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Chapter 5: Problem 95

Proof Prove that a function has an inverse function if and only if it is one- to-one.

Short Answer

Expert verified
In summary, a function \( f \) has an inverse if and only if it is one-to-one. This is because by definition of an inverse, no two different elements in the domain of \( f \) can map to the same element in the range. Similarly, if \( f \) is one-to-one, then each element in the range of \( f \) corresponds with exactly one element in the domain, so it is possible to define an inverse.

Step by step solution

01

Prove that if a function has an inverse, it is one-to-one

First, suppose that a function \( f \) has an inverse function \( f^{-1} \). For \( f \) to have an inverse, each \( y \) in the codomain of \( f \) must correspond to exactly one \( x \) in the domain of \( f \). Suppose \( f(a) = f(b) \) for some \( a \) and \( b \). This would mean that \( f^{-1}(f(a)) = f^{-1}(f(b)) \), and since \( f^{-1}(f(a)) = a \) and \( f^{-1}(f(b)) = b \) (by the definition of an inverse), we have \( a = b \). Therefore \( f \) must be one-to-one.
02

Prove that if a function is one-to-one, it has an inverse

Now suppose that \( f \) is a one-toone function from set \( A \) to set \( B \). Define a function \( g: B \rightarrow A \) by the rule that if \( b = f(a) \) for some \( a \) in \( A \), then \( g(b) = a \). Since \( f \) is one-to-one, there will be exactly one \( a \) in \( A \) such that \( b = f(a) \) for each \( b \) in \( B \), so \( g \) is well-defined. Also, we see that \( g(f(a)) = a \) for each \( a \) in \( A \) and \( f(g(b)) = b \) for each \( b \) in \( B \), so \( g \) is indeed the inverse of \( f \). This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-to-One Function
A one-to-one function, also known as an injective function, is a fundamental concept in mathematics. It ensures that every element in the domain of the function maps to a unique element in the codomain. This means that if you have two different inputs, you will always get two different outputs. In more formal terms, a function \( f \) is one-to-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
  • This property is crucial for the existence of inverse functions.
  • One-to-one functions are important for establishing a reversible and unique relationship between inputs and outputs.
Understanding the concept of one-to-one functions is a stepping stone towards grasping more complex ideas in calculus and algebra.
Function Proof
Proving whether a function is one-to-one or not involves logical reasoning and systematic demonstration. Proofs play an essential role in mathematics by confirming whether statements and properties hold true. For a function to be one-to-one, you must show that it passes the injectivity condition.
To prove injectivity, assume two points \( a \) and \( b \) in the domain where \( f(a) = f(b) \). You must then show that \( a = b \), confirming the function's one-to-one nature.
Here鈥檚 a simple guide to proving functions:
  • Start from the assumption \( f(a) = f(b) \).
  • Use definitions and properties of the function.
  • Arrive at the conclusion \( a = b \).
This methodical approach can also be applied to establish other mathematical properties.
Inverse Function Existence
An inverse function is like the reverse operation of an original function. For an inverse function \( f^{-1} \) to exist, the original function \( f \) must be one-to-one. Essentially, each output of \( f \) must map to exactly one input, which is retrievable in reverse with \( f^{-1} \).
  • This relationship ensures that \( f(f^{-1}(y)) = y \) and \( f^{-1}(f(x)) = x \) for all \( y \) in the codomain and \( x \) in the domain.
  • The existence of an inverse function allows us to solve equations by reversing calculations.
Functions without an inverse are not fully "reversible" and thus lose some utility in problem solving.
Bijective Functions
Bijective functions are functions that are both one-to-one (injective) and onto (surjective). This means every element in the codomain has a pre-image in the domain, and no two elements in the domain map to the same element in the codomain.
In short, every function point has a partner, making it a perfect pairing.
  • Bijective functions are crucial because they guarantee the existence of an inverse.
  • They establish a perfect "marriage" between domain and codomain elements, preventing any element from being unpaired or over-paired.
Learning about bijective functions provides a comprehensive understanding of the relationship between two sets, expanding your mathematical toolbox.

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Most popular questions from this chapter

Using Properties of Exponents In Exercises \(107-110\) , find the exact value of the expression. $$6^{(\ln 10) / \ln 6}$$

In Exercises 43-46, use the specified substitution to find or evaluate the integral. $$\begin{array}{l}{\int \frac{\sqrt{x-2}}{x+1} d x} \\\ {u=\sqrt{x-2}}\end{array}$$

In Exercises 61 and 62, find the particular solution of the differential equation that satisfies the initial condition. $$\begin{array}{l}{\frac{d y}{d x}=\frac{1}{4+x^{2}}} \\\ {y(2)=\pi}\end{array}$$

Finding an Indefinite Integral In Exercises \(69-76,\) find the indefinite integral. $$\int x\left(5^{-x^{2}}\right) d x$$

Finding an Equation of a Tangent Line In Exercises \(61-64,\) find an equation of the tangent line to the graph of the function at the given point. $$y=\log _{10} 2 x, \quad(5,1)$$
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