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Chapter 1: Problem 1

Prove that \(n^{2}+1 \geq 2^{n}\) when \(n\) is a positive integer with \(1 \leq n \leq 4 .\)

Short Answer

Expert verified
The inequality holds for all positive integers from 1 to 4.

Step by step solution

01

Check for n = 1

Substitute n = 1 into the inequality. Verify that the inequality holds.\(n^2 + 1\geq 2^{n}\)For n = 1:\(1^2 + 1 = 1 + 1 = 2\)\(2^{1} = 2\)Thus, \(1^2 + 1 = 2\) and \(2^{1} = 2\), so \(2 \geq 2\) holds true.
02

Check for n = 2

Substitute n = 2 into the inequality. Verify that the inequality holds.\(n^2 + 1\geq 2^{n}\)For n = 2:\(2^2 + 1 = 4 + 1 = 5\)\(2^{2} = 4\)Thus, \(2^2 + 1 = 5\) and \(2^{2} = 4\), so \(5 \geq 4\) holds true.
03

Check for n = 3

Substitute n = 3 into the inequality. Verify that the inequality holds.\(n^2 + 1\geq 2^{n}\)For n = 3:\(3^2 + 1 = 9 + 1 = 10\)\(2^{3} = 8\)Thus, \(3^2 + 1 = 10\) and \(2^{3} = 8\), so \(10 \geq 8\) holds true.
04

Check for n = 4

Substitute n = 4 into the inequality. Verify that the inequality holds.\(n^2 + 1\geq 2^{n}\)For n = 4:\(4^2 + 1 = 16 + 1 = 17\)\(2^{4} = 16\)Thus, \(4^2 + 1 = 17\) and \(2^{4} = 16\), so \(17 \geq 16\) holds true.

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Key Concepts

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

positive integers
A positive integer is any whole number greater than zero. They are the numbers you typically count with and include numbers like 1, 2, 3, and so on.

Understanding the concept of positive integers is crucial in many areas of mathematics. They form the backbone of number theory and are often the first numbers we learn to work with.

In this exercise, all our values for \(n\) are positive integers. Specifically, we are asked to consider \(n\) values between 1 and 4. This is important because positive integers have unique properties that can affect how inequalities and other mathematical rules apply to them. For instance, since there are no fractions or decimals, each calculation involves straightforward whole numbers.
mathematical induction
Mathematical induction is a powerful proof technique used to establish the truth of an infinite number of statements. It's often compared to dominoes falling because proving one statement helps you prove the next, and so on.

The process of mathematical induction involves two main steps:
  • **Base Case**: Verify the statement is true for the initial value (often \(n = 1\)).
  • **Inductive Step**: Show that if the statement holds for some positive integer \(k\), then it also holds for \(k+1\).

This technique ensures that if the base case is true and the inductive step works, the statement is true for all positive integers starting from the base case.

In our exercise, proving the base cases for \(n = 1\) through \(n = 4\) uses a similar approach but without the full induction step. However, understanding induction helps you get why checking individual cases is sometimes necessary.
base case checking
Base case checking is the first step in mathematical induction. It establishes that the statement holds for the smallest value in the domain, often \(n = 1\). This step is crucial because it serves as the foundation for proving the statement for all subsequent values.

Here's a breakdown of the steps in our exercise:
  • For \(n = 1\), we substitute 1 into the inequality \(n^2 + 1 \geq 2^n\) and verify it holds since 2 equals 2.
  • For \(n = 2\), substituting 2 results in 5 being greater than 4, showing the inequality is true.
  • For \(n = 3\), we find 10 is greater than 8, again supporting the inequality.
  • For \(n = 4\), substituting 4 yields 17 greater than 16, proving the base case for all four values.

By confirming these cases, we build confidence that the inequality holds true, laying the groundwork for more comprehensive proofs. Base case checking is an essential part of ensuring logical consistency in mathematical arguments.

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Most popular questions from this chapter

Exercises \(61-64\) are based on questions found in the book Symbolic Logic by Lewis Carroll. Let P(x), Q(x), R(x), and S(x) be the statements 鈥渪 is a baby,鈥� 鈥渪 is logical,鈥� 鈥渪 is able to manage a crocodile,鈥� and 鈥渪 is despised,鈥� respectively. Suppose that the domain consists of all people. Express each of these statements using quantifiers; logical connectives; and P(x), Q(x), R(x), and S(x). a) Babies are illogical. b) Nobody is despised who can manage a crocodile. c) Illogical persons are despised. d) Babies cannot manage crocodiles. e) Does (d) follow from (a), (b), and (c)? If not, is there a correct conclusion?

Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) Something is not in the correct place. b) All tools are in the correct place and are in excellent condition. c) Everything is in the correct place and in excellent condition. d) Nothing is in the correct place and is in excellent condition. e) One of your tools is not in the correct place, but it is in excellent condition.

Prove or disprove that you can use dominoes to tile a \(5 \times 5\) checkerboard with three corners removed.

Translate in two ways each of these statements into logical expressions using predicates, quantifiers, and logical connectives. First, let the domain consist of the students in your class and second, let it consist of all people. a) Someone in your class can speak Hindi. b) Everyone in your class is friendly. c) There is a person in your class who was not born in California. d) A student in your class has been in a movie. e) No student in your class has taken a course in logic programming.

Express the negations of these propositions using quantifiers, and in English. a) Every student in this class likes mathematics. b) There is a student in this class who has never seen a computer. c) There is a student in this class who has taken every mathematics course offered at this school. d) There is a student in this class who has been in at least one room of every building on campus.
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