Show that the modulus function $f : R \rightarrow R$ given by $f(x) = |x|$ is neither one-one nor onto,where $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

(N/A) The function $f: R \rightarrow R$ is defined as $f(x) = |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$.
To check if $f$ is one-one:
Consider $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function $f$ is not one-one.
To check if $f$ is onto:
Consider the codomain $R$. For any negative value,such as $-1 \in R$,there exists no $x \in R$ such that $f(x) = |x| = -1$,because the absolute value of any real number is always non-negative $(|x| \ge 0)$.
Therefore,$f$ is not onto.
Hence,the modulus function is neither one-one nor onto.