Show $(\mathbb{R},0,+,-)$ has infinitely many automorphisms

Question

Show $(\mathbb{R},0,+,-)$ has infinitely many automorphisms :

My idea is that if $\phi:(\mathbb{R},0,+,-) \rightarrow (\mathbb{R},0,+,-)$ is an automorphism. Then since here we don't have the order relation. For any $x_1, x_2, y_1, y_2 \in \mathbb{R}$, $(x_1,x_2 ) \rightarrow (y_1, y_2)$ will be an automorphism. In another words, any we can always map an open interval to another open interval to have automorphism.

Could someone let me know if my thinking is correct?

If not, could you let me know what went wrong, and how should I fix it?

Any hints or ideas will be appreciated!

Thanks!

Answer 1

The answer you propose in the comments is correct: For any non-zero $r\in \mathbb{R}$, the function $x\mapsto rx$ is an automorphism of $(\mathbb{R},0,+,-)$. I'll leave it to you to check the details.

On the other hand, what you've written doesn't make any sense to me. You say "$(x_1,x_2)\to (y_1,y_2)$ will be an automorphism". What function $\mathbb{R}\to \mathbb{R}$ do you have in mind here? It's not true that any open interval can be mapped to any other open interval by an automorphism. For example, $(-1,1)$ can't be mapped to $(1,2)$, since the first interval contains $0$, the second does not contain $0$, and any automorphism maps $0$ to $0$.

Answer 2

Hint: $\mathbb R$ is an infinite-dimensional vector space over $\mathbb Q$.