Exercise 1.2 · Q4: The modulus function

Basics you need

Solution

1. The two formulas define one output for each $x$

   Every real number $x$ satisfies exactly one of $x \ge 0$ and $x < 0$⁠. So exactly one of the two formulas applies, and it gives one real number: $x$ or $-x$⁠. Hence $f$ is a function from $\R$ to $\R$⁠.

   With a piecewise definition it is worth confirming that the pieces do not overlap (which could give two outputs) or leave gaps (which could give none). Here the condition "positive or 0" and the condition "negative" split $\R$ cleanly, with $0$ in the first piece.
   Evaluate a piecewise function by first choosing the piece (Basic 1).

2. Not one-one: $f(-1) = f(1)$

   Compute each value using the correct piece:

   $$\begin{aligned} -1 < 0 \ &\Longrightarrow\ f(-1) = -(-1) = 1,\\ 1 \ge 0 \ &\Longrightarrow\ f(1) = 1. \end{aligned}$$

   So $f(-1) = f(1) = 1$ although $-1 \ne 1$⁠. Hence $f$ is not one-one.

   The same happens for every $a > 0$⁠: $f(-a) = -(-a) = a = f(a)$⁠. Every positive output is shared by exactly two inputs, $a$ and $-a$⁠.

   A single pair of different inputs with equal outputs disproves one-one, because one-one is a claim about all pairs. We choose $1$ and $-1$ because they sit in different pieces and the arithmetic is easy. The general line $f(-a) = f(a)$ shows the failure is not an accident of one pair; it is built into the definition (the sign is discarded).
   One counterexample disproves a "for all" statement (Basic 3). Basics: how to prove and disprove

3. $|x|$ is never negative

   Check each piece:

   • If $x \ge 0$⁠, then $f(x) = x \ge 0$⁠.
   • If $x < 0$⁠, then $f(x) = -x > 0$⁠, by Basic 4.

   Since the two cases cover all real numbers, $f(x) \ge 0$ for every $x\in\R$⁠.

   This is the fact the onto argument needs, and it must be proved for both pieces. It would not be enough to look at the first formula $f(x) = x$⁠, which on its own can be negative; for negative $x$ that formula is not the one in use.
   Multiplying $x<0$ by $-1$ reverses the inequality: $-x>0$ (Basic 4).

4. Not onto: $-1$ is never an output

   The number $-1$ is in the codomain $\R$⁠. By Step 3, $f(x) \ge 0 > -1$ for every $x\in\R$⁠, so $f(x) \ne -1$ for every $x$⁠. No element of the domain is mapped to $-1$⁠, and $f$ is not onto.

   To disprove onto we must name a value in the codomain and show that no input gives it. Choosing a negative value makes this immediate, because Step 3 already proved that all outputs are $\ge 0$⁠. Any negative number would do; non-negative numbers would not, since every $y \ge 0$ is reached ($f(y) = y$⁠).
   Definition 6: onto means every element of the codomain is an image. Equivalently, the range must equal the codomain; here the range is $[0, \infty)$⁠, not $\R$⁠. Basics: one-one and onto

5. Conclude, and see how the sets change the answer

   $f$ is not one-one (Step 2) and not onto (Step 4), so the modulus function $f : \R\to\R$ is neither one-one nor onto.

   The verdict depends on the domain and codomain, not only on the formula $|x|$⁠:

   • With codomain $[0, \infty)$ (domain still $\R$⁠), it becomes onto: any $y \ge 0$ equals $f(y)$⁠, since $y\ge0$ uses the first piece. It is still not one-one, since $f(-1) = f(1)$⁠.
   • With domain and codomain both $[0, \infty)$⁠, only the first piece is ever used, so $f(x) = x$⁠. Then $f(x_1) = f(x_2)$ means $x_1 = x_2$ (one-one), and every $y\ge0$ equals $f(y)$ (onto). So that function is bijective.
   Removing the negative inputs removes the partner $-a$ of each $a$⁠, which repairs one-one. Shrinking the codomain to the actual range repairs onto. Seeing which change fixes which property is a good check that you understand why each property failed.

Every fact used, and where it comes from