discrete signals - Why is cos(n/6) aperiodic?

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

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.