Arithmetic with integers and rational numbers

Proof.- (by contradiction or reductio ad absurdum)

Let's suppose that $0 \neq a \in \mathbb{Z}$ and $a = (\frac{b}{c})^2$ with $b, c \in \mathbb{Z}$ . Then this implies that $ac^2 = b^2$ . Due to Fundamental theorem of Arithmetic there exist prime integers distinct from two to two $p_1, ... , p_r$ and non- negative integers $\alpha_1, ... , \alpha_r, \beta_1, ... , \beta_r, \gamma_1, ... , \gamma_r$ such that $a = \pm p_1^{\alpha_1}\cdot \cdot \cdot p_r^{\alpha_r}, \quad b = \pm p_1^{\beta_1}\cdot \cdot \cdot p_r^{\beta_r}, \quad c = \pm p_1^{\gamma_1}\cdot \cdot \cdot p_r^{\gamma_r} \Rightarrow$ $\alpha_i + 2\gamma_i = 2\beta_i, \space \forall i=1, ... , r \Rightarrow 2(\beta_i - \gamma_i) = \alpha_i \ge 0, \space \forall i=1, ... , r \Rightarrow \beta_i \ge \gamma_i, \space \forall i=1, ... , r \Rightarrow$ $c \space | \space b, \space \text{(c divides to b)}$ and this implies that $a$ would be a perfect square, i, e, $a = x^2$ would have some integer solution $\square$ q.e.d