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

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 ?

• Yes, you definitely need that $A_n,B_n\in\mathbb Q_n[X]$. I assume $\mathbb Q_n[X]$ means polynomials of degree at most $n$? – Thomas Andrews May 23 '15 at 17:17
• @ThomasAndrews yes, indeed. – Gabriel Romon May 23 '15 at 17:46
• This one is nice. Normally the proofs start by constructing polynomials $A_{n}, B_{n}$ and then reaching an integral like $J_{n}$ (see paramanands.blogspot.com/2015/08/irrationality-of-exp-x.html), but you have done the reverse. Nice idea (although the integral looks to come out of nowhere). I will put a link to this question in my blog post. +1 – Paramanand Singh Aug 23 '15 at 4:23

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 $(1)$.
• Nicely done. I was aware of $(1)$, but what made you think that bounding $J_n$ would yield the contradiction? More importantly, do you know how the guy who created this proof came up with this very function $J_n$ ? – Gabriel Romon May 24 '15 at 7:55
• This is a standard trick it is used to prove irrationality of $\pi$, $\tan r$, $e^r$. One just needs to find a suitable integral. I learned that trick from this book – Norbert May 24 '15 at 11:27
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.