Problem 4 - Algebra 2

Question

Prove that each statement is true for all positive integers. \(4^{n}-1\) is divisible by 3

Step-by-Step Solution

Answer

\(4^n - 1\) is divisible by 3 for all positive integers, as proven by induction.

Step 1: Introduce the Inductive Proof Method

To prove that \( 4^n - 1 \) is divisible by 3 for all positive integers \( n \), we will use mathematical induction. This involves two key steps: base case and inductive step.

Step 2: Base Case Verification

First, verify the base case where \( n = 1 \). Substitute \( n = 1 \) into the expression to get \( 4^1 - 1 = 4 - 1 = 3 \). Clearly, 3 is divisible by 3.

Step 3: Inductive Hypothesis

Assume that the statement is true for some positive integer \( k \), i.e., \( 4^k - 1 \) is divisible by 3. In mathematical terms, this means there exists an integer \( m \) such that \( 4^k - 1 = 3m \).

Step 4: Inductive Step

We need to prove that \( 4^{k+1} - 1 \) is divisible by 3, assuming the inductive hypothesis is true. Start by expressing \( 4^{k+1} \) as \( 4 \times 4^k \). Then, consider \( 4^{k+1} - 1 = 4 \times 4^k - 1 = (3 + 1)4^k - 1 = 3 \times 4^k + 4^k - 1 \).

Step 5: Simplify the Inductive Step

Within the expression \( 3 \times 4^k + 4^k - 1 \), the term \( 3 \times 4^k \) is clearly divisible by 3. Based on our inductive hypothesis, \( 4^k - 1 \) is also equal to \( 3m \), which means \( 4^k - 1 \) is divisible by 3. Hence, \( 4^k = 3m + 1 \), substituting back results in \( 4^{k+1} - 1 = 3 \times 4^k + 4^k - 1 = 3 \times 4^k + 3m = 3(n+1) \), proving divisibility.

Step 6: Conclusion

Since the base case holds true, and assuming the hypothesis leads to the next case being true, by the principle of mathematical induction, \( 4^n - 1 \) is divisible by 3 for all positive integers \( n \).

Key Concepts

Divisibility

Divisibility is a fundamental concept in mathematics. It means that when one number, the dividend, is divided by another, the divisor, there is no remainder. In the problem of proving that \(4^n - 1\) is divisible by 3, we need to show that \(4^n - 1\) can be expressed as a multiple of 3 for all positive integers \(n\).

• This requires checking that the expression results in an integer when divided by 3.
• Being divisible by a number, like 3, means the expression forms an exact product using that number.
• When divisibility is true across a range of numbers, like all positive integers, it suggests a predictable pattern.

Understanding divisibility gives us a simplified look at the behavior of numbers, which is crucial in proofs like the one we're tackling.

Base Case

The base case is an essential first step in mathematical induction. It involves proving that the statement is true for an initial value, generally the smallest positive integer. In the exercise, the base case verifies the statement is true for \(n = 1\).

• We substitute \(n = 1\) in \(4^n - 1\) to find \(4 - 1 = 3\), and since 3 is divisible by 3, the base case holds.
• This initial check is crucial because it provides a foundation for the rest of the inducted proof.
• If the base case fails, the entire induction method will not work, as there's no assurance the property holds for any other numbers.

Thus, establishing the base case is essential before moving forward with complex proofs.

Inductive Step

The inductive step is the heart of mathematical induction. After confirming the base case, we must show that if the statement is true for some arbitrary positive integer \(k\), it must also be true for \(k + 1\). This step extends the truth of the statement to all subsequent numbers.
Here's how the process works:

• Assume \(4^k - 1\) is divisible by 3, implying we can express it as \(4^k - 1 = 3m\) for some integer \(m\).
• Then we look at \(4^{k+1} - 1\). This can be rewritten, using our assumption, to show it is a multiple of 3.
• As seen in the steps, we simplify \(4 \times 4^k - 1\) by manipulating the terms to show they form a multiple of 3.

By proving the inductive step, we've established that the statement holds from one integer to the next, supporting our overall proof.

Positive Integers

Positive integers are numbers greater than zero that have no fractional or decimal parts. In the context of this exercise, we're concerned with proving a statement true for all positive integers \(n\). These numbers are the building blocks of our number system and are essential in various mathematical theorems and proofs.

• Positive integers are \(1, 2, 3, 4, ...\).
• Each proves the base case and inductive step, which further verifies the statement for all following integers.
• Understanding their properties is vital for setting up initial values and extending the proof through induction.

Positive integers serve as an infinite sequence in mathematics, underscoring the universality of proofs when they apply to this set.