If $\phi$ is an automorphism of the group $\Bbb R^+$ under multiplication satisfying $\phi(4)=7$, then find $\phi(2)$.

Question

Let $\Bbb R^+$ denote the real numbers. Suppose $\phi:\Bbb R^+\to\Bbb R^+$ is an automorphism of the group $\Bbb R^+$ under multiplication with $\phi(4)=7$.

Find $\phi(2)$, $\phi(8)$, and $\phi(1/4).$

I've done this problem in groups of integers modulo $n$ but I am not sure how to start here. Please explain what $\phi(4)=7$ is.

Thank you!

Answer 1

If $\phi(4)=7$, then $7=\phi(4)=\phi(2\cdot 2)=\phi(2)\phi(2) = \phi(2)^2$. Can you take it from there?

Answer 2

Hint If $a \in \Bbb R^+$ satisfies $a^2 = 4$, then because $\phi$ is a homomorphism, we have $$\phi(a)^2 = \phi(a^2) = \phi(4) .$$

It's interesting to note that we cannot, however, conclude the value of, e.g., $\phi(3)$ from the given information.

Additional hint To compute $\phi\left(\frac{1}{4}\right)$ it will be useful to recall that any homomorphism maps the identity to the identity.