Q.Show that the function $f: \mathbf{R} \to \mathbf{R}$ given by $f(x) = x^3$ is injective.

The key idea is that a strictly monotonic function is injective.

1. Take any two real numbers $a$ and $b$ such that $f(a) = f(b)$.

   This means $a^3 = b^3$.

2. Subtract and factor: $a^3 - b^3 = (a-b)(a^2 + ab + b^2) = 0$.

3. The quadratic $a^2 + ab + b^2$ is always positive for real $a, b$ (it equals $\frac12[(a+b)^2 + a^2 + b^2]$, which is zero only if $a=b=0$). So the only way the product is zero is $a-b=0$, i.e., $a=b$.

Thus $f$ is injective.

Why a Bijection Proof Works

The standard way to prove injectivity is to assume $f(a) = f(b)$ and deduce $a = b$. This is the contrapositive of the definition: "different inputs give different outputs" is logically equivalent to "equal outputs imply equal inputs." For $f(x) = x^3$, the proof is almost immediate because the cube function is strictly increasing on $\mathbb{R}$ — but we'll give a clean algebraic proof that doesn't rely on calculus.

Step-by-Step Proof

1. Set up the assumption. Let $a, b \in \mathbb{R}$ and suppose $f(a) = f(b)$. That is:

2. Rearrange to a difference of cubes. Bring everything to one side:

3. Factor using the difference of cubes identity. Recall: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. So:

4. Analyse the factors. A product equals zero if and only if at least one factor is zero. So either:

5. Show the second factor is never zero for real numbers (unless both are zero). Consider the quadratic form $a^2 + ab + b^2$. Complete the square in $a$:

This is a sum of two squares (one in $a$, one in $b$). A sum of squares is zero only when each square is zero:

• $\left(a + \frac{b}{2}\right)^2 = 0$ implies $a = -\frac{b}{2}$
• $\frac{3b^2}{4} = 0$ implies $b = 0$

Substituting $b=0$ into $a = -b/2$ gives $a=0$. So $a^2 + ab + b^2 = 0$ forces $a = b = 0$.

6. Conclude from both cases.

   • If $a - b = 0$, then $a = b$.
   • If $a^2 + ab + b^2 = 0$, then $a = b = 0$, which also gives $a = b$.

   In every possible case, $a = b$.

7. Therefore, $f$ is injective.

   We have shown: $f(a) = f(b) \implies a = b$, which is exactly the definition of an injective function.

Method: Proving injectivity of a cubic by the difference-of-cubes factorisation

To prove $f(x)=x^3$ is one-one without calculus, assume equal outputs and factor.

Steps

Step 1: Assume equal outputs

Let $f(a)=f(b)$, i.e. $a^3=b^3$, and move everything to one side: $a^3-b^3=0$.

Step 2: Factor the difference of cubes

Step 3: Show the quadratic factor cannot vanish for distinct reals

Complete the square: $a^2+ab+b^2=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge 0$, and it equals $0$ only when $a=b=0$. So the product is zero only if $a-b=0$, giving $a=b$. (Alternatively, $x^3$ is strictly increasing, hence injective.)

Common Mistakes

Mistake 1: Claiming $a^2+ab+b^2$ can equal zero for some non-zero real $a,b$.

Why it's wrong: completing the square gives $\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge 0$, which is zero only at $a=b=0$. Correct approach: use this to argue the only factor that can vanish is $a-b$, so $a=b$.

Mistake 2: Cancelling $a^3=b^3$ to $a=b$ with no justification.

Why it's wrong: a rigorous proof must show the cube map is injective, not merely assert cube roots are unique. Correct approach: factor $a^3-b^3=(a-b)(a^2+ab+b^2)$ and rule out the quadratic factor.