# Prove the irrationality of $e=\exp(1)$ with $\exp$ defined as solution of ODE

I am writing math lessons as I will teach soon on my freetime. In my course, I defined $\exp(x)$ as the solution to the O.D.E : $$f'(x)=f(x),\qquad f(0)=1$$ Is there a way to prove, maybe by contradiction, that $e$ is irrational without using series?

• You can always prove that the solution is $f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$, and then prove $f(1)$ is irrational. You usually prove irrationality of $e$ by an infinite sum, so it's the most straight up approach. – Jakobian Jun 17 '18 at 16:24
• Seems to me you could just use the normal proof. Just avoid defining $e$ as a sum and use Taylor's theorem centered at $0$ extrapolated to $1$ directly. – Robert Wolfe Jun 17 '18 at 16:56

I don't think there is; I think you'll have to prove $e=\sum_{k\ge 0}\frac{1}{k!}$ first. However, there's an easy way to do that using the ODE. We make repeated use of $f(1)=1+\int_0^1 f(t) dt$: $$f(1)=1+\int_0^1 (1+\int_0^t f(t') dt')dt+\cdots$$So $$f(1)=1+\int_0^1 dt+\int_0^1 dt\int_0^t dt' t'+\cdots=\sum_{k\ge 0}\frac{1}{k!}.$$
• @T.D. Here's the simplest way to understand it. The $k$th term after the $1$ is a $k$-tuple integral with ordered integration variables $\in [0,\,1]$, so a random point in the unit hypercube has a $1/k!$ probability of "counting". So instead of the integrand, which is equal to the constant $1$, integrating to the hypercube's volume $1$, it's only $1/k!$. – J.G. Aug 11 '18 at 12:20