Problem 33 - College Algebra

Question

Some statements are false for the first few positive integers, but true for some positive integer on. In these instances, you can prove \(S_{n}\) for \(n \geq k\) by showing that \(S_{k}\) is true and that \(S_{k}\) implies \(S_{k+1} .\) Use this extended principle of mathematical induction to prove that each statement in Exercises \(33-34\) is true. Prove that \(n^{2}>2 n+1\) for \(n \geq 3 .\) Show that the formula Fis true for \(n=3\) and then use step 2 of mathematical induction.

Step-by-Step Solution

Answer

Following the principles of mathematical induction, it has been proven that the inequality \(n^{2} > 2n+1\) holds for all positive integers larger than or equal to 3. The ground condition is true for \(n=3\), and if the inequality is true for any \(k\), it is true for the \(k+1\) integer as well.

Step 1: Prove for base case (n=3)

To form the base of induction, put \(n=3\) into the inequality. We get \(3^2 > 2*3 + 1\), which simplifies to \(9 > 7\). As this statement is true, we have verified the ground condition.

Step 2: Assume for k

Assume that the inequality holds for some arbitrary positive integer \(k\), such that \(k^2 > 2k+1\). This is known as the induction hypothesis.

Step 3: Prove for k+1

Next, we need to show that if the inequality holds for \(k\), then it will also hold for the next integer, \(k+1\). To do that, we express \((k+1)^2\) as \(k^2 + 2k + 1\) and prove that it should be greater than \(2(k+1) + 1\). Starts with \((k+1)^2 > 2(k+1) + 1\), substituting values gets \(k^2 + 2k + 1 > 2k + 3\). Since we assumed \(k^2 > 2k+1\), we can see that \(k^2 + 2k + 1 > 2k+1 + 2\) which simplifies to \(k^2 > 2k + 3\). Therefore, the inequality holds for the next integer, \(k+1\), thus the mathematical induction has been proved.

Key Concepts

Understanding Inequalities

Inequalities are expressions that describe the relative size or order of two values. In mathematics, they serve as a way to express that one quantity is larger or smaller than another. For example, the inequality \( n^2 > 2n + 1 \) indicates that for a certain range of positive integer values of \( n \), \( n^2 \) will always be greater than \( 2n + 1 \).
There are a few reasons why understanding inequalities is crucial:

• Inequalities are used to express conditions in various mathematical problems and solutions.
• They help determine the range of solutions and to predict possible outcomes in problems.
• In practical terms, they can express limits, like minimum or maximum requirements.

In the exercise, understanding the inequality \( n^2 > 2n + 1 \) allows us to see that starting from a specific point \( n = 3 \), the statement holds true.
Applying mathematical induction to proofs involving inequalities helps ensure that the inequality remains valid as \( n \) increases. This can be particularly useful in theoretical work or solving real-world applications where predictions are key.

Base Case

The base case in mathematical induction acts as the foundation on which the rest of the proof is built. Establishing a base case involves showing that a statement is true for the initial value of the sequence you are considering. In the given exercise, the base case is at \( n = 3 \).
The process of proving the base case typically involves the following:

• Substituting the base value into the inequality or equation.
• Verifying that the resulting statement is true.

For the inequality \( n^2 > 2n + 1 \) at \( n = 3 \), we substitute and find \( 3^2 > 2 \times 3 + 1 \). Simplifying gives \( 9 > 7 \), which is true. This verifies that the base case holds.
This step is crucial because it proves the specific instance from which we can extend our proof to larger instances. Without a valid base case, the entire induction process has no starting point, and therefore, cannot be applied.

Induction Hypothesis

The induction hypothesis is a critical component of a mathematical induction proof. It involves assuming that a statement is true for some positive integer \( k \) to prove it for \( k + 1 \).
In our exercise, the induction hypothesis states: Assume \( k^2 > 2k + 1 \) is true for some integer \( k \). This assumption forms the bridge to prove the next step.

• This allows us to "borrow" the truth of the statement for \( k \) and use that leverage to prove it for the next integer \( k+1 \).
• The hypothesis is not a conclusion itself, but a pivotal stepping-stone necessary for logical progression in the proof.

Mathematically, once we assume \( k^2 > 2k + 1 \), we need to show that \((k+1)^2 > 2(k+1) + 1\). This process usually involves algebraic manipulation to connect the induction hypothesis with the desired result.
This logical framework is what allows mathematical induction to "cascade" the truth from a single proven base case to an infinite number of cases, hence proving the statement universally for each case in its domain.

What Are Positive Integers?

Positive integers are the set of all natural numbers greater than zero. They are not fractions, decimals, or negatives, and are commonly denoted as \( \{1, 2, 3, \ldots\} \). In the context of mathematical induction and this exercise, they define the domain over which the inequality \( n^2 > 2n + 1 \) is to be proved.
When working with positive integers in mathematical proofs, it is important to keep in mind:

• They provide a structured, ordered set that can be dealt with in sequence, as induction requires.
• They serve as the "steps" in the building of induction proofs, moving from one integer to the next.
• Induction proofs often start from a smallest positive integer, in this case, \( n = 3 \).

The selection of positive integers ensures that the base case and induction step are both valid. Choosing positive integers as the focus allows us to use the neat properties that arise from their predictability in terms of sequence and value. This is fundamental in demonstrating universal truths across an infinite set of numbers through induction.