I was using Wolfram Alpha for things, and I came across $I_{0}(2)$. For fun I tried asking Wolfram Alpha if the number was irrational, but said it's unknown. I believe this is an error, as its irrationality should already be confirmed, as by generalizing the Fourier's proof of the irrationality of $e$, you can prove that a whole class of numbers (in which $I_{0}(2)$ appear) is irrational, specifically

$A(n) = {}_{0}F_{n}(;\overbrace{1,1,...,1}^{n};1) = \sum\limits_{k=0}^{\infty}\frac{1}{k!^{n+1}}$

is irrational, where $F$ is the Generalized Hypergeometric Function. In particular,

$A(0) = e$

$A(1) = I_{0}(2)$

For completeness, I'll add the generalized version of Fourier's proof:

If $A(n)$ is a rational number, then there exist $a_n,b_n \in \mathbb{N}^+$ such that $A(n) = \frac{a_n}{b_n}$.

Let's define the number

$x_{n} = b_{n}!^{n+1}\left(A(n) - \sum\limits_{k=0}^{b_{n}}\frac{1}{k!^{n+1}}\right)$

Under the previous assumption, we have

$x_{n} = b_{n}!^{n+1}\left(\frac{a_n}{b_n} - \sum\limits_{k=0}^{b_{n}}\frac{1}{k!^{n+1}}\right) = a_n(b_n-1)!b_{n}!^{n} - \sum\limits_{k=0}^{b_{n}}\frac{b_{n}!^{n+1}}{k!^{n+1}}$

$x_{n} \in \mathbb{Z}$ as the first term of the previous expression is obviously an integer, and every term of the series is an integer too, as $k \leq b_{n}$ for each term.

Let's prove that $x_{n} > 0$ by inserting the corresponding series in place of $\frac{a_n}{b_n}$:

$x_{n} = b_{n}!^{n+1}\left(\sum\limits_{k=0}^{\infty}\frac{1}{k!^{n+1}} - \sum\limits_{k=0}^{b_{n}}\frac{1}{k!^{n+1}}\right) = \sum\limits_{k=b_{n}+1}^{\infty}\frac{b_{n}!^{n+1}}{k!^{n+1}}$

Each term of the series is strictly positive.

Let's prove that $x_{n} < 1$, by first showing that, for all terms $k \geq b_{n}+1$, we have the upper estimate

$\frac{b_{n}!^{n+1}}{k!^{n+1}} = \left(\frac{b_{n}!}{k!}\right)^{n+1} \leq \left(\frac{1}{(b_{n}+1)^{k-b_{n}}}\right)^{n+1}$

By changing the index of summation and using the formula for infinite geometric series, we have

$\sum\limits_{k=b_{n}+1}^{\infty}\frac{b_{n}!^{n+1}}{k!^{n+1}} = \sum\limits_{k=b_{n}+1}^{\infty}\left(\frac{b_{n}!}{k!}\right)^{n+1} < \sum\limits_{k=b_{n}+1}^{\infty}\left(\frac{1}{(b_{n}+1)^{k-b_{n}}}\right)^{n+1} = \sum\limits_{j=1}^{\infty}\left(\frac{1}{(b_{n}+1)^{j}}\right)^{n+1} = \frac{1}{(b_{n}+1)^{n+1}-1}$

which is $\leq 1$. Because $x_n \in \mathbb{Z}$ and $0 < x_n < 1$, we have reached a contradiction, so $A(n)$ is irrational.

Now, I don't know which criterion Wolfram Alpha uses to test the irrationality of a number, but I'm pretty sure my proof is correct, so I guess this is just an error on their front?