Suppose 0 is not in $\displaystyle \mathbb{N}$. There is a natural injection $\displaystyle \mathbb{N}\to\mathbb{Q}^+$. Conversely, $\displaystyle \left(\frac{a}{b}\right)\mapsto(a,b)$ where $\displaystyle \frac{a}{b}$ is *in lowest terms* is an injection from $\displaystyle \mathbb{Q}^+$ to $\displaystyle \mathbb{N}\times\mathbb{N}\equiv\mathbb{N}$. Therefore, by Cantor–Bernstein–Schroeder theorem, $\displaystyle \mathbb{Q}^+\equiv\mathbb{N}$.

Alternatively, in addition to the injection $\displaystyle \mathbb{N}\to\mathbb{Q}^+$, there is a surjection $\displaystyle \mathbb{N}\times\mathbb{N}\to\mathbb{Q}^+$.