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Chapter 1: Problem 13

Prove that \(n<2^{n}\) for all \(n \in \mathbb{N}\)

Short Answer

Expert verified
By using mathematical induction, it has been shown that the inequality \(n<2^{n}\) holds true for all natural numbers \(n\). The base case was shown to hold true, and the inductive step also held true under the assumption of the inductive hypothesis. So the proposition is proven true.

Step by step solution

01

Base Case

Prove the statement for the base case, \(n = 1\). We have \(1 < 2^{1}\) or \(1 < 2\), which is true. Hence, the inequality holds for the base case.
02

Inductive Hypothesis

For some \(k \in \mathbb{N}\), assume the statement is true, i.e., \(k<2^{k}\). We'll use this assumption to prove that the statement is true for \(k + 1\).
03

Inductive Step

We prove the statement for \(n = k + 1\). So, we need to prove that \(k + 1 < 2^{k+1}\). Now, we know from the inductive hypothesis that \(k < 2^{k}\). This means \(k + 1 < 2^{k} + 1\). Because \(2^{k} + 1 < 2^{k} + 2^{k}\) or \(2^{k} + 1 < 2^{k+1}\), we have \(k + 1 < 2^{k+1}\).

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Key Concepts

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

Proof Techniques
Understanding different proof techniques is crucial in mathematics to assert the validity of conjectures and theorems. One key technique is mathematical induction, which is used to prove statements that are asserted to be true for all natural numbers. It consists of two main steps: the base case, where the statement is verified for the initial value (usually when n is 1); and the inductive step, where assuming the statement is true for some arbitrary natural number k, it is then shown to be true for k+1. This creates a domino effect, proving the statement for all subsequent natural numbers.

Inequalities
The concept of inequalities is another key topic in mathematics. Inequalities demonstrate the relative size or order of two values. They help us understand the bounds within which numbers lie, and they are represented by symbols like \<, \>, \<=, and \>=. When working with inequalities, it's important to remember that the operations performed on one side of the inequality must also be carried out on the other side to maintain the inequality's truth. Additionally, if you multiply or divide both sides of an inequality by a negative number, the direction of the inequality must be reversed. These properties are central when manipulating inequalities during proofs, such as in the exercise of proving that \(n<2^{n}\) for all natural numbers.

The manipulation of the original inequality during the inductive step showcases the elegance of using inequalities within proofs, where statements about relative magnitude can lead us to logical and concise conclusions.
Natural Numbers
The set of natural numbers, denoted by \(\mathbb{N}\), is the most basic and fundamental number system in mathematics. It includes all positive integers starting from 1 and proceeding indefinitely: 1, 2, 3, 4, and so on. Natural numbers are used for counting and ordering, making them indispensable in mathematics. A significant concept to note is that mathematical induction relies exclusively on the nature of these numbers, as it is a technique designed for proving statements that are supposed to hold true within the bounds of \(\mathbb{N}\). The concept that for every natural number, there's a subsequent number (\(k+1\) in the case of induction), contains the progression principle that induction is built upon.

Understanding natural numbers and their properties are essential for grasping the foundation of mathematical theories and their proof techniques, including those used in our exercise related to inequalities.

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Most popular questions from this chapter

Let \(f: A \rightarrow B\) and \(g: B \rightarrow C\) be functions. (a) Show that if \(g \circ f\) is injective, then \(f\) is injective. (b) Show that if \(g \circ f\) is surjective, then \(g\) is surjective.

Prove that \(3+11+\cdots+(8 n-5)=4 n^{2}-n\) for all \(n \in \mathbb{N}\).

Give an example of two functions \(f, g\) on \(\mathbb{R}\) to \(\mathbb{R}\) such that \(f \neq g\), but such that \(f \circ g=g \circ f\).

(a) Suppose that \(f\) is an injection. Show that \(f^{-1}\) o \(f(x)=x\) for all \(x \in D(f)\) and that \(f \circ f^{-1}(y)=y\) for all \(y \in R(f)\) (b) If \(f\) is a bijection of \(A\) onto \(B\), show that \(f^{-1}\) is a bijection of \(B\) onto \(A\).

Prove that \(1 / 1 \cdot 2+1 / 2 \cdot 3+\cdots+1 / n(n+1)=n /(n+1)\) for all \(n \in \mathbb{N}\).
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