/*! 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 3 If \(x, y, z \in \mathbb{R}\) an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 3

If \(x, y, z \in \mathbb{R}\) and \(x \leq z\), show that \(x \leq y \leq z\) if and only if \(|x-y|+|y-z|=|x-z|\). Interpret this geometrically.

Short Answer

Expert verified
Thus we have proved that for any real numbers x, y, and z, \(x \leq y \leq z\) holds if and only if \(|x-y| + |y-z| = |x-z|\). This can be geometrically interpreted as 'the shortest distance between two points is a straight line', where detours increase the total distance.

Step by step solution

01

If \(x \leq y \leq z\) then \(|x-y|+|y-z| = |x-z|\)

Suppose that \(x \leq y \leq z\). We know that for any complex numbers a and b, |a-b| is the distance between a and b on the real number line. Therefore, |x-y| is the distance between x and y, |y-z| is the distance between y and z. So the sum |x-y| + |y-z| represents the total distance from x to z through y. Because \(x \leq y \leq z\), there's no 'back-tracking', so this is equivalent to the direct distance from x to z, which is |x-z|. Therefore, if \(x \leq y \leq z\), then \(|x - y| + |y - z| = |x - z|\).
02

If \(|x-y|+|y-z| = |x-z|\) then \(x \leq y \leq z\)

Suppose that \(|x-y| + |y-z| = |x-z|\). If y is not between x and z, then we are 'back-tracking' while going from x to z through y, therefore the total distance |x-y| + |y-z| should be greater than the direct distance |x-z|. But since \(|x-y| + |y-z| = |x-z|\), this means there's no 'back-tracking' and y must be between x and z. Therefore if \(|x-y| + |y-z| = |x-z|\), then \(x \leq y \leq z\).
03

Geometric Interpretation

In the real number line, consider three points x, y and z. If y is between x and z, the total distance of going from x to z through y is just the direct distance from x to z. Conversely, if y is not between x and z, going from x to z through y forms a 'detour', so the total distance is greater than the direct distance from x to z. This geometrically illustrates the proposition.

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!

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 \(S_{2}=\\{x \in \mathbb{R}: x>0\\} .\) Does \(S_{2}\) have lower bounds? Does \(S_{2}\) have upper bounds? Does inf \(S_{2}\) exist? Does sup \(S_{2}\) exist? Prove your statements.

Let \(X\) and \(Y\) be nonempty sets and let \(h: X \times Y \rightarrow \mathbb{R}\) have bounded range in \(\mathbb{R}\). Let \(f: X \rightarrow \mathbb{R}\) and \(g: Y \rightarrow \mathbb{R}\) be defined by $$ f(x):=\sup (h(x, y): y \in Y\\}, \quad g(y):=\inf \\{h(x, y): x \in X\\} $$ Prove that $$ \sup \\{g(y): y \in Y\\} \leq \inf \\{f(x): x \in X\\} $$ We sometimes express this by writing $$ \sup _{y} \inf _{x} h(x, y) \leq \inf _{x} \sup _{y} h(x, y) $$ Note that Exercises 8 and 9 show that the inequality may be either an equality or a strict inequality.

Let \(S\) be a nonempty subset of \(\mathbb{R}\) that is bounded below. Prove that inf \(S=-\sup (-s: s \in S\\}\).

Modify the argument in Theorem \(2.4 .7\) to show that there exists a positive real number \(y\) such that \(y^{2}=3\)

Let \(J_{n}:=(0,1 / n)\) for \(n \in \mathbb{N}\). Prove that \(\bigcap_{n=1}^{\infty} J_{n}=\emptyset\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.