Show that the modulus function `f: R->R` , given by `f(x)=|x|` is neither one-one nor onto.

Explanation

It is seen that
`f(−1)=∣−1∣=1,f(1)=∣1∣=1`
`therefore``f(−1)=f(1)`
, but `−1!=1`

`therefore`f is not one-one

Now, consider `−1inR`

but `f(x)=∣x∣` is always non-negative.

Thus, there does not exist any element x in domain R such that
`f(x)=∣x∣=−1 `

`therefore f` is not onto.

Hence, the modulus function is neither one-one nor onto.