Show that Modulus Function $f:R \to R,$ given by $f(x) = |x|,$ is neither one-one nor onto, where $|x|$ is x, if x is positive or 0 and $|x|\;\;is\;\; - x,$ if x is negative.

$f:R \to R,\;\;given\;\;by\;\;f(x) = \;|x|$

Where $|x|\; = \left\{ \begin{array}{l}x,\;\;if\;\;x \ge 0\\ - x,\;\;if\;\;x < 0\end{array} \right.$

Injectivity

Clearly, f contains (-1, 1), (1, 1), (-2, 2), (2, 2) $\cdots$

We have, $f( - 1) = f(1) \Rightarrow \;| - 1| = |1|\; \Rightarrow 1 = 1$

But, $( - 1) \ne (1)$

Thus, negative integers are not images of any elements.
Therefore, f is not one-one.

Surjectivity
We have, for $f:R \to R,\;f(x) = \;|x|$ assumes only non-negative values. So, negative real numbers in R (co-domain) do not have their pre images in R (domain)

Therefore, f is not onto. Hence, f is neither one-one nor onto.