Q.Prove that the function $f : \mathbf{R} \to \mathbf{R}$ defined by $f(x) = 2x + 5$ is one-one, using the contrapositive.

The contrapositive of "$f(x_1)=f(x_2)\Rightarrow x_1=x_2$" is "$x_1 \ne x_2 \Rightarrow f(x_1)\ne f(x_2)$". Take $x_1 \ne x_2$; then

$2x_1 \ne 2x_2 \ \Rightarrow\ 2x_1 + 5 \ne 2x_2 + 5 \ \Rightarrow\ f(x_1) \ne f(x_2).$ …

The one-one condition is the conditional

A conditional $p \Rightarrow q$ is logically equivalent to its contrapositive $\sim q \Rightarrow \sim p$, formed by interchanging and negating the two parts. Here the contrapositive is

Step 1 — Start from the negated conclusion.

Let $x_1, x_2 \in \mathbf{R}$ with

Step 2 — Multiply both sides by $2$ (a non-zero factor preserves the inequality):

Step 3 — Add $5$ to both sides: …

Method: Proving a Function is One-One via the Contrapositive

This method proves injectivity for functions defined by a simple algebraic rule, where working directly from "$f(x_1)=f(x_2)\Rightarrow x_1=x_2$" is exactly as easy going the other way around — the contrapositive route.

Steps

Step 1: State the definition you must prove

$f$ is one-one on its domain if

Recognise this as a conditional $p \Rightarrow q$.

Step 2: Form the contrapositive

Since $p\Rightarrow q$ is logically equivalent to $\sim q \Rightarrow \sim p$, restate the goal as:

This is usually the easier direction to prove when $f$ is built from reversible algebraic operations (multiplication by a nonzero constant, addition, etc.), because you can track an inequality through each operation instead of trying to "cancel" from an equation.

Step 3: Start from $x_1 \ne x_2$ and apply $f$'s operations one at a time …

Common Mistakes

Mistake 1: Writing the converse or inverse instead of the contrapositive

Stating "if $x_1=x_2$ then $f(x_1)=f(x_2)$" (the converse) and treating it as equivalent to the one-one definition. Why it's wrong: only the true contrapositive $x_1\ne x_2\Rightarrow f(x_1)\ne f(x_2)$ is logically equivalent to the defining implication $f(x_1)=f(x_2)\Rightarrow x_1=x_2$; the converse of a true statement need not be true. Correct approach: form the contrapositive by negating and swapping both parts of the original conditional, never just negating one side.

Mistake 2: Multiplying an inequality without checking the multiplier's sign …