Example 11: $f(x)=x^2$ on $\R$⁠: neither

Basics you need

Solution

1. Not one-one: $-1$ and $1$ have the same square

   Compute both images:

   $$f(-1) = (-1)^2 = 1, \qquad f(1) = 1^2 = 1.$$

   So $f(-1) = f(1)$⁠, although $-1 \ne 1$⁠. Two different inputs have the same image, so $f$ is not one-one.

   One-one says every pair with equal images consists of equal inputs. The pair $-1, 1$ breaks this, and one such pair is a complete disproof. The choice of $\pm 1$ is not special: for every $a \ne 0$⁠, $f(-a) = a^2 = f(a)$ with $-a \ne a$⁠, so every positive output is in fact hit twice.
   $(-a)^2 = (-a)(-a) = a^2$⁠: a negative times a negative is positive. That identity is the reason squaring cannot be one-one on a set containing both $a$ and $-a$⁠.

2. Why a square is never negative

   Take any real $x$⁠. There are three cases.

   • If $x > 0$⁠: $x^2 = x \cdot x$ is a product of two positive numbers, so $x^2 > 0$⁠.
   • If $x = 0$⁠: $x^2 = 0$⁠.
   • If $x < 0$⁠: $x^2 = x \cdot x$ is a product of two negative numbers, so $x^2 > 0$⁠.

   Every real number falls in exactly one case, so

   $$x^2 \ge 0 \quad \text{for every } x \in \R.$$
   This is the answer to the book's "(Why?)" in the next step. It is a "for every $x$" statement, so we prove it for an arbitrary $x$⁠, splitting into cases that cover all real numbers.
   Proof by cases: the three cases $x > 0$⁠, $x = 0$⁠, $x < 0$ cover every real number, so proving the claim in each covers them all.

3. Not onto: $-2$ is never a square

   The codomain is $\R$⁠, so $-2$ is an element of it. Suppose some real $x$ had $f(x) = -2$⁠, that is $x^2 = -2$⁠. By Step 2, $x^2 \ge 0$⁠, while $-2 < 0$⁠. A number cannot be both $\ge 0$ and $< 0$⁠, so there is no such $x$⁠. Hence $-2$ is not the image of any element of the domain, and $f$ is not onto.

   To show an element has no preimage, failing to find one is not enough; we must show that none can exist. Assuming a preimage exists and reaching a contradiction (a number that is both $\ge 0$ and negative) does exactly that.
   On a graph, $y_0$ has a preimage exactly when the horizontal line $y = y_0$ meets the graph. Two meeting points mean "not one-one"; none means "not onto". Basics: one-one and onto

4. The range is $[0, \infty)$

   Step 2 shows every output is $\ge 0$⁠. Conversely every $y \ge 0$ is reached: $\sqrt{y}$ is a real number and $f(\sqrt{y}) = (\sqrt{y})^2 = y$⁠. So

   $$\begin{aligned} \text{Range of } f &= \{y \in \R : y \ge 0\} \\ &= [0, \infty) \ne \R. \end{aligned}$$

   The range is not the whole codomain, which again shows (by the Remark after Definition 6) that $f$ is not onto. Every negative number is missed, not just $-2$⁠.

   Knowing the exact range explains how to repair the function. If we shrink the codomain to $[0, \infty)$⁠, $f$ becomes onto; if we also shrink the domain to $[0, \infty)$⁠, it becomes one-one too, and then $\sqrt{y}$ is its inverse.
   $f$ is onto exactly when its range equals its codomain.

5. Conclude

   $f$ is not one-one (Step 1) and not onto (Step 3). So $f : \R \to \R,\ f(x) = x^2$ is neither one-one nor onto.

   Each failure was established with its own counterexample, so both halves of "neither … nor" are proved. Compare Exercise 1.2 Q2, where the same formula $x^2$ is tested on $\N$ and $\Z$⁠: on $\N$ there are no negative inputs, so squaring becomes one-one there.

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