> 
> My general question is to find, if this is possible, two real numbers $a,b$ such that $K=\Bbb Q(a,b)$ is not a simple extension of $\Bbb Q$.
>  $\newcommand{\Q}{\Bbb Q}$
> 
> 
> 

Of course $a$ and $b$ can't be both algebraic, otherwise $K$ would be a separable ($\Q$ has characteristic $0$) and finite extension, which [has to be](https://en.wikipedia.org/wiki/Primitive_element_theorem#Classical_Primitive_Element_Theorem) simple. So I tried with $\Q(\sqrt 2, e)$ but any other example would be accepted.

The field $\Q(\sqrt 2, e)$ has transcendence degree $1$ over $\Q$, but I'm not sure if this imply that it is isomorphic to $\Q(a)$ for some transcendental number $a$ (the fact that two fields have the same transcendence degree over another field shouldn't imply that the fields are isomorphic).

I'm not sure about the relation between the algebraic independence of $a$ and $b$, and the fact that $\Q(a,b)/\Q$ is a simple extension.
Notice that $\Q(\pi, e)$ is probably unknown to be a simple extension of $\Q$.

Thank you for your help!