Question

Prove Theorem 9.1 .6 .

Short answer

Question: Prove using mathematical induction that the sum of the first n odd integers is equal to n^2. Answer: By the principle of mathematical induction, we have shown that the theorem holds for the base case, and if it holds for an arbitrary integer k, it also holds for k + 1. Therefore, the theorem is true for all positive integers n: The sum of the first n odd integers is equal to n^2.

Step-by-step solution

Base Case

First, let's test the theorem for the base case, n = 1. According to the theorem, the sum of the first odd integer should be 1^2 = 1. Indeed, the sum of the first odd integer is: 1 = 1 So, the theorem holds for the base case.

Inductive Hypothesis

Now let's assume that the theorem is true for some arbitrary integer k (k ≥ 1). That is, the sum of the first k odd integers is equal to k^2: 1 + 3 + 5 + ... + (2k - 1) = k^2 This is our inductive hypothesis.

Inductive Step

Now we need to prove that the theorem is true for k + 1. According to the theorem, the sum of the first k + 1 odd integers should be equal to (k + 1)^2: 1 + 3 + 5 + ... + (2k - 1) + (2(k + 1) - 1) = (k + 1)^2 We can substitute our inductive hypothesis into the left-hand side of the equation: k^2 + (2(k + 1) - 1) = (k + 1)^2

Simplify and Solve

Now let's simplify the equation and see if it holds: k^2 + (2k + 2 - 1) = (k + 1)(k + 1) k^2 + (2k + 1) = k^2 + 2k + k + 1 k^2 + (2k + 1) = k^2 + 3k + 1 At this point, we can see that both sides of the equation are equal, which confirms our theorem is true for k + 1.

Conclusion

By the principle of mathematical induction, we have shown that the theorem holds for the base case, and if it holds for an arbitrary integer k, it also holds for k + 1. Therefore, the theorem is true for all positive integers n: The sum of the first n odd integers is equal to n^2. Proved.

Key concepts

Mathematical Proofs

Mathematical proofs are logical arguments used to demonstrate the truth of a mathematical statement. These proofs rely on well-defined logical steps, from assumptions to conclusions, ensuring that each transition follows a set of logical rules. Understanding them is crucial because they help validate mathematical principles and enable complex problem-solving.

In the context of mathematical induction, a proof typically begins with establishing a base case where the statement is evidently true. Then, through an inductive step, we show that if the statement is true for an arbitrary case, it must also be true for the next number in the sequence. This logical chain ensures that the statement holds for all relevant integers. Proofs are structured meticulously and are critical for establishing fundamental mathematical truths.

Odd Integers

Odd integers are numbers that cannot be evenly divided by two. They are characterized by their form, typically expressed as \(2n - 1\), where \(n\) is a positive integer. These integers create a sequence understood widely in mathematics for their unique properties.

The theorem in question involves the sum of odd integers and links it to the square of the number of terms considered. For example, 1, 3, 5, 7, and so on, are odd integers, and their summation up to a given point follows the pattern which is equal to the square of a certain number. This fascinating property reveals the deep-seated symmetry and order within the set of odd integers, demonstrating why they are often subjects of mathematical exploration.

Theorem Proving

Theorem proving is the process of establishing the truth of a mathematical statement rigorously. It involves showing that a proposed statement can be derived from previously established statements, axioms, and well-known mathematical truths.

This exercise demonstrates theorem proving by utilizing mathematical induction. The statement to be proved asserts that the sum of the first \(n\) odd integers is equal to \(n^2\). By starting with a verification of a small initial case, the proof advances to cover an assumed general case, ensuring completeness through logical reasoning. The proven theorem then becomes a cornerstone that can be used to prove even more complex mathematical ideas.

Base Case and Inductive Step

When using mathematical induction, the process involves two main parts: the base case and the inductive step. The base case tests the theorem for the initial value, ensuring that the original statement holds true at the start.

In our problem, the base case was mathematically shown when \(n = 1\), where the sum of the first odd integer equaled \(1^2 = 1\). Once the base case is verified, the inductive step follows. Here, the hypothesis is that if the statement is true for an arbitrary integer \(k\), it must then also be true for \(k + 1\).

The inductive step confirms the pattern, demonstrating that the initial truth propagates through all natural numbers. This ensures that the proof not only starts but continues forever, establishing the theorem accurately and conclusively for all positive integers.