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Chapter 1: Problem 41

Prove that at least one of the real numbers \(a_{1}, a_{2}, \ldots, a_{n}\) is greater than or equal to the average of these numbers. What kind of proof did you use?

Short Answer

Expert verified
At least one number must be greater than or equal to the average by proof of contradiction.

Step by step solution

01

- Define the average

The average of the numbers \(a_{1}, a_{2}, \ldots, a_{n}\) can be defined as \[ \text{Average} = \frac{a_{1} + a_{2} + \ldots + a_{n}}{n} \]
02

- Assume none of the numbers are greater than or equal to the average

Assume for the sake of contradiction that none of the numbers are greater than or equal to the average, meaning \[ a_{i} < \text{Average} \] for all \( i = 1, 2, \ldots, n \).
03

- Sum the inequalities

Sum up the inequalities \[ a_{1} + a_{2} + \ldots + a_{n} < n \times \text{Average} \]
04

- Substitute the average

Substitute the definition of the average into the inequality: \[ a_{1} + a_{2} + \ldots + a_{n} < n \times \frac{a_{1} + a_{2} + \ldots + a_{n}}{n} \]
05

- Simplify the inequality

Simplify the expression to \[ a_{1} + a_{2} + \ldots + a_{n} < a_{1} + a_{2} + \ldots + a_{n} \] which is a contradiction.
06

- Conclude the proof

Since our assumption leads to a contradiction, at least one of the numbers must be greater than or equal to the average. This proof technique is known as proof by contradiction.

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Key Concepts

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

Proof by Contradiction
To understand a proof by contradiction, let's break down its core idea. We start by assuming that the statement we want to prove is false. Then, we carefully follow logical steps to see where this assumption leads us. If this false assumption leads to an impossible or nonsensical conclusion, we call it a contradiction. Since the assumption is false, the original statement must be true.

In our exercise, we aimed to show that at least one of the numbers is greater than or equal to the average. To do this using proof by contradiction, we assumed the opposite: no number is greater than or equal to the average. Following this assumption, we arrived at a mathematical impossibility - a contradiction. Therefore, our initial assumption must be wrong, and hence our original statement is true.
Real Numbers
Real numbers are a fundamental concept in mathematics. They include all the numbers on the number line, both rational and irrational numbers. This broad category encompasses:

  • Natural numbers: 1, 2, 3, ...
  • Whole numbers: 0, 1, 2, 3, ...
  • Integers: -3, -2, -1, 0, 1, 2, 3, ...
  • Rational numbers: any number that can be expressed as a fraction, like 1/2, 3.75, etc.
  • Irrational numbers: numbers that cannot be expressed as fractions, such as \(\sqrt{2}\) and \(\pi\)

In this exercise, we dealt with real numbers in the form of \(\a_{1}, a_{2}, \ldots, a_{n}\). We used their properties to show certain relationships and to prove statements about them. The beauty of real numbers is that they allow for comprehensive mathematical operations, making them essential for almost all areas of mathematics.
Average
The average, also known as the mean, is a way of finding the central value of a set of numbers. To calculate the average, you sum all the numbers and then divide by the quantity of those numbers. The mathematical expression looks like this:\[ \text{Average} = \frac{a_{1} + a_{2} + \ldots + a_{n}}{n} \]

In the exercise, the average served as a crucial point of comparison. By setting up our proof, we compared each number \(\a_{i}\) to the average to examine their relationship. The aim was to show that at least one number is as large as or larger than the average. This led to a contradiction if assumed otherwise, confirming the average's role in balancing numerical values in a set.

Understanding averages helps in many practical situations, like calculating grades, determining central points in data sets, and more. It's a fundamental tool for analyzing a collection of values.

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Most popular questions from this chapter

For each of these sets of premises, what relevant conclusion or conclusions can be drawn? Explain the rules of inference used to obtain each conclusion from the premises. a) 鈥淚f I play hockey, then I am sore the next day.鈥� 鈥淚 use the whirlpool if I am sore.鈥� 鈥淚 did not use the whirlpool.鈥� b) 鈥淚f I work, it is either sunny or partly sunny.鈥� 鈥淚 worked last Monday or I worked last Friday.鈥� 鈥淚t was not sunny on Tuesday.鈥� 鈥淚t was not partly sunny on Friday.鈥� c) 鈥淎ll insects have six legs.鈥� 鈥淒ragonflies are insects.鈥� 鈥淪piders do not have six legs.鈥� 鈥淪piders eat dragon-flies.鈥� d) 鈥淓very student has an Internet account.鈥� 鈥淗omer does not have an Internet account.鈥� 鈥淢aggie has an Internet account.鈥� e) 鈥淎ll foods that are healthy to eat do not taste good.鈥� 鈥淭ofu is healthy to eat.鈥� 鈥淵ou only eat what tastes good.鈥� 鈥淵ou do not eat tofu.鈥� 鈥淐heeseburgers are not healthy to eat.鈥� f ) 鈥淚 am either dreaming or hallucinating.鈥� 鈥淚 am not dreaming.鈥� 鈥淚f I am hallucinating, I see elephants running down the road.鈥�

Let M(x, y) be 鈥渪 has sent y an e-mail message鈥� and T(x, y) be 鈥渪 has telephoned y,鈥� where the domain consists of all students in your class. Use quantifiers to express each of these statements. (Assume that all e-mail messages that were sent are received, which is not the way things often work.) a) Chou has never sent an e-mail message to Koko. b) Arlene has never sent an e-mail message to or tele- phoned Sarah. c) Jose has never received an e-mail message from Deborah. d) Every student in your class has sent an e-mail mes- sage to Ken. e) No one in your class has telephoned Nina. f ) Everyone in your class has either telephoned Avi or sent him an e-mail message. g) There is a student in your class who has sent every- one else in your class an e-mail message. h) There is someone in your class who has either sent an e-mail message or telephoned everyone else in your class. i) There are two different students in your class who have sent each other e-mail messages. j) There is a student who has sent himself or herself an e-mail message. k) There is a student in your class who has not received an e-mail message from anyone else in the class and who has not been called by any other student in the class. l) Every student in the class has either received an email message or received a telephone call from another student in the class. m) There are at least two students in your class such that one student has sent the other e-mail and the second student has telephoned the first student. n) There are two different students in your class who between them have sent an e-mail message to or telephoned everyone else in the class.

Use quantifiers and predicates with more than one variable to express these statements. a) Every computer science student needs a course in discrete mathematics. b) There is a student in this class who owns a personal computer. c) Every student in this class has taken at least one computer science course. d) There is a student in this class who has taken at least one course in computer science. e) Every student in this class has been in every building on campus. f ) There is a student in this class who has been in every room of at least one building on campus. g) Every student in this class has been in at least one room of every building on campus.

Translate these statements into English, where \(R(x)\) is "x is a rabbit" and \(H(x)\) is " \(x\) hops" and the domain consists of all animals. $$ \begin{array}{ll}{\text { a) } \forall x(R(x) \rightarrow H(x))} & {\text { b) } \forall x(R(x) \wedge H(x))} \\ {\text { c) } \quad \exists x(R(x) \rightarrow H(x))} & {\text { d) } \exists x(R(x) \wedge H(x))}\end{array} $$

Show that the equivalence \(p \wedge \neg p \equiv \mathbf{F}\) can be derived using resolution together with the fact that a conditional statement with a false hypothesis is true. [Hint: Let \(q=\) \(r=\mathbf{F}\) in resolution.
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