Problem 18 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Base Case

Verify the statement for the smallest natural number, which is typically 1. Substitute \( n = 1 \) into the expression \( n^3 - n + 3 \):\[1^3 - 1 + 3 = 1 - 1 + 3 = 3\]Clearly, 3 is divisible by 3. Thus, the base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary natural number \( k \); that is, assume:\[k^3 - k + 3 = 3m\]for some integer \( m \). This assumption is the inductive hypothesis.

Step 3: Inductive Step

Prove that the statement holds for \( k + 1 \). Consider the expression for \( n = k + 1 \):\[(k+1)^3 - (k+1) + 3\]Expand the expression and simplify:\[= (k^3 + 3k^2 + 3k + 1) - k - 1 + 3= k^3 + 3k^2 + 2k + 3\]Substitute the inductive hypothesis into this expression:\[= (3m) + 3k^2 + 2k + 3\]Factor out 3:\[= 3m + 3(k^2 + \frac{2k}{3} + 1)\]The term \(2k/3\) is an integer because \(2k\) must be divisible by 3 if the entire expression is divisible (consider congruence modulo 3), assuring that the result is a multiple of 3.

Step 4: Conclusion

Since both the base case and the inductive step are verified, by mathematical induction the statement "\( n^3 - n + 3 \) is divisible by 3 for all natural numbers \( n \)" is true.

Key Concepts

Divisibility

Divisibility is a fundamental concept in mathematics. It refers to the ability of one number to be divided by another without leaving a remainder. In our exercise, we are dealing with the statement that the expression \( n^3 - n + 3 \) is divisible by 3. This means that when you divide \( n^3 - n + 3 \) by 3, there should be no remainder.

• To check for divisibility, we can perform the division or understand it through modular arithmetic.
• In the context of the exercise, we prove divisibility through mathematical induction.

Let's remember the basic rule here: if your formula can be rewritten in such a way that its terms or sub-expressions result in zero (or a multiple of the divisor) when divided by 3, then it is divisible by 3. This is exactly what we confirmed for the expression for different values of \( n \).

Natural Numbers

Natural numbers are the set of positive integers starting from 1, and they do not include zero or negative numbers. These numbers form the basis of simple arithmetic operations and sequence exercises in mathematics. For our exercise, proving that the formula holds for all natural numbers \( n \) essentially means demonstrating the formula's truth for an infinite sequence of inputs.

• First, we confirm the base case, showing it works for the smallest natural number, which is 1.
• Next, we assume it works for an arbitrary natural number \( k \) and subsequently prove it for the next natural number, \( k+1 \).

Using the principle of mathematical induction, if an expression holds true at the base level and holds true moving from \( k \) to \( k+1 \), it must hold for all natural numbers. This is a powerful method to handle problems involving infinite sequences.

Algebraic Expressions

Algebraic expressions involve variables, numbers, and operations like addition, subtraction, multiplication, and division. The expression in question, \( n^3 - n + 3 \), is a simple polynomial.

• These expressions allow us to generalize mathematical rules.
• In doing so, they help us set up equations to model real-life situations or even solve theoretical problems like the one in the exercise.

Within the exercise, we use the algebraic properties of the expression to manipulate it and prove its divisibility for mathematical induction purposes.The process involves expanding and simplifying expressions, substituting variables, and finding common factors. This requires a solid understanding of algebraic manipulation, as seen when we expanded \((k+1)^3 - (k+1) + 3\) and simplified it using the assumption \( k^3 - k + 3 = 3m \). With this method, the power of algebra lies in its flexibility to handle variable and constant terms to prove general statements of number theory.