nicht angemeldet
Aloha 'oe
»» bg = f(b + g) | : (b + g)
aber nur wenn man den Wertevorrat nicht beachtet. Für [latex]b,f,g \in \mathbb{R} \backslash {0}[/latex] sind sie äquivalent.
Bei b = -g gibt's ebenfalls ein Problem.
Gruß, Volker
--
„I conclude that there are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies and the other way is to make it so complicated that there are no obvious deficiencies."
- Tony Hoare
„I conclude that there are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies and the other way is to make it so complicated that there are no obvious deficiencies."
- Tony Hoare
Auge
Gunnar Bittersmann