Show that the function $f: \mathbf{R} \rightarrow \mathbf{R}$, defined as $f(x)=x^{2}$, is neither one-one nor onto.

Step 1: Check for One-One (Injectivity)

To check if a function is one-one, we assume $f(x_1) = f(x_2)$ and try to prove that $x_1 = x_2$. For the given function $f(x) = x^2$, if $f(x_1) = f(x_2)$, then $x_1^2 = x_2^2$. This implies that $x_1 = x_2$ or $x_1 = -x_2$.

Step 2: Provide a Counterexample for One-One

Since $x_1$ can be equal to $-x_2$, the function is not one-one. For example, $f(1) = 1$ and $f(-1) = 1$. Here, two different elements in the domain (1 and -1) map to the same element in the codomain (1).

Step 3: Check for Onto (Surjectivity)

To check if a function is onto, we need to show that for every element $y$ in the codomain, there exists an element $x$ in the domain such that $f(x) = y$. For $f(x) = x^2$, we set $x^2 = y$, which means $x = \pm \sqrt{y}$.

Step 4: Provide a Counterexample for Onto

The domain and codomain are both the set of real numbers $\mathbf{R}$. If we choose a negative number from the codomain, for example, $y = -1$, then $x = \pm \sqrt{-1}$. This value of $x$ is not a real number. Therefore, there is no real number $x$ such that $f(x) = -1$. This means not every element in the codomain has a pre-image in the domain, so the function is not onto.

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