DT complex exponential periodic if rational

A discrete-time signal is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$.

For $x[n] = e^{j\omega_0 n}$:

$$\begin{aligned} x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N} \end{aligned}$$

For $x[n+N] = x[n]$ we need

$$e^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer}$$ $$\frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number}$$

So $e^{j\omega_0 n}$ is periodic only if $\omega_0/2\pi$ is a rational number. The fundamental period is the smallest positive integer $N = 2\pi m/\omega_0$, taking $m$ as the smallest integer that makes $N$ an integer (with $m$ and $N$ having no common factor). The same condition holds for $\cos(\omega_0 n+\theta)$ and $\sin(\omega_0 n+\theta)$, because they are sums of such exponentials.

This differs from continuous time, where $e^{j\omega_0 t}$ is periodic for every $\omega_0 \neq 0$ (period $2\pi/|\omega_0|$). In discrete time $n$ takes only integer values, so $N$ must be an integer.

Conversely, if $\omega_0/2\pi = m/N$ is rational, then $e^{j\omega_0 (n+N)} = e^{j\omega_0 n}e^{j2\pi m} = e^{j\omega_0 n}$, so the signal is periodic. Hence the DT complex exponential is periodic if and only if its frequency (in cycles/sample, $f_0 = \omega_0/2\pi$) is rational. Hence proved.

Example: $e^{j(2\pi/5)n}$: $f_0 = 1/5$, periodic with $N = 5$. $e^{jn}$: $f_0 = 1/2\pi$, irrational, not periodic.