Why is cos(n/6) aperiodic?

Question

This is a very common example in most Signal Processing books I have come across.

x(n) = cos($\frac{n}{6}$) is a non-periodic discrete signal because it doesn't satisfy the periodicity condition for discrete time signals i.e, it is not of the form 2$\pi$($\frac{m}{N}$).

My question is :

the coefficient of n, i.e, $\Omega_0$=$\frac{1}{6}$ here can also be expressed as $\frac{1}{6}$ = $\frac{1}{6}$ * $\frac{2\pi}{2\pi}$ = 2$\pi$$\frac{1}{12\pi}$

Now, substituting for $\pi$ = $\frac{22}{7}$ in above, we get 2$\pi$$\frac{7}{12*22}$. So, $\frac{1}{6}$ can be written as 2$\pi$($\frac{7}{264}$), which is in the form 2$\pi$($\frac{m}{N}$) with a period N=264.

I'm sure I'm missing something which may be obvious but it would be of great help if someone could point it out and explain.

Answer 1

The problem with your reasoning is that $\pi \ne \frac{22}{7}$; $\pi$ is an irrational number. There is no period $N$ for which $x[n] = x[n+N] \ \forall \ n \in \mathbb{Z}$. Hence, the sequence is not periodic.

Answer 2

The periodicity of a signal holds if we can show $x(n)=x(n+N)$, otherwise, the signal is nonperiodic. Simply start with $$ \begin{align} x(n+N) &= \cos( \frac{n}{6} + \frac{N}{6}) \\ &= \cos(\frac{n}{6})\cos(\frac{N}{6}) - \sin(\frac{n}{6})\sin(\frac{N}{6}) \end{align} $$ In order for $x(n)=x(n+N)$ to hold, $\cos(\frac{N}{6})=1$ and $\sin(\frac{N}{6})=0$. We search for the smallest value of $N$. This is true if $\frac{N}{6}=2\pi \implies N = 12\pi$. Indeed, if $N=12\pi$, we get $$ \begin{align} x(n+N) &= \cos( \frac{n}{6} + \frac{N}{6}) \\ &= \cos(\frac{n}{6})(1) - \sin(\frac{n}{6})(0) \\ &= \cos(\frac{n}{6}) \\ &= x(n) \end{align} $$ But $N$ must be a positive integer, therefore, the signal is nonperiodic.