Exercise 1.2 · Q5: The signum function

Basics you need

Solution

1. Check that $f$ is a genuine function on $\R$

   By trichotomy (Basic 2), each real $x$ is positive, zero or negative, and only one of these. So exactly one of the three formulas applies to each $x$⁠, and it gives exactly one value. A few sample values:

   $$f(5) = 1, \qquad f(0.001) = 1, \qquad f(0) = 0, \qquad f(-0.001) = -1, \qquad f(-7) = -1.$$
   A piecewise rule is a function only if every input falls in exactly one piece. If the pieces overlapped (say "$x \ge 0$" and "$x = 0$"), the input 0 would get two different values; if there were a gap, some input would get none. Trichotomy rules out both problems. Notice also that the size of $x$ never matters, only its sign: $f(0.001)$ and $f(5)$ are both 1.
   A function from $X$ to $Y$ assigns to every element of $X$ exactly one element of $Y$⁠. Basics: what a function is

2. Not one-one: two different inputs, one output

   Both 1 and 2 are positive, so the first piece applies to both:

   $$f(1) = 1 \qquad\text{and}\qquad f(2) = 1.$$

   So $f(1) = f(2)$ although $1 \ne 2$⁠. Two different inputs share an output, so $f$ is not one-one. (It is many-one, in the textbook's word from Definition 5.)

   "One-one" says: for every pair of inputs, equal outputs force equal inputs. A single pair that breaks this destroys the claim, so the pair 1, 2 is a complete proof. Any two positive numbers would do just as well, and so would any two negative numbers, such as $f(-1) = f(-3) = -1$⁠. In fact every output except 0 is shared by infinitely many inputs.
   To disprove a "for all" statement, one counterexample is enough. Basics: proving and disproving

3. Find the range by checking every case

   Take any $x\in\R$⁠. By trichotomy there are exactly three cases:

   • If $x > 0$⁠, then $f(x) = 1$⁠.
   • If $x = 0$⁠, then $f(x) = 0$⁠.
   • If $x < 0$⁠, then $f(x) = -1$⁠.

   So every output is one of $1, 0, -1$⁠, and each of the three really occurs (for example $f(1)=1$⁠, $f(0)=0$⁠, $f(-1)=-1$⁠). Hence

   $$\text{Range of } f = \{-1,\ 0,\ 1\}.$$
   To show that $f$ misses some value, we need to know exactly which values it does take. Because the cases cover all real numbers, this list of three values is complete: no input can produce anything else. Without the case check we would only have examples, and examples cannot prove that a value is never reached.
   Proof by cases: if every possibility falls into one of the listed cases, a conclusion checked in each case holds for all $x$⁠. Basics: proving and disproving

4. Not onto: the value 2 is never reached

   The codomain is $\R$⁠, and $2\in\R$⁠. Is there an $x\in\R$ with $f(x) = 2$⁠? By Step 3, $f(x)$ is always $1$⁠, $0$ or $-1$⁠, and none of these equals 2. So no real $x$ satisfies $f(x) = 2$⁠: the element 2 of the codomain is not the image of any element of the domain. Hence $f$ is not onto.

   The same argument works for every real number other than $-1, 0, 1$⁠, for example $\tfrac12$⁠, $-5$ or $\pi$⁠. The range $\{-1,0,1\}$ is a tiny part of the codomain $\R$⁠.

   "Onto" says every element of the codomain is hit. To disprove it we pick one element and show it is never hit. Showing "never" needs a general reason, not trial and error, and Step 3 supplies it: the full list of possible outputs does not contain 2. This is also the textbook's Remark after Definition 6 in action: $f$ is onto exactly when its range equals the codomain, and here $\{-1,0,1\} \ne \R$⁠.
   Onto means range = codomain (Remark after Definition 6). To disprove onto, exhibit one codomain element with no preimage. Basics: one-one and onto

5. Conclude

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

   The two failures are independent and each needs its own reason: one is about inputs colliding, the other about outputs being missed. Note that changing the codomain to $\{-1,0,1\}$ would make $f$ onto, but it would still not be one-one, because 1 and 2 would still collide.

Every fact used, and where it comes from