Let V be the set of real numbers regarded as a vector space over the field of rational numbers. Prove that V is infinite-dimensional.
I'm trying to prove by contradiction. I select a basis in this vector space but I do not know how to start. Hint?
Thank you.
Proof 2: Here is a proof using basic field theory. Let . Now is a root of over , and this polynomial is the minimal polynomial by Eisenstein criterion. Define then is a sub-space of . The claim is that is dimensional with basis . First if for then proving that is a smaller degree polynomial than which is impossible. Hence all of these elements are distinct. If they are lineary dependent then there exist non-zero such that and hence is a non-zero polynomial having as a root, which is a contradiction because is the minimal polynomial. Q.E.D.
This proves it because if then there shall be a bound on the size of the dimension of a sub-space, which the above argument shows is clearly impossible.
Proof 3: This is from a classmate. The real number is transcendental over . Thus, it is an infinite degree extension.
(The proof uses two facts from field theory: (i) there exist real transcendental number (such at ), (ii) finite field extensions are algebraic extensions*. Thus, if then all elements would be algebraic numbers, but that is clearly a contradiction.)