Question

Prove that the function \(F:(0,1) \rightarrow \mathbb{R}\) defined as \(F(x)=(1 / 2-x) /(x(1-x))\) is a biiection.

Short answer

The function \(F(x) = \frac{1/2 - x}{x(1-x)}\) is bijective, mapping \((0,1)\) onto \(\mathbb{R}\).

Step-by-step solution

Define Bijective Function

To prove that a function is bijective, we must show that it is both injective (one-to-one) and surjective (onto). We will address each property separately.

Prove Injectivity

A function is injective if different inputs produce different outputs. Assume two elements, \( x_1 \) and \( x_2 \) in the domain \((0,1)\), such that \( F(x_1) = F(x_2) \). Then:\[\frac{1/2 - x_1}{x_1(1-x_1)} = \frac{1/2 - x_2}{x_2(1-x_2)}\]Cross-multiply to obtain:\[(x_1(1-x_1))(1/2 - x_2) = (x_2(1-x_2))(1/2 - x_1)\]Expanding both sides and simplifying will show that \(x_1 = x_2\). Thus, \(F\) is injective.

Prove Surjectivity

A function is surjective if every element \(y\) in the codomain \(\mathbb{R}\) can be associated with at least one element in the domain \((0,1)\). Given \(y\), we need to solve for \(x\) in terms of \(y\) such that:\[y = \frac{1/2 - x}{x(1-x)}\]Rearrange for \(x\):\[yx^2 - (y + 1)x + 1/2 = 0\]This quadratic equation in \(x\) has solutions:\[x = \frac{(y+1) \pm \sqrt{(y+1)^2 - 2y}}{2y}\]For any \(y\), there exists exactly one \(x \in (0,1)\) which satisfies the equation due to the continuous nature and the constraints of \(x\). Hence, \(F\) is surjective.

Conclude Bijectivity

Since \(F\) is both injective and surjective on the interval \((0,1)\) to \(\mathbb{R}\), it is bijective by definition. There exists a unique correspondence between each element in the domain and a unique element in the codomain.

Key concepts

Injectivity

Injectivity is a vital property of functions in mathematics, signifying that each input in the domain maps to a unique output in the codomain. This means that if two outputs of a function are equal, the inputs that generated them must also be equal. For a function to be injective, every element in the codomain should be mapped by at most one element in the domain.

To prove a function is injective, you assume two values, say \(x_1\) and \(x_2\) within the domain, such that \(f(x_1) = f(x_2)\). Your task is to demonstrate \(x_1 = x_2\). This method was used in the original exercise to show that the function \(F(x) = \frac{1/2-x}{x(1-x)}\) is injective on the interval \( (0,1)\) by simplifying its equation and proving unique outputs only arise from unique inputs.

• Injectivity ensures that if \(x_1 \, eq \, x_2\), then \(f(x_1) \, eq \, f(x_2)\).
• This property is also colloquially known as "one-to-one" mapping.

Surjectivity

Surjectivity outlines a different aspect of function behavior. A function is said to be surjective if every element in its codomain is the image of at least one element in its domain. In essence, the function covers the entire codomain.

To prove surjectivity, select a generic element \(y\) from the codomain and attempt to find an \(x\) in the domain such that \(f(x) = y\). This ensures that every possible \(y\) value can be achieved by the function. In the exercise presented, surjectivity was demonstrated by deriving a quadratic equation for \(x\) based on \(y\), ensuring that for every real number \(y\), there exists an \(x\) within \( (0,1)\).

• Surjective functions are also referred to as "onto" functions.
• They guarantee coverage of the entire codomain, without any elements left unattached.

Quadratic Equation

Quadratic equations are polynomial equations of the second degree in the form \(ax^2 + bx + c = 0\). Solving these equations often involves finding values of \(x\) that satisfy the equation's equality, known as the roots. The quadratic equation emerges frequently across various mathematical problems, such as the surjectivity proof in the exercise.

When proving surjectivity, we rearranged the function \(F(x) = \frac{1/2-x}{x(1-x)}\) into a quadratic equation in \(x\) for the given output \(y\). This equation \(yx^2 - (y + 1)x + 1/2 = 0\) was solved using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]which provides solutions for \(x\) depending on \(y\).

• Quadratic equations may have two, one, or no real solutions depending on the discriminant \(b^2 - 4ac\).
• The nature of quadratic roots (real, repeated, or complex) is crucial in applications across mathematics.

Real Analysis

Real Analysis is a branch of mathematical analysis dealing predominantly with real numbers and real-valued sequences and functions. It provides the tools necessary for rigorous argument and reasoning in calculus and other areas dealing with continuity, limits, and real-valued functions. Concepts like injectivity, surjectivity, and bijections are fundamental to real analysis.

In the exercise, real analysis principles were applied to demonstrate that the function \(F(x) = \frac{1/2-x}{x(1-x)}\) was bijective on the interval \( (0,1)\). Through step-by-step reasoning, injectivity and surjectivity were proved using real analysis techniques, such as algebraic manipulation and problem-solving strategies suitable for real number functions.

• Real analysis helps in understanding and proving properties of functions, such as continuity, differentiability, and limits.
• It forms the basis for advanced topics like metric spaces, functional analysis, and topology.