Does there exist an onto group homomorphism from $(\mathbb R^*, .)$ (the multiplicative group of non-zero real numbers) onto $(\mathbb Q^*, .)$ (the multiplicative group of non-zero rational numbers)?

Let $f:\Bbb{R}^*\to\Bbb{Q}^*$ be one such homomorphism, then there exists $a \in \Bbb{R}^*$ such that $f(a)=2$. Let $w = \sqrt[3]{a} \in \Bbb{R}^*$ be a cube root of $a$ and cube roots always exist in $\mathbb{R}^*$, then $(f(w))^3 = f(w^3) = f(a) = 2$. This shows that $f(w)$ is a cube root of $2$ and also $f(w) \in \Bbb{Q}^*$. Which is a contradiction to the fact that $2$ has no cube root in $\mathbb{Q}^*$.

**Solution :****We prove by the method of contradiction.**

Let $f:\Bbb{R}^*\to\Bbb{Q}^*$ be one such homomorphism, then there exists $a \in \Bbb{R}^*$ such that $f(a)=2$. Let $w = \sqrt[3]{a} \in \Bbb{R}^*$ be a cube root of $a$ and cube roots always exist in $\mathbb{R}^*$, then $(f(w))^3 = f(w^3) = f(a) = 2$. This shows that $f(w)$ is a cube root of $2$ and also $f(w) \in \Bbb{Q}^*$. Which is a contradiction to the fact that $2$ has no cube root in $\mathbb{Q}^*$.

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