Solution: Let $f$ be a continuous function satisfying the property that for any $x,y \in \mathbb{R} $, \begin{equation}\label{eq:20Nov2023-1} x - y \in \mathbb{Q} \implies f(x) - f(y) \in \mathbb{Q} \tag{$\mathcal{P}$} \end{equation} For any rational $q$, we define the function \[ g_q(x) \coloneqq f(x + q) - f(x). \] Since $f$ is continuous and satisfying the property \eqref{eq:20Nov2023-1}, the function $g$ will also be continuous and takes only rational values. This implies, the function $g$ will be constant. Now we set \begin{gather*} f(1) - f(0) = g_1(0) = a \\ f(0) = b \\ f\left( \frac{1}{n} \right) - f(0) = g_{\frac{1}{n}}(0) = r,\ n \in \mathbb{N} . \end{gather*}

For any $n\in \mathbb{N} $, since \[ f\left( x + \frac{1}{n} \right) - f(0) = g_{\frac{1}{n}}(x) = g_{\frac{1}{n}}(0) = r, \] so for any $m \in \mathbb{N} $ \begin{align*} f\left( \frac{m}{n} \right) - f(0) & = \left[ f\left( \frac{1}{n} \right) - f(0) \right] + \left[ f\left( \frac{2}{n} \right) - f\left( \frac{1}{n} \right) \right] \\ & \ + \dots + \left[ f\left( \frac{m}{n} \right) - f\left( \frac{m-1}{n} \right) \right] \\ & = g_{\frac{1}{n}}(0) + g_{\frac{1}{n}}\left( \frac{1}{n} \right) + \dots + g_{\frac{1}{n}}\left( \frac{m-1}{n} \right) \\ & = m r. \end{align*}

Similarly, \[ f\left( -\frac{m}{n} \right) - f(0) = -mr. \] To determine the function, we need to find the value of $r$. If $m = n$, then \[ nr = f\left( \frac{n}{n} \right) - f(0) = f(1) - f(0) = a \implies r = \frac{a}{n}. \] Therefore, \[ f\left( \frac{m}{n} \right) - f(0) = mr = m\cdot \frac{a}{n} = a \cdot \frac{m}{n}. \] Combining with the assumption that $f(0) = b$, we proved that for any rational $x$, \[ f(x) = ax + b. \] Since, $f$ is continuous and rationals are dense in reals, for any real number $x$ we have \[ f(x) = ax + b. \]