amparvardi wrote: Let $f(x) = x^n$ where $n$ is a fixed positive integer and $x =1, 2, \cdots .$ Is the decimal expansion $a = 0.f (1)f(2)f(3) . . .$ rational for any value of $n$ ? The decimal expansion of a is defined as follows: If $f(x) = d_1(x)d_2(x) \cdots d_{r(x)}(x)$ is the decimal expansion of $f(x)$, then $a = 0.1d_1(2)d_2(2) \cdots d_{r(2)}(2)d_1(3) . . . d_{r(3)}(3)d_1(4) \cdots .$ Hi, $a$ is never rational. Indeed, if $k \geq 1$, then $f(10^k)$ is $1$ one followed by $kn$ zeros. Hence, in the decimal expansion of $a$, you have infinitely many ones, and sequences of consecutive zeros of arbitrarily large length. This prevents $a$ from being rational.

Official Solution : No. If $a$ were rational, its decimal expansion would be periodic from some point. Let $p$ be the number of decimals in the period. Since $f(10^{2p})$ has $2np$ zeros, it contains a full periodic part; hence the period would consist only of zeros, which is impossible.