1. ## 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.

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.

3. Proof 2: Here is a proof using basic field theory. Let $\displaystyle \alpha = \sqrt[n]{2}$. Now $\displaystyle \alpha$ is a root of $\displaystyle x^n - 2$ over $\displaystyle \mathbb{Q}[x]$, and this polynomial is the minimal polynomial by Eisenstein criterion. Define $\displaystyle V = \{ a_0 + a_1\alpha + ... + a_{n-1}\alpha^{n-1} | a_i \in \mathbb{Q} \}$ then $\displaystyle V$ is a sub-space of $\displaystyle \mathbb{R}$. The claim is that $\displaystyle V$ is $\displaystyle n$ dimensional with basis $\displaystyle \{ 1,\alpha, ... ,\alpha^{n-1} \}$. First if $\displaystyle \alpha^i = \alpha^j$ for $\displaystyle 0\leq i < j\leq n-1$ then $\displaystyle \alpha^{i-j} - 1=0$ proving that $\displaystyle x^{i-j}-1$ is a smaller degree polynomial than $\displaystyle x^n - 2$ which is impossible. Hence all of these elements are distinct. If they are lineary dependent then there exist non-zero $\displaystyle a_i$ such that $\displaystyle a_0+a_1\alpha+...+a_{n-1}\alpha^{n-1} = 0$ and hence $\displaystyle f(x) = a_0+a_1x+...+a_{n-1}x^{n-1}$ is a non-zero polynomial having $\displaystyle \alpha$ as a root, which is a contradiction because $\displaystyle x^n - 2$ is the minimal polynomial. Q.E.D.

This proves it because if $\displaystyle [\mathbb{R}:\mathbb{Q}]<\infty$ then there shall be a bound on the size of the dimension of a sub-space, which the above argument shows is clearly impossible.

4. Proof 3: This is from a classmate. The real number $\displaystyle \pi$ is transcendental over $\displaystyle \mathbb{Q}$. Thus, it is an infinite degree extension.

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