Problem 19 - Precalculus Mathematics for Calculus

Question

Show that \(8^{n}-3^{n}\) is divisible by 5 for all natural numbers \(n\)

Step-by-Step Solution

Answer

By induction, \(8^n - 3^n\) is divisible by 5 for all natural \(n\).

Step 1: Base Case

We will start with the base case where \(n = 1\). We need to show that \(8^1 - 3^1\) is divisible by 5. Calculate \(8^1 - 3^1 = 8 - 3 = 5\). As 5 is divisible by 5, the base case holds.

Step 2: Inductive Hypothesis

Assume that for some natural number \(k\), \(8^k - 3^k\) is divisible by 5. This means there exists an integer \(m\) such that \(8^k - 3^k = 5m\).

Step 3: Inductive Step

We need to prove that \(8^{k+1} - 3^{k+1}\) is divisible by 5. First, express \(8^{k+1}\) and \(3^{k+1}\) in terms of 8 and 3 raised to the power \(k\): \(8^{k+1} = 8 \cdot 8^k\) and \(3^{k+1} = 3 \cdot 3^k\). Substitute these into the expression to get: \(8^{k+1} - 3^{k+1} = 8 \cdot 8^k - 3 \cdot 3^k\).

Step 4: Simplify Using Inductive Hypothesis

Substitute \(8^k - 3^k\) with \(5m\) from the inductive hypothesis: \(8 \cdot 8^k - 3 \cdot 3^k = 8(8^k) - 3(3^k)\). This can be rewritten as \([8^k - 3^k] + 7 \cdot 8^k - 2 \cdot 3^k\). Since \(8^k - 3^k = 5m\), further simplifying gives \(7 \times 8^k - 2 \times 3^k = 5p\) (where \(p\) is some integer). This results in the entire expression being divisible by 5, as it consists of two terms individually divisible by 5.

Step 5: Conclusion

By induction, since both the base case is true and the inductive step holds, \(8^n - 3^n\) is divisible by 5 for all natural numbers \(n\).

Key Concepts

Divisibility

Divisibility refers to the ability of one number to be divided by another without leaving a remainder. In the context of this exercise, we aim to demonstrate that the expression \(8^n - 3^n\) is divisible by 5, meaning \((8^n - 3^n) \div 5\) results in a whole number.
Mathematically, if a number \(a\) is divisible by another number \(b\), there exists an integer \(q\) such that \(a = bq\). For this exercise, our goal is to confirm that for any natural number \(n\), there is an integer \(m\) such that \((8^n - 3^n) = 5m\). This indicates that 5 is a factor of \(8^n - 3^n\).
Understanding divisibility is crucial in proving mathematical statements because it's a foundational concept in various areas, including number theory and algebra. It helps us verify the relationships and behaviors of numbers under different operations.
When exploring divisibility through mathematical induction, we strategically verify it for a starting number, then proceed to show that if it holds for one number, it will logically hold for the next.

Base Case

The base case is the first step in mathematical induction, where we verify the proposition for the smallest number in the set, usually \(n = 1\). This step ensures that the statement holds true at the beginning of the sequence.
In our exercise, the base case requires us to evaluate if \(8^1 - 3^1\) is divisible by 5. We compute this as follows:

• Calculate \(8^1 - 3^1\), which equals 8 - 3 = 5.
• Since 5 is divisible by 5, the base case is satisfied.

The base case serves as an essential foundation to build upon. Just like laying the first brick of a structure, this step confirms the validity of the statement as a starting point. If this step fails, the entire argument may collapse, emphasizing the importance of a strong base case.

Inductive Step

The inductive step is where we extend the truth of our proposition from one natural number to the next, proving its validity for all subsequent numbers. It involves two major components: the inductive hypothesis and the induction process itself.
The exercise incorporates this notion by assuming the proposition holds for some natural number \(k\). This is called the inductive hypothesis, which is:\(8^k - 3^k = 5m\), where \(m\) is an integer.
We must show that if the hypothesis is true for \(k\), it will be true for \(k+1\). This is done by evaluating \(8^{k+1} - 3^{k+1}\) and verifying its divisibility by 5:

• Express \(8^{k+1}\) and \(3^{k+1}\) as \(8 \cdot 8^k\) and \(3 \cdot 3^k\), respectively.
• Substitute these into the expression: \(8^{k+1} - 3^{k+1} = 8(8^k) - 3(3^k)\).
• Substitute \(8^k - 3^k = 5m\) into the expression, leading to an arrangement that validates divisibility.

The inductive step connects each stage of the progression, ensuring that if true for \(k\), it must be true for \(k+1\), thus covering all natural numbers.

Natural Numbers

Natural numbers are a set of positive integers starting from 1 and going upwards: 1, 2, 3, 4, and so forth. These numbers are foundational in mathematics and are often the domain of propositions in mathematical induction.
This exercise deals with proving the divisibility of the expression \(8^n - 3^n\) across all natural numbers \(n\). Natural numbers are crucial here as they help us ensure the proposition holds not just for an isolated case but for an ongoing sequence that includes every possible instance of \(n\).
When employing induction with natural numbers, we leverage their inherent property of having a starting point (the smallest natural number) and being unbounded (extending indefinitely). This allows us to validate a proposition extensively and concretely. As we progress sequentially through all natural numbers, we confirm the applicability and truth of our mathematical proposition universally within this set.