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Iam looking for the solution for following two questions:

Let $G$ be a ssubgroup of $\mathbb{R}$, Let $a = \inf\,\{ g \in G : g > 0 \}$ Then,

  1. If $a=0,$ for all $x,y \in \mathbb{R}$ there is a $g$ such that $x < g < y.$

  2. If $a \neq 0$ then $G$ isomorphic to $\mathbb{Z}.$

Any help will be appreciated.

Thank you.

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  • 1
    $\begingroup$ Have you made any progress? $\endgroup$ – tomasz Feb 17 '16 at 10:51
  • $\begingroup$ No. Please let me suggest some direction. $\endgroup$ – Arup Feb 17 '16 at 10:51
  • $\begingroup$ The inequalities in (1) must be weak, I think. $\endgroup$ – DonAntonio Feb 17 '16 at 11:36
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1.- Say $\;y-x=\epsilon\;$ , and we know there exists $\;g\in G\;$ with $\;0<g<\epsilon\;$ . Assume $\;y>x>0\;$ (otherwise work with $\;-g\in G\;$) , and take $\;n\in\Bbb N\;$ such that $\;ng\le y-x\;$ .

Since $\;(n+1)g-ng=g<\epsilon=y-x\;$ , it then must be that either $\;(n+1)g=y\;$ and then trivially $\;y\le(n+1)g<x\;$ , or else $\;y<(n+1)g<x\;$

2.-Suppose $\;x=\min\{g\in G\;;\;g>0\}\;$ , and let $\;g\in G\;$ . Again, suppose $\;g>0\;$ and take $\;n\in\Bbb N\;$ such that

$$\;nx\le g<(n+1)x\implies 0\le g-nx<x\stackrel{\text{minimality of}\;x}\implies g-nx=0$$

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