Miscellaneous · Q2: $f(x)=x^3$ is injective

Basics you need

Solution

1. Assume equal outputs

   Let $x_1$ and $x_2$ be any real numbers with $f(x_1) = f(x_2)$⁠, that is,

   $$x_1^3 = x_2^3.$$

   We must show that this forces $x_1 = x_2$⁠.

   Injective is an "if … then" statement about every pair of inputs, so the proof must start from the "if" part, equal outputs, for inputs that could be any real numbers. Checking examples such as $2^3 = 8$ and $3^3 = 27$ would not cover the infinitely many other pairs.
   Definition of injective (Basic 1): $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$⁠.

2. Bring everything to one side and factorise

   Subtract $x_2^3$ from both sides and use the difference-of-cubes identity (Basic 2) with $a = x_1,\ b = x_2$⁠:

   $$0 = x_1^3 - x_2^3 = (x_1 - x_2)\left(x_1^2 + x_1x_2 + x_2^2\right).$$
   The factor $x_1 - x_2$ is exactly what we want to show is zero. Factorising makes it appear explicitly, so the question becomes: can the other factor be zero instead? We cannot just take cube roots of both sides here without first knowing that a real number has only one real cube root, which is essentially the statement we are trying to prove.
   $a^3 - b^3 = (a-b)(a^2+ab+b^2)$⁠, checked by expanding the right-hand side (Basic 2).

3. Two cases from the zero-product rule

   A product of two real numbers is zero only when one of them is zero (Basic 3). So

   $$x_1 - x_2 = 0 \qquad\text{or}\qquad x_1^2 + x_1x_2 + x_2^2 = 0.$$

   Case 1: $x_1 - x_2 = 0$⁠. Then $x_1 = x_2$ immediately.

   Case 2: $x_1^2 + x_1x_2 + x_2^2 = 0$⁠. This needs more work, done in the next step.

   The two cases cover every possibility, so if each one leads to $x_1 = x_2$⁠, the proof is complete (Basic 5). Case 1 is the outcome we want; Case 2 is the only way the proof could fail, so we must examine it carefully rather than ignore it.
   Zero-product rule (Basic 3): $PQ = 0 \Rightarrow P = 0$ or $Q = 0$⁠.

4. Case 2: complete the square

   Treat $x_1^2 + x_1x_2$ as the start of a square in $x_1$⁠. Since $\left(x_1 + \frac{x_2}{2}\right)^2 = x_1^2 + x_1x_2 + \frac{x_2^2}{4}$⁠,

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

   Both terms on the right are $\ge 0$ (a square, and $\frac34$ times a square). A sum of two non-negative numbers is 0 only when both are 0 (Basic 4), so

   $$\begin{gathered} \frac34 x_2^2 = 0 \ \Rightarrow\ x_2 = 0, \\ \text{and then}\quad \left(x_1 + 0\right)^2 = 0 \ \Rightarrow\ x_1 = 0. \end{gathered}$$

   So in Case 2, $x_1 = x_2 = 0$⁠, and in particular $x_1 = x_2$.

   An expression like $x_1^2 + x_1x_2 + x_2^2$ has a cross term $x_1x_2$ that can be negative, so it is not obvious that it is never negative. Completing the square removes the cross term and writes the expression as a sum of squares, whose sign is obvious. This shows the second factor is positive except when both numbers are 0.
   $t^2 \ge 0$ for every real $t$⁠, with equality only at $t = 0$ (Basic 4). Completing the square: $a^2 + ab = \left(a + \frac{b}{2}\right)^2 - \frac{b^2}{4}$⁠.

5. Conclude, and compare with $x^2$

   In both cases $x_1 = x_2$⁠. Since $x_1, x_2$ were arbitrary, $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ for all real $x_1, x_2$⁠, and $f(x) = x^3$ is injective.

   Compare squaring. For $g(x) = x^2$ the same method gives $x_1^2 - x_2^2 = (x_1 - x_2)(x_1 + x_2)$⁠, and the second factor can be zero with $x_1 \ne x_2$⁠: take $x_1 = 2,\ x_2 = -2$⁠. Indeed $g(2) = g(-2) = 4$⁠, so $x^2$ is not one-one on $\R$⁠. Cubing keeps the sign: $2^3 = 8$ but $(-2)^3 = -8$⁠.

   Showing where the same method breaks for $x^2$ makes clear what the proof really used: the second factor $x_1^2 + x_1x_2 + x_2^2$ is never zero unless both numbers are zero, while $x_1 + x_2$ is zero for every pair of opposite numbers.
   One counterexample ($2 \ne -2$ but $2^2 = (-2)^2$⁠) disproves one-one for $x^2$ (Basic 5).

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