I was surprised that the numbers $\pi$, $\ln{(2)}$, $\zeta{(2)}$, and $\zeta{(3)}$ can be shown to be irrational in what seems to be "three-lined proofs" (as identified here on Overflow: Establishing zeta(3) as a definite integral and its computation.).
The idea goes that we suppose, for $j \in \mathbb{N}$, we have a family of non-zero integrals such that $$\int_0^1 x^j f(x)\ dx = P_j \xi+ Q_j $$ where $P_j, Q_j \in \mathbb{Q}$, $f(x)$ is arbitrary, and $\xi$ is some number. If $\xi$ is rational, then $\int_0^1 x^j f(x)\ dx = A_j/B_j $ yields a rational expression. Multipying $a_{nj} \in \mathbb{Z}$ to both sides, and taking the sum from $j = 0$ to $n$, we have
$$\sum_{j=0}^n a_{nj} \int_0^1 x^j f(x)\ dx = \sum_{j=0}^n a_{nj} \frac{A_j}{B_j} = \frac{E_n}{F_n} $$ where $E_n, F_n \in \mathbb{Z}$.
Beuker treated the $a_{nj}$ coefficients as the coefficients of the shifted Legendre polynomials in his proof for the irrationality of $\zeta{(3)}$, where
$$ P_n{(x)} = \frac{1}{n!} \frac{d^n}{dx^n} (x^n (1-x)^n) = \sum_{j=0}^n a_{nj} \, x^j$$
This leads to
$$\int_0^1 \bigg( \sum_{j=0}^n a_{nj} \, x^j \bigg) f(x)\ dx = \int_0^1 P_n{(x)}\,f(x) \ dx = \frac{E_n}{F_n}$$
If we can show, as $n$ gets larger,
$$ \left| F_n \int_0^1 \frac{1}{n!} (x^n (1-x)^n)\, \frac{d^n}{dx^n}f(x) \ dx \right| = |E_n|\to 0 $$
we get a contradiction, since at "some point", $|E_n|$ is an integer between $0$ and $1$. Thus, $\xi$ must be irrational. Of course, the hard part is finding a suitable function $f(x)$.
Since $f(x)$ has been found for the proofs of the above numbers, has it been shown the number $e$ can be proved irrational using the above method (implying has such an $f(x)$ been found)?
Best Answer
One has $$\int_0^1 x^k e^x dx = A_ke+B_k \to 0$$ as $k\to \infty$, $A_k,B_k\in \mathbb{Z}$, by integration-by-parts and induction. If $e=a/b$, then $(A_k a+B_kb)/b \to 0$, so $A_ke+B_k=0$ for $k$ large, a contradiction.