Q.Show that the Modulus Function $f: \mathbf{R} \rightarrow \mathbf{R}$, 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.

Concept: Relation Properties (One-One and Onto Functions)

Step 1 – One-one (injective) check

For $f(x) = |x|$, two different inputs can give the same output.

Example: $f(2) = 2$ and $f(-2) = 2$, so $2 \neq -2$ but $f(2) = f(-2)$.

Hence $f$ is not one-one.

Step 2 – Onto (surjective) check

The codomain is $\mathbf{R}$, but the range of $f$ is only $[0, \infty)$ (all non-negative reals). …

We need to check two properties: injectivity (one‑one) and surjectivity (onto).

A function is one‑one if different inputs always give different outputs.

A function is onto if every possible output in the codomain is actually reached by some input.

The modulus function takes any real number and returns its non‑negative distance from zero. That single fact is the key to both failures.

1. Checking one‑one (injectivity) For $f$ to be one‑one, $f(a) = f(b)$ must imply $a = b$. Take $a = 2$ and $b = -2$. Then

So $f(2) = f(-2)$ but $2 \neq -2$.

This is a direct counterexample: two different numbers give the same output.

Hence $f$ is not one‑one.

2. Checking onto (surjectivity) …

Method: Disproving one-one and onto for an even / absolute-value function

An absolute-value (or even) function collapses opposite inputs together and never produces negative outputs, which breaks both properties — disprove each with a single counterexample.

Steps

Step 1: Break one-one using the symmetry $f(-a)=f(a)$

Pick any $a\neq 0$; then $|{-a}|=|a|$ gives two distinct inputs with equal output, e.g. $|{-2}|=|2|=2$, so $f$ is not injective.

Step 2: Break onto using the sign restriction …

Common Mistakes

Mistake 1: Claiming $|x|$ could be one-one on all of $\mathbf{R}$.

Why it's wrong: the symmetry $|{-a}|=|a|$ pairs every non-zero input with its negative, e.g. $|{-2}|=|2|=2$. Correct approach: give one such pair to disprove injectivity.

Mistake 2: Overlooking that $|x|$ never outputs a negative number. …