1. **Problem:** Examine the injectivity and surjectivity of the function $f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x^2$. Determine if $f$ is bijective. 2. **Injectivity:** A function is injective if $f(a) = f(b)$ implies $a = b$. 3. For $f(x) = x^2$, suppose $f(a) = f(b)$, then $a^2 = b^2$. 4. Since $a,b \in \mathbb{N}$ (natural numbers, usually positive integers), $a^2 = b^2$ implies $a = b$ because natural numbers are non-negative and squaring is strictly increasing on $\mathbb{N}$. 5. Therefore, $f$ is injective. 6. **Surjectivity:** A function is surjective if for every $y \in \mathbb{N}$, there exists $x \in \mathbb{N}$ such that $f(x) = y$. 7. Here, $f(x) = x^2$. For $y$ to be in the image, $y$ must be a perfect square. 8. Not all natural numbers are perfect squares (e.g., 2, 3, 5 are not squares). 9. Hence, $f$ is not surjective onto $\mathbb{N}$. 10. **Bijectivity:** A function is bijective if it is both injective and surjective. 11. Since $f$ is injective but not surjective, $f$ is not bijective. **Final answer:** The function $f(x) = x^2$ from $\mathbb{N}$ to $\mathbb{N}$ is injective but not surjective, so it is not bijective.