/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Let \(I:=[0,1]\) and let \(f: I ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 9

Let \(I:=[0,1]\) and let \(f: I \rightarrow \mathbb{R}\) be defined by \(f(x):=x\) for \(x\) rational, and \(f(x):=1-x\) for \(x\) irrational. Show that \(f\) is injective on \(I\) and that \(f(f(x))=x\) for all \(x \in I\). (Hence \(f\) is its own inverse function!) Show that \(f\) is continuous only at the point \(x=\frac{1}{2}\).

Short Answer

Expert verified
The function \(f\) is injective, it is a self-inverse function, and it is continuous at the point \(x = 1/2\) only.

Step by step solution

01

Injectivity

To show that \(f\) is injective (one-to-one function), we must prove that if \(f(x) = f(y)\), then \(x = y\). Assume \(f(x) = f(y)\). There are four cases to consider: (a) \(x\) and \(y\) are both rational, (b) \(x\) and \(y\) are both irrational, (c) \(x\) is rational and \(y\) is irrational, and (d) \(x\) is irrational and \(y\) is rational. For case (a), if \(x, y \in \mathbb{Q}\), then \(f(x) = x\) and \(f(y) = y\), so \(x = y\). Similarly, for case (b), if \(x, y \notin \mathbb{Q}\), then \(f(x) = 1 - x\) and \(f(y) = 1 - y\), so \(1 - x = 1 - y\) which implies \(x = y\). For case (c) or (d), the equation \(f(x) = f(y)\) cannot hold since the left-hand side will be rational while the right-hand side will be irrational (or vice versa), which is a contradiction. Hence, \(f\) is injective.
02

Self-inverse function

To show that \(f\) is a self-inverse function, we have to prove that \(f(f(x)) = x\) for all \(x \in I=[0,1]\). Let's consider \(f(f(x))\): If \(x\) is rational, then \(f(x) = x\) and \(f(f(x)) = f(x) = x\); if \(x\) is irrational, then \(f(x) = 1 - x\) and \(f(f(x)) = f(1 - x)\). Notice that \(1 - x\) is rational if and only if \(x\) is rational, so \(f(1 - x) = 1 - (1 - x) = x\). In both cases, we have \(f(f(x)) = x\).
03

Point of continuity

To show that \(f\) is continuous only at the point \(x = 1/2\), remember that a function is continuous at a point if the limit as \(x\) approaches that point equals the function's value at that point. By the definition of \(f\), \(f(1/2) = 1/2\), irrespective of whether \(x\) is rational or irrational. So we only need to prove the limit. To do that, consider that for \(x \neq 1/2\), the values of \(f(x)\) when \(x\) is rational are different from when \(x\) is irrational. Thus, no matter how small the interval around \(x \neq 1/2\) is, the limit does not exist. However, for \(x = 1/2\), both rational and irrational values are equal to \(1/2\). Hence, \(f\) is continuous at \(x = 1/2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Injective Function
An injective function is a type of function that has a unique output for each input. It is also called a one-to-one function. To understand this concept, imagine a scenario where no two different inputs lead to the same output. This property ensures that the function maintains a sense of individuality among values in its domain.

Let's apply this to our exercise. For the given function, we need to confirm that whenever we pick two different values from the interval \[0,1\], say \(x\) and \(y\), and they produce the same output, that is, if \(f(x) = f(y)\), it implies that \(x = y\). Through careful analysis of all possible cases - both numbers being rational, both irrational, or one of each - we conclude that the function meets this criterion of injectivity. By ruling out the possibility of different inputs leading to the same output, we establish that the given function is indeed injective.
Self-inverse Function
A self-inverse function is a function that serves as its own inverse. This means when you apply the function twice, \(f(f(x))\), you'll get your original input \(x\) back. In simple terms, running the process forwards and then backwards gets you right to where you started.

In our exercise, we see that the designed function \(f\) takes rational numbers to themselves and maps irrational numbers to their reflective point across \(x = \frac{1}{2}\). When we apply the function again, we essentially

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(I \subseteq \mathbb{R}\) be an interval and let \(f: I \rightarrow \mathbb{R}\) be increasing on \(I\). Suppose that \(c \in I\) is not an endpoint of \(I\). Show that \(f\) is continuous at \(c\) if and only if there exists a sequence \(\left(x_{n}\right)\) in \(I\) such that \(x_{n}c\) for \(n=2,4,6, \ldots ;\) and such that \(c=\lim \left(x_{n}\right)\) and \(f(c)=\lim \left(f\left(x_{n}\right)\right)\)

Let \(\delta\) be a gauge on \(I:=[a, b]\) and suppose that \(I\) does not have a \(\delta\) -fine partition. (a) Let \(c:=\frac{1}{2}(a+b)\). Show that at least one of the intervals \([a, c]\) and \([c, b]\) does not have a \(\delta\) - fine partition. (b) Construct a nested sequence \(\left(I_{n}\right)\) of subintervals with the length of \(I_{n}\) equal to \((b-a) / 2^{n}\) such that \(I_{n}\) does not have a \(\delta\) -fine partition. (c) Let \(\xi \in \cap_{n=1}^{\infty} I_{n}\) and let \(p \in \mathbb{N}\) be such that \((b-a) / 2^{p}<\delta(\xi)\). Show that \(I_{p} \subseteq[\xi-\delta(\xi)\), \(\xi+\delta(\xi)]\), so the pair \(\left(I_{p}, \xi\right)\) is a \(\delta\) -fine partition of \(I_{p}\).

Show that the function \(f(x):=1 / x\) is uniformly continuous on the set \(A:=[a, \infty)\), where \(a\) is a positive constant.

Let \(I:=[a, b]\) and let \(f: I \rightarrow \mathbb{R}\) and \(g: I \rightarrow \mathbb{R}\) be continuous functions on \(I\). Show that the set \(E:=\\{x \in I: f(x)=g(x)\\}\) has the property that if \(\left(x_{n}\right) \subseteq E\) and \(x_{n} \rightarrow x_{0}\), then \(x_{0} \in E\)

Show that the function \(f(x):=2 \ln x+\sqrt{x}-2\) has root in the interval \([1,2]\), Use the Bisection Method and a calculator to find the root with error less than \(10^{-2}\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.