Exercise 1.2 · Q2: Injectivity and surjectivity of $x^2$ and $x^3$

Basics you need

Solution

1. Two factorisations used in every part

   For all real numbers $x_1, x_2$⁠, expanding the right-hand sides gives

   $$\begin{aligned} x_1^2 - x_2^2 &= (x_1 - x_2)(x_1 + x_2),\\ x_1^3 - x_2^3 &= (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2). \end{aligned}$$

   The second bracket of the cube identity can be rewritten by completing the square:

   $$x_1^2 + x_1x_2 + x_2^2 = \left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3x_2^2}{4}.$$

   (Check: $\left(x_1 + \frac{x_2}{2}\right)^2 = x_1^2 + x_1x_2 + \frac{x_2^2}{4}$⁠, and $\frac{x_2^2}{4} + \frac{3x_2^2}{4} = x_2^2$⁠.) Both terms on the right are squares times non-negative numbers, so this bracket is always $\ge 0$⁠.

   To test injectivity we assume $f(x_1) = f(x_2)$⁠, which means $x_1^2 - x_2^2 = 0$ or $x_1^3 - x_2^3 = 0$⁠. Factoring pulls out $(x_1 - x_2)$⁠. If we can show the other bracket is not zero, the zero-product rule forces $x_1 - x_2 = 0$⁠. If the other bracket can be zero for different inputs, that tells us exactly where to look for a counterexample. Doing this algebra once, for all reals, lets every part reuse it.
   Zero-product rule (Basic 4) and squares are non-negative (Basic 5).

Part (i): $f : \N\to\N$⁠, $f(x) = x^2$

1. It is a function from $\N$ to $\N$

   If $x\in\N$⁠, then $x^2 = x\cdot x$ is a product of natural numbers, so $x^2\in\N$⁠. Each input has exactly one square.

   Outputs must lie in the codomain before onto can even be asked. Natural numbers are closed under multiplication.

2. Injective? Yes

   Let $x_1, x_2\in\N$ with $x_1^2 = x_2^2$⁠. Then by Step 1 of the Solution,

   $$(x_1 - x_2)(x_1 + x_2) = 0.$$

   Since $x_1 \ge 1$ and $x_2 \ge 1$⁠, we have $x_1 + x_2 \ge 2$⁠, so $x_1 + x_2 \ne 0$⁠. By the zero-product rule, $x_1 - x_2 = 0$⁠, that is, $x_1 = x_2$⁠. So $f$ is injective.

   The only way two different numbers can have the same square is $x_1 = -x_2$ (the second bracket vanishing). In $\N$ every number is positive, so $x_1 + x_2$ can never be zero, and that escape route is closed.
   If $ab=0$ and $b\ne0$⁠, then $a=0$ (Basic 4).

3. Surjective? No: $2$ is not a square of a natural number

   The number $2$ is in the codomain $\N$⁠. Look for $x\in\N$ with $x^2 = 2$⁠:

   • $x = 1$ gives $1^2 = 1 \ne 2$⁠;
   • every other natural number has $x\ge 2$⁠, so $x^2 \ge 2^2 = 4 > 2$⁠.

   Every natural number falls into one of these two cases, and neither gives $2$⁠. So $2$ has no preimage, and $f$ is not surjective.

   "Not onto" needs one specific codomain value and a proof that no input reaches it. Trying a few inputs is not a proof; splitting $\N$ into "$x = 1$" and "$x\ge2$" covers every natural number, so the argument is complete. (The squares $1, 4, 9, 16, \dots$ skip $2, 3, 5, 6, 7, 8, \dots$⁠; any of these would do.)
   If $a \ge b > 0$ then $a^2 \ge b^2$ (Basic 5).

4. Verdict for part (i)

   Injective but not surjective.

Part (ii): $f : \Z\to\Z$⁠, $f(x) = x^2$

1. Injective? No: $-1$ and $1$ have the same square

   Both $-1$ and $1$ are integers, and

   $$f(-1) = (-1)^2 = 1 = 1^2 = f(1), \qquad\text{while } -1 \ne 1.$$

   So $f$ is not injective.

   The factorisation $(x_1 - x_2)(x_1 + x_2) = 0$ shows equal squares happen when $x_1 + x_2 = 0$⁠, that is, $x_2 = -x_1$⁠. In part (i) this was impossible, but $\Z$ contains negative numbers, so any pair $x, -x$ with $x\ne0$ is a counterexample. One such pair is enough to disprove a "for all" claim.
   One counterexample disproves injectivity (Basic 3). Basics: how to prove and disprove

2. Surjective? No: $-1$ is not a square

   The number $-1$ is in the codomain $\Z$⁠. For every integer $x$⁠, $x^2 \ge 0$ (Basic 5), so $x^2 \ne -1$⁠. No integer is mapped to $-1$⁠, and $f$ is not surjective.

   Choosing a negative target makes the proof one line, because squares are never negative. (The value $2$ would also work: the argument of part (i) applies to $|x|$⁠, since $x^2 = |x|^2$ and $|x|$ is $0$⁠, $1$ or at least $2$⁠.)

3. Verdict for part (ii)

   Neither injective nor surjective. Compare part (i): same formula, but allowing negative inputs destroys injectivity.

Part (iii): $f : \R\to\R$⁠, $f(x) = x^2$

1. Injective? No

   $-1, 1\in\R$⁠, $-1\ne1$⁠, but $f(-1) = 1 = f(1)$⁠. So $f$ is not injective.

   Enlarging the domain from $\Z$ to $\R$ keeps the old counterexample available: $-1$ and $1$ are still inputs. A pair that breaks injectivity on a smaller domain breaks it on any larger domain too.

