Problem 24 - Precalculus: Mathematics for Calculus

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) $$100 n \leq n^{2} \text { for all } n \geq 100$$

Step-by-Step Solution

Answer

By induction, the inequality holds for all \(n \geq 100\).

Step 1: Base Case Verification

We need to verify that the inequality \(100n \leq n^2\) holds true for \(n = 100\). Substitute 100 into the inequality:\[ 100(100) \leq 100^2 \]\[ 10000 \leq 10000 \]This is true, so the base case holds for \(n = 100\).

Step 2: Inductive Hypothesis

Assume that the inequality \(100k \leq k^2\) is true for some natural number \(k \geq 100\). This is our inductive hypothesis.

Step 3: Inductive Step

We need to prove that the inequality \(100(k+1) \leq (k+1)^2\) also holds. Start with:\[ 100(k+1) = 100k + 100 \]Consider the expression for \((k+1)^2\):\[ (k+1)^2 = k^2 + 2k + 1 \]

Step 4: Comparing Inequality Sides

From the inductive hypothesis, we know \(100k \leq k^2\). We can add \(100\) and \(2k + 1\) to both sides:\[ 100k + 100 \leq k^2 + 100 \]\[ k^2 + 100 \leq k^2 + 2k + 1 \]

Step 5: Verifying Results

To verify if the step is correct, simplify \(k^2 + 100 \leq k^2 + 2k + 1\):Subtract \(k^2\) from both sides:\[ 100 \leq 2k + 1 \]Re-arrange to find \(k\):\[ 99 \leq 2k \]\[ 49.5 \leq k \]Since \(k \geq 100\), this inequality holds, so \(100(k+1) \leq (k+1)^2\) is true.

Step 6: Conclusion

Since the base case holds, and assuming the inequality true for \(k\) implies it is also true for \(k+1\), by mathematical induction, the inequality \(100n \leq n^2\) is true for all natural numbers \(n \geq 100\).

Key Concepts

Understanding Inequality in Mathematical Induction

In mathematics, an inequality is a statement that expresses the relative size or order of two values. It is a condition of being unequal, and a typical example is the inequality presented as \(100n \leq n^2\). This specific inequality essentially compares two expressions: one linear (\(100n\)) and one quadratic (\(n^2\)). For this exercise, we are examining whether for all natural numbers \(n\) greater than or equal to 100, the quadratic expression is larger or equal to the linear one.

In simple terms:

• The process is about proving mathematically that no matter what large number \(n\) gets chosen as long as \(n\) is \(\geq 100\), the equation holds true.
• It highlights the shift from linear growth to quadratic growth, with quadratic terms growing faster than linear terms as \(n\) increases.

The inequality forms the basis for establishing a pattern or property across a range of numbers, particularly focusing on natural numbers beginning from 100 going upwards.

Exploring Natural Numbers in Induction

Natural numbers are simply the set of positive integers beginning from 1 and continuing infinitely. They are the most basic numbers we use to count (like 1, 2, 3, and onward). However, for this specific problem, we start from 100 instead of 1 purely due to the nature of the inequality itself, as it's not valid for numbers smaller than 100.

When performing induction with natural numbers, it's crucial to:

• Understand that they naturally form a sequence that is unending.
• They serve as a foundation in mathematical proofs where a property is to be established for all numbers within this infinite set, starting from a specific point (in this case, \(n \geq 100\)).
• Ensure that induction is suitable to the sequence we are considering, especially when upper or lower bounds like those given in this problem are involved.

Through this focus on natural numbers, mathematical induction allows us to traverse this infinite sequence and confirm properties for every relevant number effectively and convincingly.

The Role of Inductive Hypothesis in Mathematical Induction

Mathematical induction is a systematic way of proving that a statement holds true for all natural numbers. An important component of this method is the inductive hypothesis. The hypothesis involves assuming that a statement, usually an equation or inequality, is true for a particular natural number \(k\).

The steps are structured as follows:

• Base Case: Start by proving the statement for the initial number in our natural number sequence (in our problem, this is \(n = 100\)).
• Inductive Hypothesis: Assume the statement is true for an arbitrary natural number \(k\). Here, it's crucial as it becomes our jumping-off point to prove the succeeding case.
• Inductive Step: Demonstrate that if the statement holds for \(k\), it must also be true for \(k+1\). This step relies heavily on the validity of the inductive hypothesis to establish a chain of truths through the natural numbers.

This thoughtful process naturally extends the truth from one number to every other number beyond it in the sequence, setting the foundation for proofs of statements involving infinitely continuing numbers.