Let x_c(t) be any continuous-time periodic signal with period T. It is sampled uniformly with a sampling period Ts where T_s ̸ = T, resulting in the discrete se quence x[n] = x_c(nT_s), where n is an integer. Which one of the following statements is correct about x[n]?

The correct option is: x[n] will be periodic if and only if T/T_s is a rational number.

To understand why this is correct, let us analyze the condition for periodicity of a discrete-time signal.
A discrete-time signal $x\left[n\right]$ is periodic if there exists a positive integer $N$ such that for all integers $n$:

$x[n+N]=x\left[n\right]$

Given that the continuous-time signal $x_c\left(t\right)$ is periodic with a fundamental period $T$, we have:

$x_c(t+mT)=x_c\left(t\right)$

for any integer $m$.
The discrete sequence is obtained by uniform sampling with sampling period $T_s$:

$x\left[n\right]=x_c\left(n{T}_s\right)$

Applying the periodicity condition to the discrete sequence $x\left[n\right]$:

$x[n+N]=x_c\left(\right(n+N\left)T_s\right)=x_c(n{T}_s+N{T}_s)$

For this to equal $x\left[n\right]=x_c\left(n{T}_s\right)$, the time shift $N{T}_s$ must be an integer multiple of the continuous-time period $T$. That is, there must exist an integer $k$ such that:

$N{T}_s=kT$

Rearranging this relationship, we get:

$\frac{T}{T_s}=\frac{N}{k}$

Since both $N$ (the discrete period) and $k$ are integers, their ratio $\frac{N}{k}$ is by definition a rational number.
Therefore, the discrete-time sequence $x\left[n\right]$ is periodic if and only if the ratio of the continuous-time period $T$ to the sampling period $T_s$ is a rational number. If $\frac{T}{T_s}$ is irrational, the samples never repeat at the exact same phases of the continuous signal, making $x\left[n\right]$ non-periodic.