Automorphism, relation 'greater than' Disclaimer: I'm very new to group theory and trying to understand it.
My book states the following:
 The image  $\mathbb{R} \rightarrow  \mathbb{R}: x \rightarrow -x $ is an automorphism for $ \mathbb{R} $ relative to the order relations $\leq$ and $\geq$.
The book doesn't state proof however and I'm trying to understand it better. My question is, how do you proof this?
Is it enough to say the following:
$
-x + -y \leq -(x + y): x,y \in\mathbb{R} $ is true and therefore it is an automorphism. 
 A: I'd rather say that $\phi: \mathbb R \ni x \mapsto -x \in \mathbb R$ is a homomorphism of ordered sets $(\mathbb R, \le)$ and $(\mathbb R, \ge)$. Strictly speaking, they are different structures, so I wouldn't call it an automorphism, but maybe the author of your book uses a slightly different definition.
In general, $\phi: A \rightarrow B$ is an homomorphism of $(A, \sim_A)$ and $(B, \sim_B)$ if
$$ \forall x,y\in A: \big((x \sim_A y) \Rightarrow (\phi(x) \sim_B \phi(y))\big)$$
In this case $\phi: \mathbb R \rightarrow \mathbb R$ is a homomorphism of $(\mathbb R, \le)$ and $(\mathbb R, \ge)$ if 
$$ \forall x,y\in \mathbb R: \big((x \le y) \Rightarrow (\phi(x) \ge \phi(y))\big)$$
For $\phi(x) = -x$ it's true, so it's a homomorphism (of ordered sets).
$\phi$ also happens to be an endomorphism of group $(\mathbb R, +)$, and for that you'd need to check the condition
$$ \forall x,y,z\in \mathbb R: \big((x + y = z) \Rightarrow (\phi(x) + \phi(y) = \phi(z))\big)$$
or equivalently
$$ \forall x,y\in \mathbb R: \phi(x) + \phi(y) = \phi(x+y)$$
It's also invertible ($\phi^{-1} = \phi$) so it's an automorphism. But in this case there's no need to consider the ordering at all.
