I am aware of the standard proofs using a cardinality argument or using the fact that there infinitely many prime numbers (see this post for instance). Using induction looked promising at first, but unfortunately I got stuck. What I am trying to prove is the following:
There exist no numbers $x_1,\ldots,x_n \in \mathbb{R}$, with $n \in \mathbb{N}$, such that $\text{span}_\mathbb{Q}(x_1,\ldots,x_n) = \mathbb{R}$.
Assume the contrary to (hopefully) obtain contradictions:
For the base case $n=1$ we show that $\text{span}_\mathbb{Q}(x_1) = \mathbb{R}$ is impossible. If $x_1 \in \mathbb{Q}$, we get that $\sqrt{2} \notin \text{span}_\mathbb{Q}(x_1)$, which is a contradiction. If $x_1 \in \mathbb{R \setminus Q}$, we get that $1 \notin \text{span}_\mathbb{Q}(x_1)$, since we would have a linear combination $1=\lambda_1x_1$, and the LHS is rational, whereas the RHS is not, contradiction.
For the induction hypothesis, assume that we have proven that $\text{span}_\mathbb{Q}(x_1,\ldots,x_n) = \mathbb{R}$ is impossible for any $x_1,\ldots,x_n \in \mathbb{R}$. We want to show that $\text{span}_\mathbb{Q}(x_1,\ldots,x_n,x_{n+1}) = \mathbb{R}$ is impossible.
Again, I assumed the contrary: For any $y \in \mathbb{R}$ there exist $\lambda_1,\ldots,\lambda_{n+1}$ such that $y=\sum_{i=1}^{n+1}\lambda_ix_i$. This is equivalent to $$ \lambda_1x_1+\ldots+\lambda_nx_n = y-\lambda_{n+1}x_{n+1}. \tag1 $$ It seemed like it would be helpful to have a term on the LHS that is not dependent on $x_{n+1}$ anymore. I thought about setting $y=0$, to obtain $$ \lambda_1x_1+\ldots+\lambda_nx_n =-\lambda_{n+1}x_{n+1}, $$ but I'm not sure if this is helpful.
My questions:
- Is a proof of this theorem by induction possible in the first place?
- If yes, how can I apply the induction hypothesis to equation $(1)$?
