/*! 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 264 A collection of \(n\) intervals ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A collection of \(n\) intervals on the \(x\) -axis is such that every pair of intervals have a point in common. Prove that all \(n\) intervals must then have at least one point in common.

Short Answer

Expert verified
All \(n\) intervals have at least one point in common since the maximum start point and minimum end point still form a valid intersection.

Step by step solution

01

Understand the Problem

We have a set of intervals on the x-axis such that every pair of intervals shares a common point, and we need to prove that there is a single point that all intervals share.
02

Define Intervals

Let's denote the intervals as \(I_1, I_2, \ldots, I_n\), where each interval \(I_i\) is defined as \([a_i, b_i]\) on the x-axis.
03

Apply the Intersection Property

We know that for any two intervals \(I_i\) and \(I_j\), there exists at least one point \(x\) such that \(x \in I_i\) and \(x \in I_j\). This holds true for any pair from the collection.
04

Consider the Intersection of All Intervals

If every pair of intervals overlaps, it suggests there might be an overlapping region common to all. Formally, consider \(x_0 = \max(a_1, a_2, \ldots, a_n)\) and \(x_1 = \min(b_1, b_2, \ldots, b_n)\).
05

Check Validity of Common Point

For \(x_0\) to be less than or equal to \(x_1\), each interval must include this overlap, i.e., for all \(1 \leq i \leq n\), \(x_0 \leq b_i\) and \(x_1 \geq a_i\). Since each \(x_0\) is a starting point and \(x_1\) is an ending point within all intervals, \([x_0, x_1]\) is the intersection of all intervals.
06

Conclude with the Common Point

The interval \([x_0, x_1]\) represents the common intersection of all intervals. Thus, all \(n\) intervals share at least one point, satisfying the requirement.

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Key Concepts

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

Common Point
When dealing with intervals on the x-axis, a **common point** refers to a specific location that all intervals simultaneously cover. In this problem, we're tasked with proving that such a shared point exists for a collection of intervals. Imagine each interval as a horizontal line on a graph, stretching from one number to another. If every two intervals share at least one point of overlap, then intuitively, all of them may overlap at a single point too.
  • A common point is a single x-value present in every interval.
  • This occurs when all intervals intersect somewhere along the line.
Finding this common point relies on understanding where individual interval intersections overlap. By examining the maximum of all the starting points and the minimum of all the ending points across intervals, we can pinpoint the dense core where intersections occur.
X-axis Intervals
Intervals on the **x-axis** are segments defined by two points, the start, and end. These segments are represented by \([a_i, b_i]\) for each interval \(I_i\). Here, \(a_i\) is where the interval starts on the x-axis, and \(b_i\) is where it ends. These intervals can be thought of as little highway stretches on a number line. Each one covers a specific range of x-values.
  • Intervals help in visualizing ranges of inclusion on a line graph.
  • The size and position of an interval depend on its start and endpoint.
Understanding how these intervals are positioned along the x-axis allows us to assert the presence of a shared x-axis section among them.
Pairwise Intersection
In terms of intervals, a **pairwise intersection** means that for every possible pair of intervals, there is at least one x-value that they both include. This concept is like having a pair of roads that cross or meet each other at some point. If we consider any two intervals in our set, their overlap guarantees a spot that they both contain, offering a bridge to uniting all intervals together.
  • At least one shared point must exist for each pair of intervals.
  • This repeated sharing is crucial to finding a universal common point.
By observing this repeated intersection across all pairs, we can hypothesize the existence of a single intersection point that all intervals must share.
Elementary Proof
An **elementary proof** is a basic yet complete demonstration of how a statement or theorem holds true based on logical reasoning and existing mathematical principles. In the context of intervals, we need to show that if every pairwise intersection exists, there must indeed be a common point for all intervals.The process involves:
  • Listing each interval's start and endpoint \([a_i, b_i]\).
  • Determining the maximum starting point \(x_0 = \max(a_1, a_2, \ldots, a_n)\) and the minimum endpoint \(x_1 = \min(b_1, b_2, \ldots, b_n)\).
  • Checking that this condition \(x_0 \leq x_1\) holds, signifying all intervals overlap at or between \(x_0\) and \(x_1\).
These steps provide a logical conclusion that confirms the presence of an overall common point across all intervals, validating our elementary proof.

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Most popular questions from this chapter

(Gray codes) There are \(2^{n}\) sequences of length \(n\) consisting of 0 s and 1 s. Prove that, for each \(n \geqslant 2,\) these sequences can be arranged in a cyclic list such that any two neighbouring sequences (including the last and the first) differ in exactly one coordinate position.

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