Question

Show that \(\mathbb{Q}\) has only the trivial derivative.

Short answer

The only derivative on \( \mathbb{Q} \) is the trivial one: \( D(x) = 0 \) for all \( x \).

Step-by-step solution

Understanding the Derivative Function

A derivative on a field is a map \( D: \mathbb{Q} \rightarrow \mathbb{Q} \) that satisfies two properties: for all \( a, b \in \mathbb{Q} \), it holds that \( D(a+b) = D(a) + D(b) \) and \( D(a\cdot b) = a \cdot D(b) + b \cdot D(a) \). Our task is to prove that the only function that satisfies these conditions for the rationals is the zero function, i.e., \( D(x) = 0 \) for all \( x \in \mathbb{Q} \).

Proving the Zero Function is a Derivative

Show that the zero function \( D(x) = 0 \) satisfies the properties of a derivative. Clearly:1. \( D(a+b) = 0 = 0 + 0 = D(a) + D(b) \).2. \( D(a\cdot b) = 0 = a\cdot 0 + b\cdot 0 = a \cdot D(b) + b \cdot D(a) \).This shows the zero function fulfills the conditions of a derivative.

Establishing Uniqueness of the Trivial Derivative

Consider any derivative \( D \) with the properties stated in Step 1. We must show \( D(x) = 0 \) for all \( x \in \mathbb{Q} \). Start by observing that for any rational number \( q \), \( q \cdot 1 = q \). Applying the product property to \( D(q \cdot 1) = q \cdot D(1) + 1 \cdot D(q) \) yields \( D(q) = q \cdot D(1) \).

Determining \(D(1) = 0\)

Set \( q = 1 \) in the equation \( D(q) = q \cdot D(1) \), and we have \( D(1) = 1 \cdot D(1) = D(1) \). Suppose \( D(1) eq 0 \), then for any \( q \in \mathbb{Q} \), \( D(q) = q \cdot D(1) \) implies \( D(q) eq 0 \) for \( q eq 0 \). However, this would violate the additive property of the derivative because \( D(q - q) = D(0) = 0 \) and \( D(q - q) = D(q) + D(-q) \). As a result, \( 0 = D(q) - D(q) = 0 \), making it necessary that \( D(1) = 0 \).

Conclusion

From \( D(q) = q \cdot D(1) \) and \( D(1) = 0 \), we deduce \( D(q) = 0 \times q = 0 \) for all \( q \in \mathbb{Q} \). Thus, the only derivative on \( \mathbb{Q} \) is the zero (trivial) derivative \( D(x) = 0 \).

Key concepts

Zero Function

The zero function is a special type of function where every element in its domain is mapped to zero in its range. In mathematical terms, if we have a function \( f: X \rightarrow Y \), then the zero function is defined as \( f(x) = 0 \) for all \( x \) in \( X \). This function is noteworthy because it retains certain properties across different mathematical contexts.
In the case of derivatives on rational numbers \( \mathbb{Q} \), the zero function plays a crucial role. As demonstrated in the exercise, the zero function is the only derivative that satisfies the required properties of linearity and the product rule in this field.
By applying the zero function in step 2 of the solution, it becomes clear how it maintains the essential criteria by producing zero for every input. This consistent output is due to the additive and multiplicative properties of zero, supporting the notion that the zero function is the only trivial, or 'uninteresting,' derivative in the rationals.

Rational Numbers

Rational numbers are a set of numbers that can be expressed as the quotient or fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b eq 0 \). Represented by \( \mathbb{Q} \), the set of rational numbers includes all positive and negative fractions, whole numbers, and the number zero.
The significance of rational numbers in this context lies in proving that the zero function is the trivial derivative for this set. Rationals are closed under addition, subtraction, multiplication, and division (except by zero), making them an important field to study in calculus and algebra.

• Rational numbers are dense, meaning between any two rationals, there exists another rational number.
• They form a field because they satisfy all the field axioms, including associativity, distributivity, and the existence of additive and multiplicative identities and inverses.

Understanding the properties of rational numbers and their operation rules is vital in applying the concepts of derivatives and demonstrating why only the zero function works as a derivative in \( \mathbb{Q} \).

Properties of Derivative

The derivative of a function at a point provides the rate at which the function's value changes as its input changes. In the context of a field like \( \mathbb{Q} \), a derivative must satisfy specific mathematical properties:

• Linearity: For any two rational numbers \( a, b \), the derivative \( D \) must satisfy \( D(a + b) = D(a) + D(b) \). This means that the derivative of a sum is the sum of the derivatives.
• Product Rule: For any two rational numbers \( a, b \), the product rule states \( D(a \cdot b) = a \cdot D(b) + b \cdot D(a) \). This states that the derivative of a product is structured around individual derivatives of each factor.

These properties are foundational in calculus and ensure the derivative behaves predictably and consistently across mathematical operations.
In the solution's context, applying these properties helps demonstrate why the zero function remains the only valid derivative on \( \mathbb{Q} \). The proof hinges on showing that if the derivative of \( 1 \) isn't zero, contradictions arise with these properties, implying every derivative must conform to the zero function. Thus, understanding these properties is crucial for grasping more advanced topics in calculus and algebra.