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Magazine Chapter 10: Problem 15
Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ n<2^{n} $$
Short Answer
Expert verified
The statement is true for all positive integers by induction.
Step by step solution
01
Understanding the Base Case
We begin by proving the base case for the smallest positive integer, which is often \(n = 1\). Substitute \(n = 1\) into the inequality to verify: \(1 < 2^{1}\). Calculating this gives us \(1 < 2\), which is true.
02
Establishing the Inductive Hypothesis
Assume that the statement \(n < 2^n\) holds true for some positive integer \(k\), such that \(k < 2^k\). This assumption is called the inductive hypothesis.
03
Proving the Inductive Step
Using the inductive hypothesis \(k < 2^k\), we need to prove that it holds true for \(k + 1\) as well, i.e., \(k+1 < 2^{k+1}\). Start by expressing \(2^{k+1}\) as \(2\times2^k\).
04
Demonstrating the Inductive Step
From the inductive hypothesis, we know \(k < 2^k\). Therefore, \(k+1 \leq k + 1< 2^k + 2^k = 2\times2^k = 2^{k+1}\). Hence, \(k+1 < 2^{k+1}\) is true.
05
Conclusion
Since the base case is true and the statement holds for \(k+1\) assuming it holds for \(k\), by mathematical induction, the statement \(n < 2^n\) is true for every positive integer \(n\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inequalities
Inequalities are mathematical expressions showing the relationship between two values, where one is larger or smaller than the other. They are represented by signs such as \(<\), \(>\), \(\leq\), and \(\geq\). In our specific exercise, the inequality \(n < 2^{n}\) is to be proven true for every positive integer \(n\). This particular expression highlights that the powers of 2 increase more rapidly than linear growth, represented by \(n\).
- The inequality tells us that as \(n\) grows, \(2^n\) grows much faster. This is fundamental in understanding why the statement holds true as \(n\) becomes larger.
- Inequalities enable us to compare different functions or numbers in terms of their size, which is crucial in various fields like optimization and analysis.
Base Case
The base case is the foundation of a mathematical induction proof. It is the initial step where we show the statement is true for the smallest possible value, typically \(n = 1\). For this inequality, substituting \(n = 1\) verifies it as \(1 < 2^1\) or \(1 < 2\), which is indeed true.
- Establishing the base case ensures that the statement holds at least for the beginning number, providing a starting point for our logical chain.
- Without a valid base case, the induction process cannot properly demonstrate the truth of the statement for all subsequent values.
Inductive Hypothesis
The inductive hypothesis is a key component in the method of induction. It involves assuming that a statement, like our inequality \(n < 2^n\), is true for some positive integer \(k\). This assumption is not a conclusion but a step in building our overall proof structure.
- By assuming the statement is true for \(k\), we can use this as a stepping stone to show it is also true for \(k + 1\).
- This logical assumption effectively allows us to "bridge the gap" between the base case and demonstrating the statement's truth for all natural numbers.
Inductive Step
The inductive step is where we apply our inductive hypothesis to prove that if the statement holds for a particular number \(k\), then it must also hold for \(k + 1\). In this exercise, we use the hypothesis \(k < 2^k\) to show that \(k + 1 < 2^{k+1}\).
- We begin by expressing \(2^{k+1}\) as \(2 \times 2^k\).
- From the inductive hypothesis, we have \(k < 2^k\), allowing us to state \(k + 1 < k + 2^k\), which further simplifies to \(k + 1 < 2^{k+1}\).
