Problem 19 - Precalculus: Mathematics for Calculus

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) \(8^{n}-3^{n}\) is divisible by 5 for all natural numbers \(n\)

Step-by-Step Solution

Answer

The expression \( 8^n - 3^n \) is divisible by 5 for all natural numbers \( n \).

Step 1: Base Case Verification

First, we verify the base case, where \( n=1 \). Calculate \( 8^{1} - 3^{1} = 8 - 3 = 5 \). Since 5 is divisible by 5, the formula holds true for \( n=1 \).

Step 2: Induction Hypothesis

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

Step 3: Induction Step

We need to prove that \( 8^{k+1} - 3^{k+1} \) is also divisible by 5. Start from the expression:\[8^{k+1} - 3^{k+1} = 8 \cdot 8^k - 3 \cdot 3^k\]

Step 4: Use Induction Hypothesis

Using the induction hypothesis \( 8^k = 5m + 3^k \), substitute it in to \(8^{k+1} - 3^{k+1}\):\[8^{k+1} = 8(5m + 3^k) = 40m + 8 \cdot 3^k\]So,\[8^{k+1} - 3^{k+1} = 40m + 8 \cdot 3^k - 3 \cdot 3^k = 40m + 5 \cdot 3^k\]

Step 5: Conclusion of Induction Step

From the expression in Step 4, \( 8^{k+1} - 3^{k+1} = 40m + 5 \cdot 3^k \), which is clearly divisible by 5, since both terms are multiples of 5.

Step 6: Final Conclusion

Since both the base case and the induction step have been successfully proved, by the principle of mathematical induction, \( 8^n - 3^n \) is divisible by 5 for all natural numbers \( n \).

Key Concepts

Natural Numbers

Natural numbers are the set of positive integers starting from 1. They are the numbers we often use in counting things, like 1, 2, 3, and so on. In mathematics, natural numbers play a crucial role as they form the basis of various operations and expressions.
Mathematically, the set of natural numbers is denoted by \( \mathbb{N} \). It is essential to understand that natural numbers do not include zero or any negative numbers when considering them in the context of mathematical induction exercises like the one at hand.
When dealing with natural numbers in proofs or equations, it is vital to confirm the truth of statements for all values that fall within this infinite set. In our problem, we start checking the divisibility for the smallest natural number, which is 1.

Divisibility

Divisibility is a fundamental concept in arithmetic and number theory, which assesses whether one number can be divided by another without leaving a remainder. In simpler terms, if a number \( a \) divides another number \( b \), \( b \) can be expressed as \( b = a \times k \), where \( k \) is an integer.
For the exercise, the goal is to show that the difference between \( 8^n \) and \( 3^n \) is divisible by 5 for all natural numbers \( n \).
To verify divisibility, we thoroughly compute expressions or use logical deductions as seen in the solution. When you replace terms like \( 8^k - 3^k = 5m \) in the step-by-step process, it means you have successfully shown that \( 8^k - 3^k \) divides perfectly by 5, proving divisibility.

Exponents

Exponents are a way of expressing repeated multiplication of a number by itself. For instance, \( 8^2 \) is 8 multiplied by itself, which is 64. In mathematical notation, an exponent of a base number like 8 in \( 8^n \) stands for the product 8 multiplied by itself \( n \) times.
Understanding exponents is crucial when dealing with expressions like \( 8^n - 3^n \), as our problem involves calculating powers of numbers. Working with exponents requires applying rules such as the product of powers property, where \( a^{m+n} = a^m \times a^n \), and the power of a power rule, \( (a^m)^n = a^{m \times n} \).
Mastering these principles helps in simplifying complex expressions and is central to proving statements in mathematical induction, as seen in transforming \( 8^{k+1} - 3^{k+1} \) to reach a conclusion.

Inductive Reasoning

Inductive reasoning in mathematics is a logical process where one determines the truth of an infinite number of cases through a finite number of steps. Mathematical induction is a powerful technique, often used to prove statements about natural numbers.
The process usually involves two main steps. First, we test the base case: for the smallest natural number, usually \( n=1 \), we confirm the statement is true. This is what we did in Step 1 of the solution above.
The second step, called the induction step, involves assuming that a statement holds for an arbitrary natural number \( k \), and then proving it for \( k+1 \). This covers an infinite sequence by linking each number's truth value to the previous one. This chain-like method proves a statement holds for all natural numbers once both steps are validated.
Inductive reasoning forms the backbone of the proof in the exercise, ensuring that the claim "\( 8^n - 3^n \) is divisible by 5 for all natural numbers \( n \)" is logically sound and complete.