Does there exists any non-trivial group homomorphism from $\mathbb{R}$ to $\mathbb{Q}$

Question

If $f:\mathbb{R}\longrightarrow \mathbb{Q}$ then it should be of the form $f(q)=qf(1)$, then how it will work for irrationals?

Answer 1

The answer depends on the axioms of set theory. Using the axiom of choice, $\mathbb{R}$ has a basis as a vector space over $\mathbb{Q}$, so you can write down examples of nontrivial linear maps $f : \mathbb{R} \to \mathbb{Q}$ by defining then arbitrarily on such a basis.

On the other hand, it's consistent with the negation of the axiom of choice that every homomorphism $\mathbb{R} \to \mathbb{Q}$ is continuous, which means it is necessarily zero since its image must be connected. See this MO thread for some details.