Dimensional Vector Space Over Q?

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SUMMARY

The set of all real numbers (R) is definitively not a finite-dimensional vector space over the field of rational numbers (Q). This conclusion is supported by the argument that if R were finite-dimensional, it would imply that R is countable, which contradicts the established fact that R is uncountable. The discussion references Halmos' "Finite-Dimensional Vector Spaces" and emphasizes the importance of understanding cardinal numbers to grasp the complexities of this topic. The dimension of R over Q is not only infinite but also uncountable.

PREREQUISITES
  • Understanding of vector spaces and their properties
  • Familiarity with cardinal numbers and their implications in set theory
  • Knowledge of Hamel bases and their role in vector space theory
  • Basic concepts of countability and uncountability in mathematics
NEXT STEPS
  • Study the concept of Hamel bases in vector spaces
  • Learn about cardinality and its applications in set theory
  • Explore the differences between finite and infinite-dimensional vector spaces
  • Investigate the implications of Schauder bases in functional analysis
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in advanced vector space theory and cardinality concepts will benefit from this discussion.

  • #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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