# Infinite-D problem

• Jan 26th 2008, 02:34 PM
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.
• Jan 26th 2008, 02:44 PM
ThePerfectHacker
Quote:

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.
• Jan 26th 2008, 02:51 PM
ThePerfectHacker
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.
• Jan 29th 2008, 02:34 PM
ThePerfectHacker
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.)