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Chapter 2: Problem 21

(a) Prove there is no \(n \in \mathbb{N}\) such that \(0

Short Answer

Expert verified
There is no natural number \(n\) such that \(0<n<1\) due to the Well-Ordering Property of the natural numbers, which guarantees that the smallest element in any non-empty set of natural numbers is 1. Also, it's impossible for any natural number to be both even and odd simultaneously, because the definitions of even and odd numbers are mutually exclusive.

Step by step solution

01

Prove there is no \(n \in \mathbb{N}\) such that \(0

Firstly, we know the Well-Ordering Property states that every non-empty set of natural numbers contains a least element. In this case, for any natural number \(n\), \(n\) must be greater than or equivalent to 1 because 1 is the least natural number by definition. So, there are no natural numbers \(n\) that can satisfy the condition \(0<n<1\). Alternatively, as \(n\) is an integer and the only integer that lies in the interval (0, 1) is 0. But 0 does not belong to \(\mathbb{N}\), so the statement is proved right.
02

Introduction to even and odd numbers

An even number is defined as a natural number \(n\) that can be written in the form \(n = 2k\) for some integer \(k\), and an odd number can be written in the form \(n = 2k + 1\) for some integer \(k\). These definitions are mutually exclusive; a number cannot be written in both forms at the same time.
03

Prove that no natural number can be both even and odd

Assume, for contradiction, that there is a number \(n\) that is both even and odd. Then we can write it in two ways: \(n = 2k\) and \(n = 2m + 1\) where \(k\) and \(m\) are integers. If we equate these two expressions, we have \(2k = 2m + 1\). This simplifies to \(2(k - m) = 1\). The left-hand side of this equation is an even number, as it is the multiplication of 2 by an integer, while the right-hand side is 1, which is an odd number. As we stated before, a number cannot be both even and odd at the same time. Consequently, the assumption that a number \(n\) can be both even and odd leads to a contradiction, which completes the proof.

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Key Concepts

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

Natural Numbers
Natural numbers are the set of positive integers starting from 1, 2, 3, and so on. They are sometimes denoted by the symbol \( \mathbb{N} \), representing a form of counting numbers. These numbers do not include zero, negative numbers, or any fractions and decimals. Because of their whole nature, they help us in forming the foundation of mathematics.

A key property of natural numbers is the **Well-Ordering Property**. This principle states that any non-empty set of natural numbers will have a least element. For instance, if you take any group of natural numbers, there will always be a smallest number in that group. Understanding this property is crucial for various proofs, including those that question whether a number less than one can ever exist in the set of natural numbers.
Even and Odd Numbers
For natural numbers, distinction into even and odd is fundamental. An **even number** is any number that can be expressed in the form \( n = 2k \), where \( k \) is an integer. This means it can be divided exactly by 2 without leaving a remainder. Examples include 2, 4, and 6.

On the flip side, an **odd number** is expressed as \( n = 2k + 1 \), also where \( k \) is an integer. These numbers leave a remainder of 1 when divided by 2. Examples include 1, 3, and 5.

It鈥檚 important to note that a number cannot be both even and odd as they have distinctly different definitions. Understanding these definitions helps us in proving mathematical statements that rely on the evenness or oddness of numbers.
Contradiction in Proofs
The method of proof by contradiction is a powerful tool in mathematics. This approach involves assuming that the opposite of what you want to prove is true, then deriving a logical contradiction from this assumption. Such contradictions indicate that the original assumption is false, therefore proving that the statement you set out to prove must be true.

This logical framework is widely used to prove the impossibility of certain scenarios, like how no natural number can be both even and odd. In this case, assuming there exists a number \( n \) that is both even and odd leads to a contradiction because it would need to satisfy two mutually exclusive conditions.

Contradictions in mathematical proofs help in eliminating various possibilities, making them highly effective in verifying the veracity of mathematical claims.

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Most popular questions from this chapter

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

Find all \(x \in \mathbb{R}\) that satisfy the inequality \(4<|x+2|+|x-1|<5\).

Find all values of \(x\) that satisfy the following equations: (a) \(x+1=|2 x-1|\), (b) \(2 x-1=|x-5|\).

Let \(S \subseteq \mathbb{R}\) and suppose that \(s^{*}:=\sup S\) belongs to \(S .\) If \(u \notin S\), show that \(\sup (S \cup\\{u\\})=\sup \left\\{s^{*}, u\right\\}\)

Let \(S \subseteq \mathbb{R}\) be nonempty. Prove that if a number \(u\) in \(\mathbb{R}\) has the properties: (i) for every \(n \in \mathbb{N}\) the number \(u-1 / n\) is not an upper bound of \(S\), and (ii) for every number \(n \in \mathbb{N}\) the number \(u+1 / n\) is an upper bound of \(S\), then \(u=\sup S\). (This is the converse of Exercise 2.3.9.)
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