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

Concept: Onto Function Definition — A function $f: A \to B$ is onto (surjective) if every element in the codomain $B$ has at least one pre-image in the domain $A$.

Reasoning:

1. Not one-one: For $f(x) = x^2$, we have $f(1) = 1$ and $f(-1) = 1$, so two distinct inputs give the same output. Hence, it is not injective. …

Why this approach works

The question asks us to check two properties: injectivity (one-one) and surjectivity (onto). For a function to be one-one, each output must come from exactly one input. For it to be onto, every possible output in the codomain must actually be produced by some input. The square function is the classic counterexample for both — it's symmetric (so not one-one) and never negative (so not onto). Let's verify each property step by step.

1. Checking one-one (injectivity)

A function $f: A \to B$ is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2$ in the domain.

Take $x_1 = 2$ and $x_2 = -2$. Then:

So $f(2) = f(-2)$ but $2 \neq -2$. This directly violates the definition.

Thus $f$ is not one-one.

2. Checking onto (surjectivity)

A function $f: A \to B$ is onto if for every $y \in B$, there exists some $x \in A$ such that $f(x) = y$.

Here the codomain is $\mathbb{R}$, the set of all real numbers. But $f(x) = x^2$ is always non-negative:

So any negative number, say $y = -1$, has no preimage. There is no real $x$ such that $x^2 = -1$. …

Method: Disproving one-one and onto with counterexamples

Use this when you expect a function to fail both properties and must show it cleanly.

Steps

Step 1: Disprove one-one with two inputs sharing an output

Find $x_1\ne x_2$ with $f(x_1)=f(x_2)$. For $f(x)=x^2$, take $x_1=2,\ x_2=-2$: both give $4$. One such pair is enough.

Step 2: Disprove onto with an unreachable target …

Common Mistakes

Mistake 1: Believing $x^2=y^2$ forces $x=y$

Why it's wrong: it only gives $x=\pm y$, so $f(2)=f(-2)=4$ and $f$ is not one-one. Correct approach: use the pair $2,-2$ as a counterexample to injectivity.

Mistake 2: Thinking $x^2$ is onto $\mathbb{R}$ because it "hits many values" …