For each function $f:\mathbb{N}\to \mathbb{N}$ we define the real number, in decimal notation $A(f)=0.f(0)f(1)f(2)f(3)\ldots $. Show that, if $f(x) =x^2$, then $A(f)=.0149162536\ldots$ is irrational.
We tried comparing this to the square root of a prime number irrationality proof that we know, but we do not have a function to undo. Similarly we tried to compare this to the proof that proves $\pi$ is irrational, but that did not help either.
The decimal $0.01491625\dots$ is neither terminating nor ultimately periodic. One way to see this is that there are arbitrarily long strings of $0$'s in the decimal expansion, because it contains as substrings the squares of $10,100,1000,\dots$.
Any rational number has a terminating decimal expansion, or an ultimately periodic one. So our number is irrational.