Problem 25 - Algebra for College Students

Question

\(9^{n}-1\) is divisible by 8 for all positive integer values for \(n\).

Step-by-Step Solution

Answer

Yes, \(9^n - 1\) is divisible by 8 for all positive integers \(n\).

Step 1: Verify for Base Case

To prove that \(9^n - 1\) is divisible by 8 for all positive integers, we first check the base case when \(n = 1\). Calculate \(9^1 - 1 = 8\). Since 8 is divisible by 8, the base case holds.

Step 2: Assume for Inductive Hypothesis

Assume for some integer \(k\), \(9^k - 1\) is divisible by 8. This means \(9^k - 1 = 8m\) for some integer \(m\).

Step 3: Prove for \(n = k + 1\)

To prove for \(n = k + 1\), consider \(9^{k+1} - 1\). We can express this as \(9 \cdot 9^k - 1\). Rewrite it as \((9 \cdot 9^k - 9) + 8 = 9(9^k - 1) + 8\).

Step 4: Apply Inductive Hypothesis

By the inductive hypothesis, we know \(9^k - 1 = 8m\). Substitute this into the expression: \(9(8m) + 8 = 72m + 8 = 8(9m + 1)\). This shows that \(9^{k+1} - 1\) is divisible by 8, completing the inductive step.

Step 5: Conclusion

Since both the base case and the inductive step have been proven, by the principle of mathematical induction, \(9^n - 1\) is divisible by 8 for all positive integers \(n\).

Key Concepts

Divisibility

Divisibility tells us when one number can be divided by another without a remainder. In this exercise, you need to show that \(9^n - 1\) is always divisible by 8 for any positive integer \(n\). When we say a number is divisible by 8, it means that if you divide it by 8, you get an integer result with no leftover.

To test divisibility by 8, consider:\(9^n - 1\) as \((9^n) - 1\). Check the calculated result from each exponent, subtract 1, then see if it divides evenly by 8.

For the base case, \(9^1 - 1 = 8\), and since 8 divided by 8 is 1 (a whole number), the divisibility holds for this example. Understanding divisibility is crucial in proving many mathematical assumptions, such as the one in this exercise.

Exponentiation

Exponentiation is the mathematical operation of raising a number to a power. In simple terms, multiplying a number by itself a certain number of times. For this exercise, the operation is represented by \(9^n\). When \(n = 1\), it simplifies to 9; when \(n = 2\), it becomes 81; and so forth.

Exponentiation has various properties, such as \(a^m \cdot a^n = a^{m+n}\) and \((a^m)^n = a^{m \cdot n}\). These can help simplify complex expressions.
In solving mathematical proofs involving exponentiation, understanding these principles aids in picking apart the algebra to see how expressions can be rearranged or reduced, like illustrating \(9^{k+1} = 9 \cdot 9^k\). This aids the proof of supplementing the inductive step.

Base Case

The base case is the starting point of any proof using mathematical induction. It confirms that the statement holds true for the initial value, which in this problem is \(n = 1\).

Checking the base case involves replacing \(n\) with the smallest value: \(9^1 - 1 = 8\). Since 8 is divisible by 8, the base case is verified.

• This simple check is fundamental to induction, as it sets the ground truth for the hypothesis you assume over successive steps.
• Without a valid base case, further induction steps would be as flawed as building a house on a foundation of sand.

Inductive Step

The inductive step carries the proof forward from the base case. It involves assuming that the statement is true for a particular \(k\), and then proving it for \(k + 1\).

In this exercise:

• Assume \(9^k - 1\) is divisible by 8, meaning \(9^k - 1 = 8m\) for some integer \(m\).
• Consider the expression for \(k + 1\): \(9^{k+1} - 1 = 9 \cdot 9^k - 1\).
• Rearrange and substitute using the inductive hypothesis: \(9(9^k - 1) + 8 = 72m + 8\).

The rewriting shows that the expression can again be factored as 8 times an integer, \(8(9m + 1)\), thus completing the proof that \(9^{n} - 1\) is divisible by 8 for all \(n\). It's like climbing a ladder: confirm the first step is sound (base case), then show that you can consistently climb to the next rung (inductive step).