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Given $b\in \mathbb{R}_{>1}$ is there $U\subseteq\mathbb{R}_{\ge 0}$ such that $U+bU=\mathbb{R}_{\ge 0}$ and $(U-U)\cap b(U-U)=\{0\}$ (or equivalently: $u+bv=u'+bv' \implies u=u', v=v'$)? Here is ...