$e$ in Roth's Theorem THEOREM 1.8 of the book _Making Transcendence Transparent_ by Burger says: ![enter image description here]( then it says: ![enter image description here]( But $e$ is not algebraic how it satisfies Roth's Theorem ?

The theorem states that IF $\alpha$ is algebraic, THEN [something holds].

It does not state that IF [the same something holds for some number $\beta$] THEN $\beta$ is algebraic.