Suppose $G$ an additive group of real numbers, that is $x,y\in G\implies x-y\in G$ with $G\neq \{0\}$. Consider $G^+$ the set of positive elements of $G$.
If $G\subset \mathbb{R}$ is an additive group such that $\inf{G^+}=0$, then $G$ is dense in $\mathbb{R}$.
Proof. If $G^+$ admits an infimum, then $\inf{G^+}\leq g$ for all $g\in G^+$. This also implies that for every $\varepsilon>0$, there exists $g\in G^+$ such that $\inf{G^+}\leq g < \inf{G^+}+\varepsilon$.
As $\inf{G^+}=0$, then we know that for every $\varepsilon>0$ there exists $g\in G^+$ such that $0\leq g < \varepsilon$. No matter how close we get to $0$, there is a $g\in G^+$ even closer.
For $G$ to be dense, there is a $g\in G$ as element of any interval $(a,b)\subset\mathbb{R}$, that is, $g\in (a,b)$ or even $(a,b)\cap G\neq \varnothing$ for any $a,b\in\mathbb{R}$. Let $b=a+\varepsilon$.
We must understand the behavior of $G$ next to its infimum. There's always a $g\in G$ between $0$ and $\varepsilon$, for any $\varepsilon>0$. This behavior is interestingly comparable to the Archimedean property in $\mathbb{R}$, but instead of $\mathbb{N}$ we have $G$ with its own internal behavior.
If we sum a given $a\in\mathbb{R}$ to the inequality we get $0\leq g<\varepsilon \implies a\leq a+g<a+\varepsilon$. For the proof to be concluded, we only need to find a way to prove that $a+g\in G$.
Now, how do I prove that $a+g\in G$? I thought I could also find some $K\in\mathbb{Z}$ such that $a<Kg<a+\varepsilon$, and $Kg\in G$ because $g\in G\implies g+\cdots+g\in G$. Some hints, please.
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Update. As there's a $g\in G^+$ such that $g<\varepsilon$ for any $\varepsilon>0$, let $a\in\mathbb{R}$ exist such that $\varepsilon<a$. Because $\mathbb{R}$ is Archimedean, we know there's an $n\in\mathbb{N}$ such that $g<\varepsilon<a\implies a<ng$. Now we need to prove that $ng<a+\varepsilon$ and we do that with the well-ordering principle, creating a set $A=\{ng>a:n\in\mathbb{N}\}$. This set has a smallest element $n_0\in\mathbb{N}$, so $(n_0−1)g<a$. If it was that $(n_0−1)g<a<a+\varepsilon<ng$, we would arrive at an absurd, $g>\varepsilon$, so $ng<a+\varepsilon$.
Best Answer
It is not necessarily true that $a+g\in G$, for that of course would imply $(a+g) - g = a\in G$. Your second idea is better, and ties in nicely with what you remarked about the Archimedean property.
You've already shown that you can always find $g$ small enough in $G$. No matter how small the length $b-a$ of $(a,b)$ is, show that if $g\in G$ is small enough then some multiple of $g$ must lie in $(a,b)$.