
Originally Posted by
particlejohn
Fix $\displaystyle b>1 $. Prove that $\displaystyle b^{x+y} = b^{x}b^{y} $ for $\displaystyle x,y \in \bold{R} $. We know that $\displaystyle b^{x}b^{y} = \sup B(x) \sup B(y) \geq b^{l}b^{w} = b^{l+w} $ for rational $\displaystyle l \leq x $ and $\displaystyle w \leq y $. Here, $\displaystyle l+w \leq x+y $, where $\displaystyle l+w \in \bold{Q} $. This implies that $\displaystyle \sup B(x) \sup B(y) $ is an upper bound of $\displaystyle B(x+y) $.
Is this correct? We want to prove this property for all reals.
Note: If $\displaystyle x $ is real, define $\displaystyle B(x) $ to be the set of all numbers $\displaystyle b^t $, where $\displaystyle t $ is rational and $\displaystyle t \leq x $.