Onto homomorphism from $\mathbb{R}^{*}$ to $\mathbb{Q}^{*}$ (CSIR)

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)?

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 $\mathbb{Q}^*$.

SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO SUBSCRIBE TO MY BLOG



No comments:

Post a Comment

Featured Post

NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...

Popular Posts