Linear function from $\mathbb R$ over $\mathbb Q$ to $\mathbb R$ over $\mathbb Q$

Question

Let $ V $ be a vector space of $ \mathbb R$ over the field $\mathbb Q$ and $x$ is an irrational number. I want to prove or disprove that there is a linear function $f$ from $ V $ to $ V $ that satisfies $f(1) = 1 $ and $f(x) = -1$. Since $1$ and $x$ are independent vectors in $V$ I tend to think that such function exists but could not find a way to prove or disprove it.

Answer 1

Assuming the axiom of choice it is shown that every vector space has a basis. Since $1,x$ are linearly independent we can extend it to a basis $B$ of $V$.

Taking hints from comments of Berci, define a function $f$ on $V$ such that $f(1)=1, f(x) = -1$ and $f(v) =0$ for all other $v \in B$ and $$f(y) = \sum_{v_i \in B}a_if(v_i) $$ where $y=\sum_{v_i \in B}a_iv_i$ and $a_i \in \mathbb{Q}$. Note that for a given $y$ , $a_i$ are unique so the function $f$ is well-defined. I let you verify that $f$ is a linear function.