Since you would like a multiplicative group of rationals bounded between $\frac{1}{d}$ and $b$, you can prove its non-existence in an easy way: assume your group exists and is non-trivial and let $g\neq1$ be one of its elements. Consider $g^n$. If $0 < g < 1$, then $g^n\to0$; if $g>1$, then $g^n\to\infty$. On the other hand, $g^n$ must belong to your group for all $n$, showing that such bounds you want cannot exist. Analogue procedure shows that a bounded additive group of real numbers cannot exist.

Since you want a multiplicative group of rationals bounded between $\frac{1}{d}$ and $b$, you can prove its non-existence in an easy way: assume your group exists and is non-trivial and let $g\neq1$ be one of its elements. Consider $g^n$. If $0 < g < 1$, then $g^n\to0$; if $g>1$, then $g^n\to\infty$. On the other hand, $g^n$ must belong to your group for all $n$, showing that such bounds you want cannot exist. Analogue procedure shows that a bounded additive group of real numbers cannot exist.

Since you would like a multiplicative group of rational bounded between $\frac{1}{d}$ and $b$, you can prove its non-existence in a very trivial way: assume your group exists and is non-trivial and let $g\neq1$ be one of its elements. Consider $g^n$. If $0 < g < 1$, then $g^n\to0$; if $g>1$, then $g^n\to\infty$. On the other hand, $g^n$ must belong to your group for all $n$, showing that such bounds you want cannot exist. Analogue procedure shows that a bounded additive group of real numbers cannot exist.

Since you would like a multiplicative group of rationals bounded between $\frac{1}{d}$ and $b$, you can prove its non-existence in an easy way: assume your group exists and is non-trivial and let $g\neq1$ be one of its elements. Consider $g^n$. If $0 < g < 1$, then $g^n\to0$; if $g>1$, then $g^n\to\infty$. On the other hand, $g^n$ must belong to your group for all $n$, showing that such bounds you want cannot exist. Analogue procedure shows that a bounded additive group of real numbers cannot exist.