Q.Show that the Signum Function $f: \mathbf{R} \rightarrow \mathbf{R}$, given by $f(x) = \begin{cases} 1, \text{ if } x > 0 \\ 0, \text{ if } x = 0 \\ -1, \text{ if } x < 0 \end{cases}$ is neither one-one nor onto.

Concept: One-One and Onto

A function is one-one if distinct inputs give distinct outputs, and onto if every element in the codomain has a preimage.

Step 1 — Not one-one:

Take $x = 1$ and $x = 2$. Both are $>0$, so $f(1) = 1$ and $f(2) = 1$. Two different inputs map to the same output, so $f$ is not injective.

Step 2 — Not onto: …

The signum function is a classic example of a function that fails both injectivity and surjectivity. Let’s see why, step by step.

1. What does “one-one” (injective) mean?

A function $f: A \to B$ is one-one if different inputs always give different outputs. Equivalently: if $f(x_1) = f(x_2)$, then we must have $x_1 = x_2$.

Now look at the signum function. Take any two positive numbers, say $x = 2$ and $x = 5$.

$f(2) = 1$ and $f(5) = 1$. So $f(2) = f(5)$ but $2 \neq 5$.

That’s a direct violation of the one-one condition.

Thus, $f$ is not one-one.

2. What does “onto” (surjective) mean?

A function $f: A \to B$ is onto if every element of the codomain $B$ is actually an output of the function — i.e., for every $y \in B$, there exists some $x \in A$ such that $f(x) = y$.

Here, the codomain is $\mathbf{R}$, the set of all real numbers. But the signum function only ever outputs three values: $-1$, $0$, and $1$.

Pick any number not in $\{-1, 0, 1\}$, say $y = 5$ or $y = -0.5$ or $y = \pi$. There is no real $x$ for which $f(x)$ equals that number. So the function misses almost all real numbers. …

Method: Showing a finite-range function is neither one-one nor onto

When a function on an infinite domain takes only a few values, both properties fail — many inputs must repeat outputs, and most codomain values are never reached.

Steps

Step 1: Not one-one — because infinitely many inputs map to a small set of outputs, pick two distinct inputs giving the same value, e.g. two positive numbers both mapping to $1$. …

Common Mistakes

Mistake 1: Thinking three distinct outputs ($-1,0,1$) make the signum function one-one.

Why it's wrong: one-one is about outputs not repeating, but every positive input maps to $1$ and every negative to $-1$, so outputs repeat infinitely. Correct approach: give two positive inputs, e.g. $f(1)=f(2)=1$.

Mistake 2: Treating the small range as if it were the codomain. …