Showing $r\in\mathbb Q\setminus \{0\}\implies e^r\notin \mathbb Q$

Question

For a given $n>0$, let $\displaystyle J_n:x\to \frac{1}{n!}\int_{-x}^x(x^2-t^2)^ne^tdt$

a. Prove that there exists $A_n,B_n\in \mathbb R_n[X]$ such that $\forall x\in \mathbb R^+, J_n(x)=A_n(x)e^x+B_n(x)e^{-x}$

b. Show that $r\in\mathbb Q\setminus \{0\}\implies e^r\notin \mathbb Q$

I'm able to take care of a., and I proved that one can even assume $A_n,B_n\in \mathbb Z_n[X]$ which is stronger than what is asked.

But I'm stuck with b. I'm sure one has to go for contradiction: suppose there is some $r\neq 0$ such that $e^r\in \mathbb Q$. Then $\forall n\in \mathbb N^*, J_n(r) \in \mathbb Q$. How can I proceed further ?

Answer 1

Let $r=s/t$ where $s\in\mathbb{Z}$ and $t\in\mathbb{N}$. Assume $e^r=p/q$ where $p,q\in\mathbb{N}$. Then $$ pqt^nJ_n(r)= p^2 t^n A_n\left(\frac{s}{t}\right)+ q^2 t^n B_n\left(\frac{s}{t}\right)\in\mathbb{Z} \tag{1} $$ Note that $$ 0<J_n(x)\leq\frac{x^{2n}}{n!}\int_{-x}^xe^tdt=\frac{2 x^{2n}\sinh x}{n!} $$ So $$ 0<pqt^nJ_n(r)\leq \frac{2 pq t^n r^{2n}\sinh r}{n!}\rightarrow 0 \qquad(n\to\infty) $$ This contradicts $(1)$.

Answer 2

I'm not sure what you are allowed to assume. Do you know that $e$ is transcendental and can you use that? If so consider $e^{\frac{p}{q}}=l$ where $l\in \mathbb{Q}$. What happens when you look at the polynomial $x^p-l^q$ ?

Looking at the first part of your question I'm assuming it's unlikely you are allowed to use the fact that $e $ is transcendental since it makes the question just an exercise in understanding the definition.