Our professor defined exponentiation for real numbers in the following way: $$b^x = \text{sup}\{b^r; r \in \mathbb{Q},\ r \leq x \}$$
Then he said that the rules of exponentiation that work for rational numbers work for real numbers as well. He said that he recommends us to try to prove it at home ourselves.
I have been trying to prove that $$\forall x,y \in \mathbb{R}:\ b^x \cdot b^y = b^{x+y}$$
for multiple hours but still didn't figure out how to do it. The professor hinted that we should use the fact that $\forall \varepsilon>0\ \exists r \in \mathbb{Q}, r \leq x$ such that $b^x - \varepsilon < b^r < b^x$, but I don't know how it is useful in any way.
One thing that I tried was to say that $b^x - \varepsilon < b^r \leq b^x$ and $b^y - \varepsilon < b^s \leq b^y$ and then $(b^y - \varepsilon)(b^x-\varepsilon)<b^r \cdot b^s \leq b^x\cdot b^y$. Here we can use the rule of exponentiation for rational numbers: $b^r \cdot b^s = b^{r+s}$ but I have no idea how to proceed further.