1. The problem asks whether the function $f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x^2$ is one-one (injective). 2. A function is one-one if for every pair of distinct inputs $x_1$ and $x_2$, the outputs are distinct, i.e., if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. 3. Consider $f(x_1) = f(x_2)$, which means: $$x_1^2 = x_2^2$$ 4. Taking square roots on both sides, we get: $$x_1 = x_2 \quad \text{or} \quad x_1 = -x_2$$ 5. Since the domain is $\mathbb{N}$ (natural numbers), which are positive integers, the case $x_1 = -x_2$ is not possible. 6. Therefore, $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in \mathbb{N}$. 7. Hence, the function $f(x) = x^2$ is one-one (injective) when defined on natural numbers. **Final answer:** $f$ is one-one on $\mathbb{N}$.