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Chapter 1: Problem 12

Prove the Principle of Mathematical Induction \(1.2 .3\) (second version).

Short Answer

Expert verified
The proof relies on demonstrating the principle holds for the base case \(n = 1\), and then showing by induction that if the principle holds for \(n = k\), it also holds for \(n = k + 1\). The principle of mathematical induction, second version, hence, holds true for all positive integers.

Step by step solution

01

Initial Step

To apply the Principle of Mathematical Induction, the first step is to prove the base case. The base case for this problem is typically \(n = 1\). This step involves demonstrating that when our variable \(n\) is equal to 1, our principle holds true.
02

Inductive Step

The second step is inductive. During this step, we make the inductive hypothesis. This means that we assume our principle works for \(n = k\) for some arbitrary positive integer, \(k\). This is not the same as proving the principle is true for every positive integer, but we just assume it's true for a sample positive integer, \(k\). This is the assumption step.
03

Apply the assumption

The third step involves proving that if the statement is true for \(n = k\), then it must also be true for \(n = k + 1\). Here, we attempt to demonstrate that the principle being true for the first \'k\' elements implies that it is also true for the \'(k + 1)\'th element.
04

Conclusion

When we have demonstrated that the principle holds for both the base case, and that the inductive step holds true (if it's true for an integer \(n = k\), then it's also true for \(n = k + 1\)), we can conclude that the principle holds true for all positive integers.

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Key Concepts

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

Base Case
When utilizing the Principle of Mathematical Induction, the process begins with the **Base Case**. This is the simplest form of the problem, usually starting with the smallest positive integer, often \( n = 1 \). At this initial level, you need to verify that the statement or property in question holds true for this specific value.

Imagine you're trying to understand a pattern. Before extrapolating it to all cases, checking with the first few numbers provides a foundation. If the base simple case doesn鈥檛 hold true, the entire principle cannot be applied. This makes this step essential.

If you were proving something about odd numbers, for the base case at \( n = 1 \), demonstrate: is 1 an odd number according to your property? If yes, then this supports your argument to move forward. In essence, the Base Case acts as the small domino that starts the fall of all others in the sequence.
Inductive Hypothesis
The next critical step is the **Inductive Hypothesis**. After confirming the Base Case, we assume the property or statement is valid for some arbitrary positive integer \( n = k \). It might feel counterintuitive to "assume" without proof, but this is just a strategic part of the induction process.

Consider this step the bridge between single instances and a generalized conclusion. You're hypothesizing that based on the base case and expected sequence, this assertion holds for the case \( n = k \). This isn't blind faith but a logical assumption grounded in observed patterns. This hypothesis forms the backbone for the next part of the method鈥攖he Inductive Step.

The beauty here is in trusting the logical sequence, almost like assuming the reliability of a friend based on past interactions, before putting them to the test in new situations.
Inductive Step
Now comes the crucial **Inductive Step**. With our hypothesis in place, the goal is to prove that if the statement holds true for \( n = k \), it should also hold for \( n = k + 1 \). It's like climbing a ladder鈥攊f you can safely get to one rung, can you also reach the next one?

To do this, we turn the problem into logical algebraic expressions. Assume you proved it true for \( n = k \), now using analytical manipulation, show it's true for \( n = k + 1 \).

This step is vital鈥攊t ensures that the property holds not just for the first few numbers but universally, extending that assurance to all positive integers.
  • Verify some terms algebraically.
  • Use smart substitutions based on your hypothesis.
  • Conclude with confidence once \( n = k + 1 \) works out.
Once this logical throughway is proven, you have completed your induction, showing that the property or principle holds for all applicable numbers.

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Most popular questions from this chapter

Prove that \(1^{3}+2^{3}+\cdots+n^{3}=\left[\frac{1}{2} n(n+1)\right]^{2}\) for all \(n \in \mathbb{N}\).

For each \(n \in \mathbb{N}\), let \(A_{n}=\\{(n+1) k: k \in \mathbb{N}\\}\) (a) What is \(A_{1} \cap A_{2}\) ? (b) Determine the sets \(\cup\left\\{A_{n}: n \in \mathbb{N}\right\\}\) and \(\cap\left\\{A_{n}: n \in \mathbb{N}\right\\}\).

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

Prove that a set \(T_{1}\) is denumerable if and only if there is a bijection from \(T_{1}\) onto a denumerable set \(T_{2}\)

Show that if \(f: A \rightarrow B\) and \(E, F\) are subsets of \(A\), then \(f(E \cup F)=f(E) \cup f(F)\) and \(f(E \cap F) \subseteq f(E) \cap f(F)\)
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