Problem 20 - Precalculus: Mathematics for Calculus

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) \(3^{2 n}-1\) is divisible by 8 for all natural numbers \(n\)

Step-by-Step Solution

Answer

The formula \(3^{2n} - 1\) is divisible by 8 for all natural numbers \(n\).

Step 1: Base Case

To begin, test the base case, which is the smallest natural number, usually \(n = 1\). Substitute \(n = 1\) into the expression \(3^{2n} - 1\):\[3^{2 \cdot 1} - 1 = 3^2 - 1 = 9 - 1 = 8\]Since 8 is divisible by 8, the base case holds true.

Step 2: Inductive Hypothesis

Assume that for some arbitrary natural number \(k\), the expression \(3^{2k} - 1\) is divisible by 8. This is known as the inductive hypothesis:\[3^{2k} - 1 = 8m \quad \text{for some integer } m.\]

Step 3: Inductive Step

Under the inductive hypothesis, we need to prove that \(3^{2(k+1)} - 1\) is also divisible by 8. Start by expressing \(3^{2(k+1)} - 1\) in terms of \(3^{2k}\):\[3^{2(k+1)} - 1 = 3^{2k + 2} - 1 = 3^{2k} \cdot 3^2 - 1.\]Rewrite as:\[=(3^{2k} \cdot 9) - 1 = 9 imes 3^{2k} - 1.\]

Step 4: Factor and Substitute

Using the inductive hypothesis, substitute \(3^{2k} = 8m + 1\) into the equation derived above:\[9 imes 3^{2k} - 1 = 9(8m + 1) - 1.\]Distribute and simplify:\[= 72m + 9 - 1 = 72m + 8.\]

Step 5: Conclude Divisibility by 8

Notice that \(72m + 8\) can be rewritten as:\[= 8(9m + 1).\]Since \(8(9m + 1)\) is clearly divisible by 8, the expression \(3^{2(k+1)} - 1\) is divisible by 8. Thus, the inductive step holds true.

Step 6: Conclusion

Since the base case is true and the inductive step has been verified, by mathematical induction, the formula \(3^{2n} - 1\) is divisible by 8 for all natural numbers \(n\).

Key Concepts

Divisibility

Divisibility is a concept in mathematics that helps determine whether one number can be divided by another with no remainder. In the exercise, the expression \(3^{2n} - 1\) was analyzed to check if it is divisible by 8 for all natural numbers \(n\). Here, divisibility by 8 means that when you divide \(3^{2n} - 1\) by 8, you get a whole number with zero remainder.
An easy way to check divisibility is by performing division and seeing if the remainder is zero. In the case of \(3^{2n} - 1\), showing that the expression can be rewritten as a product that includes 8 ensures divisibility. In mathematical induction, divisibility is often part of proving statements for all natural numbers.

Base Case

The base case in mathematical induction is the first step where you verify the statement for the smallest value in the domain, usually \(n = 1\). To confirm the statement holds true in this step, you substitute \(n = 1\) into the formula \(3^{2n} - 1\).
When doing this, you calculate:

• Substitute 1: \(3^{2 \cdot 1} - 1 = 3^2 - 1 = 9 - 1 = 8\)
• Confirm 8 is divisible by 8, establishing the statement is true for \(n = 1\).

Without a true base case, the entire proof would fall apart, so this step is crucial before moving on to the next.

Inductive Hypothesis

The inductive hypothesis is the second step in mathematical induction. It involves assuming that the statement is true for some arbitrary natural number \(k\). This assumption forms the foundation for the next steps.
In the exercise, you assume \(3^{2k} - 1\) is divisible by 8 for a particular \(k\), meaning:

• \(3^{2k} - 1 = 8m\) for some integer \(m\).

This assumption doesn't prove the statement itself but sets up conditions to extend the proof to the next number. The correctness of the proof hinges on skillfully using this assumption in the following inductive step.

Inductive Step

The inductive step is where you use the inductive hypothesis to prove the statement for \(k + 1\). This part ensures that if the statement holds for one number, it will hold for the next one as well, thereby proving it for all natural numbers.
In the exercise, you must prove \(3^{2(k+1)} - 1\) is divisible by 8. Following are the actions performed in this step:

• Express \(3^{2(k+1)} - 1\) as \(9 \times 3^{2k} - 1\).
• Substitute the inductive hypothesis: \(3^{2k} = 8m + 1\).
• Simplify to show \(9(8m + 1) - 1 = 72m + 8\), which equals \(8(9m+1)\), confirming divisibility by 8.

Thus, the inductive step affirms that if the formula holds for \(n = k\), it also holds for \(n = k+1\). This step solidifies the inductive reasoning, completing the full proof by induction.