Q.Prove that the function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = 2x$, is one-one and onto.

Concept: One-One (Injective) and Onto (Surjective) functions.

Step 1 – One-One:

Assume $f(x_1) = f(x_2)$. Then $2x_1 = 2x_2$. Dividing both sides by 2 gives $x_1 = x_2$. Hence $f$ is one-one.

Step 2 – Onto:

For any $y \in \mathbb{R}$, we need an $x \in \mathbb{R}$ such that $f(x) = y$. Choose $x = \frac{y}{2}$. Then $f(x) = 2 \cdot \frac{y}{2} = y$. Since $y$ is arbitrary, every real number has a preimage. Hence $f$ is onto.

The idea is simple: a function is a rule that pairs inputs with outputs. "One-one" (injective) means no two different inputs share the same output — think of it as each output having a unique "parent". "Onto" (surjective) means every possible output in the codomain actually gets used — nothing is left out. For $f(x)=2x$, the rule is just doubling. Doubling is reversible: if you know the output, you can always halve it to get back the input. That reversibility is the heart of both properties.

Let's walk through the proof step by step.

1. Proving one-one (injective)

   We need to show: if $f(a) = f(b)$, then $a = b$.

   Assume $f(a) = f(b)$. That means $2a = 2b$.

   Divide both sides by 2 (which is allowed since $2 \neq 0$), and we get $a = b$.

   That's it — the implication holds for any $a, b \in \mathbb{R}$. So $f$ is one-one.

   Tip

   The key move is that multiplying by 2 is an invertible operation. If the function were $f(x)=x^2$, we couldn't divide both sides so cleanly because $a^2=b^2$ doesn't force $a=b$ (e.g., $a=2, b=-2$). The linearity and non-zero coefficient make this trivial.

2. Proving onto (surjective)

   We need to show: for every $y \in \mathbb{R}$ (the codomain), there exists some $x \in \mathbb{R}$ (the domain) such that $f(x) = y$.

   Given any real number $y$, we want $2x = y$. Solve for $x$: $x = y/2$.

   Since $y$ is real, $y/2$ is also real — so $x$ lies in the domain $\mathbb{R}$.

   Check: $f(x) = 2 \cdot (y/2) = y$. So for every $y$, we've found a pre-image $x = y/2$. Hence $f$ is onto.

   Watch out

   A common mistake is to think "onto" means the function covers all outputs it can produce — but that's not the definition. Onto means every element of the codomain (here $\mathbb{R}$) is hit, not just the range. For $f(x)=2x$, the range is indeed all of $\mathbb{R}$, so it's onto. But for $f(x)=x^2$ with codomain $\mathbb{R}$, negative numbers are never outputs — so it's not onto.

3. Conclusion

   Since $f$ is both one-one and onto, it is a bijection. The function $f(x)=2x$ is a simple linear map with slope 2 and intercept 0 — it's a straight line through the origin. Every horizontal line $y=c$ cuts the graph exactly once (one-one), and every horizontal line does cut it (onto). That geometric intuition matches the algebraic proof.

Method: Proving a function is a bijection (one-one and onto) on $\mathbb{R}$

Use this to show a real function is both injective and surjective, hence invertible.

Steps

Step 1: Prove one-one

Assume $f(x_1)=f(x_2)$ and derive $x_1=x_2$. For $f(x)=ax+b$ with $a\ne0$, cancel to get $x_1=x_2$.

Step 2: Prove onto by solving for a preimage

Take an arbitrary $y$ in the codomain and solve $f(x)=y$ for $x$. Show the solution lies in the domain. For $f(x)=2x$, solving gives $x=\tfrac{y}{2}\in\mathbb{R}$, and $f\!\left(\tfrac{y}{2}\right)=y$.

Step 3: Conclude bijection

Both properties holding means $f$ is a bijection. The key surjectivity trick — "produce a preimage $x$ as a formula in $y$ and check it lands in the domain" — works for any invertible-looking function.

Common Mistakes

Mistake 1: Assuming a one-one proof also settles onto

Why it's wrong: injectivity and surjectivity are independent; on $\mathbb{R}$ you must separately produce a preimage for every $y$. Correct approach: solve $2x=y$ to get $x=\tfrac{y}{2}\in\mathbb{R}$ and verify $f\!\left(\tfrac{y}{2}\right)=y$.

Mistake 2: Confusing "range" with "codomain" when checking onto

Why it's wrong: onto means the range equals the declared codomain. Correct approach: here the range is all of $\mathbb{R}$ (every real is $2x$ for some real $x$), which matches the codomain, so $f$ is onto — unlike $x^2$, whose range misses the negatives.