Problem 16 - Precalculus: Mathematics for Calculus

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) \(5^{n}-1\) is divisible by 4 for all natural numbers \(n\)

Step-by-Step Solution

Answer

By induction, \(5^n - 1\) is divisible by 4 for all \(n\).

Step 1: Base Case

Firstly, verify the base case when \( n = 1 \). Substitute \( n = 1 \) into the expression \( 5^n - 1 \) to get \( 5^1 - 1 = 4 \). Since 4 is divisible by 4, the base case is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for \( n = k \), i.e., assume that \( 5^k - 1 \) is divisible by 4 for some natural number \( k \). This means there exists an integer \( m \) such that \( 5^k - 1 = 4m \).

Step 3: Inductive Step

Show that if the statement is true for \( n = k \), it must also be true for \( n = k+1 \). Consider \( 5^{k+1} - 1 \) and rewrite it as \( 5 \times 5^k - 1 \).

Step 4: Simplifying the Inductive Step

Replace \( 5^k - 1 \) using the inductive hypothesis: \( 5 \times (5^k - 1) + 5 - 1 \). This becomes \( 5(4m + 1) + 4 = 20m + 5 \).

Step 5: Show Divisibility

Rewrite \( 5(4m + 1) + 4 \) as \( 20m + 5 \), which simplifies to \( 4(5m + 1) \), showing that it is divisible by 4.

Step 6: Conclusion

Since both the base case and the inductive step have been verified, by the principle of mathematical induction, \( 5^n - 1 \) is divisible by 4 for every natural number \( n \).

Key Concepts

Base Case

Mathematical induction always starts by checking the base case. The base case is the simplest instance of a statement we're trying to prove. For our exercise, we're proving that \(5^n - 1\) is divisible by 4 for all natural numbers \(n\). To start, we see if this holds true when \(n = 1\).
We substitute \(n = 1\) into the expression to get \(5^1 - 1 = 4\).
Since 4 is divisible by 4, the base case is satisfied, and this gives us a solid foundation to continue with the induction.
Remember, the base case is crucial because if it doesn't work, the entire inductive proof falls apart. It's like the first link in a chain.

Inductive Hypothesis

Once the base case is confirmed, we move to the inductive hypothesis. This is where we assume the statement holds for some arbitrary natural number \(k\).
For our problem, we assume that \(5^k - 1\) is divisible by 4, meaning there exists an integer \(m\) such that \(5^k - 1 = 4m\).
This assumption allows us to consider the next step, like imagining the next link after the current one in a chain.
The inductive hypothesis doesn't prove anything directly, but sets the stage to show that if one link (\(n = k\)) works, the next link (\(n = k+1\)) must also work.

Inductive Step

The inductive step is where we demonstrate that if the statement is true for \(n = k\), it must also be true for \(n = k+1\).
Let's look at \(5^{k+1} - 1\) in our problem and rewrite it using our hypothesis:

• Since \(5^k - 1 = 4m\), then \(5^{k+1} = 5 \times 5^k\)
• So, \(5^{k+1} - 1 = 5 \times (5^k) - 1\)

Using the assumption, substitute \(5^k - 1\) with \(4m\):

• \(5 \times (4m + 1) - 1 = 20m + 5 - 1 = 4(5m + 1)\)

This calculation shows that \(5^{k+1} - 1\) is divisible by 4.
The inductive step is like climbing a staircase, assuring that we can move from any step \(k\) to \(k+1\).

Divisibility

Divisibility is a key element in our exercise. To say a number is divisible by 4 means it can be expressed as a multiple of 4.
In this problem, we need to ensure \(5^n - 1\) results in something like 4 times an integer.
For the base case \(n = 1\), this is easy: \(4\) is clearly a multiple of 4.
For the induction step, starting from the hypothesis \(5^k - 1 = 4m\), we extend it to \(5^{k+1} - 1 = 4(5m + 1)\).
Both parts of the induction - the base and the step - confirm divisibility, creating a solid proof through the entirety of natural numbers.
Understanding divisibility here involves seeing how the numbers rearrange into these multiples - that's what binds every natural number \(n\) to this divisible property.