Question

Let $f : \mathbb{Q} \to \mathbb{Q}$ be a function such that $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{Q}$, and $f(1) = 10$. Which one of the following statements is Correct ?

• A. $f$ is bijective.
• B. $f$ is injective but not surjective.
• C. $f$ is surjective but not injective.
• D. $f$ is neither injective nor surjective.

Hint:

Over the rational numbers \(\mathbb{Q}\)., any additive function \(f(x+y) = f(x)+f(y)\) is strictly linear, i.e., \(f(x) = f(1)x\).
Since \(f(1) \neq 0\)., the linear map \(f(x) = 10x\) is a bijection on \(\mathbb{Q}\).
This property does not automatically hold over \(\mathbb{R}\) without assuming continuity or monotonicity.

Answer 1

The Correct Option is A

•

Step 1 : Understanding the Question:
We are given a function \(f\) defined on the rational numbers \(\mathbb{Q}\).
The function satisfies Cauchy's functional equation: \(f(x + y) = f(x) + f(y)\) for all \(x, y \in \mathbb{Q}\).
We are also given the value of the function at 1: \(f(1) = 10\).
We need to determine if \(f\) is injective, surjective, or bijective.

•

Step 2 : Key Formula or Approach:
Cauchy's functional equation \(f(x+y) = f(x) + f(y)\) over the domain of rational numbers \(\mathbb{Q}\) has a unique family of solutions:
\[ f(x) = kx \quad \text{for all } x \in \mathbb{Q} \] where \(k = f(1)\). is a constant.
We will prove this general form and then analyze its injectivity and surjectivity.

•

Step 3 : Detailed Explanation:
Let us prove that \(f(x) = kx\) for all rational numbers \(x\):
- For \(x = y = 0\):
\[ f(0) = f(0) + f(0) \implies f(0) = 0 \] - For any positive integer \(n \in \mathbb{N}\):
\[ f(n) = f(1 + 1 + \dots + 1) = n \cdot f(1) = 10n \] - For negative integers, let \(y = -x\):
\[ f(0) = f(x) + f(-x) \implies f(-x) = -f(x) \] Thus, \(f(n) = 10n\) holds for all \(n \in \mathbb{Z}\).
- Now, let \(x = \frac{p}{q}\) be a rational number, where \(p \in \mathbb{Z}\) and \(q \in \mathbb{N}\).
Using the additive property:
\[ f(p) = f\left( q \cdot \frac{p}{q} \right) = q \cdot f\left( \frac{p}{q} \right) \] Substitute \(f(p) = 10p\):
\[ 10p = q \cdot f\left( \frac{p}{q} \right) \implies f\left( \frac{p}{q} \right) = 10 \cdot \frac{p}{q} \] Thus, \(f(x) = 10x\) for all \(x \in \mathbb{Q}\).
Now, let us analyze the properties of \(f(x) = 10x\):
- Injectivity:
Let \(x_1, x_2 \in \mathbb{Q}\) such that \(f(x_1) = f(x_2)\).
\[ 10x_1 = 10x_2 \implies x_1 = x_2 \] Thus, \(f\) is injective.
- Surjectivity:
Let \(y \in \mathbb{Q}\) be any element in the codomain.
We need to find \(x \in \mathbb{Q}\) such that \(f(x) = y\).
\[ 10x = y \implies x = \frac{y}{10} \] Since \(y\) is rational, \(\frac{y}{10}\) is also a rational number.
Thus, every element in the codomain has a pre-image in the domain, which means \(f\) is surjective.
Since \(f\) is both injective and surjective, \(f\) is bijective.

•

Step 4 : Final Answer:
The function \(f\) is bijective.
This corresponds to option (A).