The number is not rational.
For any fixed real number , and any natural number , let where is the factorial of , is the definite_integral, and is the sin_function.
Exercise: .
Now, so is an integer.
Also which by a simple calculation is .
Expand the integrand and then integrate by parts repeatedly:
The first integral term is
The second integral term is which is which is
Therefore
Therefore, if and are integers, then so is inductively, because is an integer and is an integer.
But also as , because is in modulus at most and hence
For larger than , this expression is getting smaller with , and moreover it gets smaller faster and faster as increases; so its limit is .
We claim that as , for any .
Indeed, we have which, for , is less than . Therefore the ratio between successive terms is less than for sufficiently large , and so the sequence must shrink at least geometrically to .
Suppose (for contradiction) that is rational; then it is for some integers .
Now is an integer (indeed, it is ), and is certainly an integer, so by what we showed above, is an integer for all .
But as , so there is some for which for all ; hence for all sufficiently large , is . We already know that and , neither of which is ; so let be the first integer such that for all , and we can already note that .
Then whence or or .
Certainly because is the denominator of a fraction; and by whatever definition of we care to use. But also is not because then would be an integer such that for all , and that contradicts the definition of as the least such integer.
We have obtained the required contradiction; so it must be the case that is irrational.