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Math Help - Infinite-D problem

  1. #1
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    Infinite-D problem

    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.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    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 1: Q is contable while R is not.
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  3. #3
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    Proof 2: Here is a proof using basic field theory. Let \alpha = \sqrt[n]{2}. Now \alpha is a root of x^n - 2 over \mathbb{Q}[x], and this polynomial is the minimal polynomial by Eisenstein criterion. Define V = \{ a_0 + a_1\alpha + ... + a_{n-1}\alpha^{n-1} | a_i \in \mathbb{Q} \} then V is a sub-space of \mathbb{R}. The claim is that V is n dimensional with basis \{ 1,\alpha, ... ,\alpha^{n-1} \}. First if \alpha^i = \alpha^j for 0\leq i < j\leq n-1 then \alpha^{i-j} - 1=0 proving that x^{i-j}-1 is a smaller degree polynomial than x^n - 2 which is impossible. Hence all of these elements are distinct. If they are lineary dependent then there exist non-zero a_i such that a_0+a_1\alpha+...+a_{n-1}\alpha^{n-1} = 0 and hence f(x) = a_0+a_1x+...+a_{n-1}x^{n-1} is a non-zero polynomial having \alpha as a root, which is a contradiction because x^n - 2 is the minimal polynomial. Q.E.D.

    This proves it because if [\mathbb{R}:\mathbb{Q}]<\infty then there shall be a bound on the size of the dimension of a sub-space, which the above argument shows is clearly impossible.
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  4. #4
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    Proof 3: This is from a classmate. The real number \pi is transcendental over \mathbb{Q}. Thus, it is an infinite degree extension.


    (The proof uses two facts from field theory: (i) there exist real transcendental number (such at \pi), (ii) finite field extensions are algebraic extensions*. Thus, if [\mathbb{R}:\mathbb{Q}] < \infty then all elements would be algebraic numbers, but that is clearly a contradiction.)
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