Problem 14 - Precalculus: Mathematical for Calculus

Question

Show that \(5^{n}-1\) 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 natural numbers \(n\).

Step 1: Base Case

To prove divisibility for all natural numbers, we start with the smallest natural number, which is 1. Substitute \(n = 1\) into the expression \(5^n - 1\). This gives us \(5^1 - 1 = 4\). Since 4 is divisible by 4, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for some natural number \(k\), meaning \(5^k - 1\) is divisible by 4. This is called the inductive hypothesis. Mathematically, this can be expressed as \(5^k - 1 = 4m\) for some integer \(m\).

Step 3: Inductive Step

Now, we need to show that if the statement is true for \(k\), then it is also true for \(k+1\). Starting with the expression for \(k+1\), we have \(5^{k+1} - 1\). Express this as \(5 \cdot 5^k - 1\).

Step 4: Use Inductive Hypothesis

Using the inductive hypothesis \(5^k = 4m + 1\), substitute into \(5^{k+1} - 1\):\[5^{k+1} - 1 = 5 \cdot (4m + 1) - 1 = 20m + 5 - 1 = 20m + 4\]The term \(20m + 4\) is clearly divisible by 4.

Step 5: Conclusion

Since both the base case and inductive step have been proven, by mathematical induction, \(5^n - 1\) is divisible by 4 for all natural numbers \(n\).

Key Concepts

Divisibility

Divisibility refers to how one number can be divided by another without leaving a remainder. In simple terms, if you divide a number and get a whole number, it is divisible by that divisor. For example, when we divide 8 by 4, we get 2—this is a whole number, indicating that 8 is divisible by 4.

In the context of the exercise, we explored whether the expression \(5^n - 1\) is divisible by 4, meaning after dividing \(5^n - 1\) by 4, there must be no remainder. The key is understanding if the result of the division is an integer every time we substitute a natural number into \(n\).

To prove divisibility, mathematical induction serves as an efficient method. We start from proving a base case (which is often the smallest possible scenario) and then show it holds true for the next situation using previously assumed results.

Natural Numbers

Natural numbers are the simplest numbers used most frequently in basic counting, starting from 1 and going on infinitely (1, 2, 3, etc.). They do not include zero, negative numbers, or fractions.

In mathematical exercises, when a statement is said to hold "for all natural numbers \(n\)," it means from the number 1 onwards. This concept is important when applying methods like induction, as it gives a starting point, denoting where the problem begins.

• Mathematically: \( n \in \mathbb{N} = \{1, 2, 3, \ldots\} \)
• Excludes: Zero, fractions, and negative numbers

Natural numbers play a critical role in this problem since proving divisibility of \(5^n - 1\) starts from the smallest natural number and assumes the pattern holds, hence facilitating an inductive proof.

Base Case

A base case is the initial step in a proof by induction. It's the foundation on which the rest of the proof is built. In mathematical induction, before assuming something is true for 'any' case, we first prove that it's true for the first case.

For our present example, the base case involves confirming that \(5^1 - 1\) is divisible by 4. Since \(5^1 - 1 = 4\), and we know 4 is divisible by 4, the base case holds.

• Significance: It ensures that the statement is true for at least one scenario.
• Step One: Substitute the smallest natural number (usually 1) into the given expression and verify divisibility.

Once the base case is confirmed, this positive outcome backs up and lends credibility to further steps where assumptions are made to build a solid pattern.