2. Surjective? No: $-1$ is still unreached

   $-1\in\R$ is in the codomain, but $x^2 \ge 0$ for every real $x$⁠, so $x^2 \ne -1$⁠. Hence $f$ is not surjective.

   Notice what changed from part (ii). Over $\R$⁠, the value $2$ is reached (by $x = \sqrt2$⁠), and in fact every $y \ge 0$ is reached (by $\sqrt y$⁠). So over $\R$ only the negative values are missed, but missing even one is enough.

   The horizontal line test pictures both properties: one-one means no horizontal line at a codomain height meets the graph twice; onto means every such line meets it at least once. The textbook settles exactly this function in Example 11, using $f(-1) = f(1)$ and the value $-2$⁠, which has no real square root. Any negative number works.
   Squares of real numbers are $\ge 0$ (Basic 5). Example 11: $x^2$ on $\R$

3. Verdict for part (iii)

   Neither injective nor surjective.

Part (iv): $f : \N\to\N$⁠, $f(x) = x^3$

1. Injective? Yes

   Let $x_1, x_2\in\N$ with $x_1^3 = x_2^3$⁠. By Step 1 of the Solution,

   $$(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0.$$

   Since $x_1, x_2 \ge 1$⁠, each of $x_1^2$⁠, $x_1x_2$⁠, $x_2^2$ is at least $1$⁠, so the second bracket is at least $3$⁠, and in particular not zero. Hence $x_1 - x_2 = 0$⁠, so $x_1 = x_2$ and $f$ is injective.

   We need an argument for all natural $x_1, x_2$⁠, and the factorisation supplies it: the difference of the cubes is $(x_1-x_2)$ times a bracket that is plainly positive for natural numbers. "It is obvious that different numbers have different cubes" is the claim we are asked to prove, so we cannot use it as a reason.
   If $ab = 0$ and $b \ne 0$ then $a = 0$ (Basic 4).

2. Surjective? No: $2$ is not a cube of a natural number

   $2$ is in the codomain $\N$⁠. For $x\in\N$⁠: if $x = 1$⁠, $x^3 = 1 \ne 2$⁠; if $x\ge2$⁠, then $x^3 \ge 2^3 = 8 > 2$⁠. Every natural number is in one of these cases, so no $x\in\N$ has $x^3 = 2$⁠, and $f$ is not surjective.

   The cubes of natural numbers are $1, 8, 27, 64, \dots$⁠, which skip $2$⁠. The two-case split covers all of $\N$⁠, which is what makes this a proof rather than a few trials.
   If $a\ge b>0$ then $a^3 \ge b^3$ (Basic 5): multiply $a \ge b$ by positive numbers three times.

3. Verdict for part (iv)

   Injective but not surjective.

Part (v): $f : \Z\to\Z$⁠, $f(x) = x^3$

1. Injective? Yes, even with negative inputs

   Let $x_1, x_2\in\Z$ with $x_1^3 = x_2^3$⁠. Then

   $$(x_1 - x_2)\,Q = 0, \qquad\text{where } Q = x_1^2 + x_1x_2 + x_2^2 = \left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3x_2^2}{4}.$$

   There are two cases.

   • $Q \ne 0$⁠. Then the zero-product rule gives $x_1 - x_2 = 0$⁠, so $x_1 = x_2$⁠.
   • $Q = 0$⁠. $Q$ is a sum of two non-negative terms, so both are zero: $\frac{3x_2^2}{4} = 0$ gives $x_2 = 0$⁠, and then $\left(x_1 + 0\right)^2 = 0$ gives $x_1 = 0$⁠. So again $x_1 = x_2$⁠.

   In both cases $x_1 = x_2$⁠, so $f$ is injective.

   In part (iv) the bracket $Q$ was obviously positive because all numbers were positive. With negative integers allowed, $x_1x_2$ can be negative, so we need a better reason: completing the square shows $Q$ is a sum of non-negative terms and vanishes only when $x_1 = x_2 = 0$⁠. Unlike the square, the cube keeps the sign ($(-x)^3 = -x^3$⁠), so the pair $x, -x$ does not collide.
   A sum of non-negative numbers is zero only if each of them is zero; here both terms are non-negative because squares are $\ge 0$ (Basic 5).

2. Surjective? No: $2$ is not the cube of an integer

   $2$ is in the codomain $\Z$⁠. Split the integers into two cases:

   • $x\in\{-1, 0, 1\}$⁠: the cubes are $-1, 0, 1$⁠, none equal to $2$⁠;
   • $|x| \ge 2$⁠: then $|x^3| = |x|^3 \ge 2^3 = 8$⁠, so $x^3$ is $\ge 8$ or $\le -8$⁠, and cannot be $2$⁠.

   So no integer has cube $2$⁠, and $f$ is not surjective.

   The case split $|x|\le1$ or $|x|\ge2$ covers every integer, so the argument is complete. It also explains part (iv), which is the special case of positive inputs. Over $\R$ the answer would change: every real $y$ has a real cube root $\sqrt[3]{y}$⁠, so $x^3 : \R\to\R$ is onto (and one-one, by the same factorisation), hence bijective. The integers simply have gaps between consecutive cubes.
   $|x^3| = |x|^3$⁠, and $|x|\ge2 \Rightarrow |x|^3 \ge 8$ (Basic 5). Miscellaneous Q2: $x^3$ is injective

3. Verdict for part (v)

   Injective but not surjective. Compare part (ii): over $\Z$⁠, the square fails one-one but the cube passes, because cubing keeps the sign.

Every fact used, and where it comes from