$e$ in Roth's Theorem

Question

THEOREM 1.8 of the book Making Transcendence Transparent by Burger says: then it says:

But $e$ is not algebraic how it satisfies Roth's Theorem ?

Answer 1

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.

Answer 2

Consider:

• For every algebraic number $\alpha$ we have that $q^2 \ge 0$.
• We can show that $e^2 \ge 0$.

Is there a problem? No. Likewise there is no issue with the statement you quote.

In fact, almost all real numbers satisfy the property in question. It can still be interesting to establish it for specific ones.