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

To determine the nature of the function \( f : \mathbb{Q} \to \mathbb{Q} \) defined by \( f(x + y) = f(x) + f(y) \) for all \( x, y \in \mathbb{Q} \) and \( f(1) = 10 \), let's proceed step-by-step through this problem.

The given functional equation \( f(x + y) = f(x) + f(y) \) is a well-known equation called Cauchy's functional equation. Its general solution over the rationals \( \mathbb{Q} \) is linear, and can be expressed as:

\(f(x) = cx\)

where \( c \) is a constant that can be determined from given conditions.

We have the condition:

\(f(1) = 10\)

Substitute into the general solution:

\(c \times 1 = 10\)

Thus, \( c = 10 \). Therefore, the function is:

\(f(x) = 10x\)

Let us analyze the different properties:

Injectivity:

A function is injective (one-to-one) if different inputs give different outputs. For our function:

\(f(x_1) = f(x_2) \Rightarrow 10x_1 = 10x_2 \Rightarrow x_1 = x_2\)

Hence, the function is injective.

Surjectivity:

A function is surjective (onto) if every possible output is an image of some input. Here, we need to find \( x \) in \( \mathbb{Q} \) such that:

\(f(x) = y \Rightarrow 10x = y \Rightarrow x = \frac{y}{10}\)

Since \( \frac{y}{10} \) is a rational number for any rational \( y \), the function is surjective.

Given that the function is both injective and surjective, we conclude that it is bijective.

Therefore, the correct answer is: \( f \) is bijective.