Prove that: $ [g+h] \geq \min{[g],[h]}$
Definitons: Take any ordered abelian field $(G, \leq)$. Two elements $g,h \in G$ are archimedean equivalent if $\exists \ n,n' \in \Bbb{N}$ : $n|g| \geq |h|$ and $n'|h| \geq |g|$. $[g]$ denotes the equivalence class of $g$.
Attempt of proof: Suppose that $ \min{ [g], [h] }= [h]$
$[h] = \{g: \exists n_1, n_1' $such that$ \ n_1|h| \geq |g|\ $and$ \ n_1'|g| \geq |h|\} $
$[h+g] =\{ k: \exists n_2, n_2' \ $such that$ \ n_2|g+h| \geq |k| \ $and$ \ n_2'|k| \geq |g+h| \}$
I have to show that $k>g$ so I have to show that $ng<k$ for all $n \in \Bbb{N}$. From my assumption $[h] < [g]$ so it implies $|h| > |g|$ and $[g] \neq[h]$. I am not sure what to do know. I have tried to multiply these inequalities by an inverse element for example of $n_1, n_1'$. I can also use the fact: $[g] \geq [h]$ for all $h \in G$ iff $g=0$.
Assume that $ [g+h] < \min([g],[h]) $. Let $ [g] = \min([g], [h]) \Longrightarrow [g] > [h] \Longleftrightarrow \forall_{n \in \mathbb{N}} \; |g| > n|h|$.
From assumption we have $ \; [g + h] < [g] \Longrightarrow |g| < |g + h| \; \text{and} \; [g + h] \neq [g] $.
We know that $$ |g| < |g + h| \Longleftrightarrow \forall_{n \in \mathbb{N}} \; n|g| < |g + h| \leq |g| + |h| \Longrightarrow (n - 1)|g| < |g + h| - |g| \leq |h| \\ \Longrightarrow (n - 1)|g| < |h| $$ So we have $$\forall_{n \in \mathbb{N}} \; (n - 1)|g| < |h| \quad \wedge \quad |g| > n|h| $$ Hence it doesn't true then $ [g+h] \geq \min([g],[h]) $. $\square$