Question

Prove by induction for \(n \geq 0\) : (a) 3 divides \(n^{3}+2 n\) (b) 5 divides \(n^{5}-n\) (c) 6 divides \(n^{3}-n\) (d) 6 divides \(n^{3}+5 n\)

Short answer

Each expression is divisible as proven by induction.

Step-by-step solution

Base Case for Part (a)

For part (a), take \(n = 0\). Then, we have \(0^3 + 2 \times 0 = 0\), which is divisible by 3. This establishes the base case for part (a).

Inductive Step for Part (a)

Assume 3 divides \(k^3 + 2k\) for some integer \(k\). We need to show that 3 divides \((k+1)^3 + 2(k+1)\). Expand: \((k+1)^3 + 2(k+1) = k^3 + 3k^2 + 3k + 1 + 2k + 2\). We can write this as \(k^3 + 2k + 3k^2 + 3k + 3\). Since 3 divides \(k^3 + 2k\) by the inductive hypothesis and 3 clearly divides \(3k^2 + 3k + 3\), 3 divides the whole expression.

Base Case for Part (b)

For part (b), take \(n = 0\). Then, we have \(0^5 - 0 = 0\), which is divisible by 5. This establishes the base case for part (b).

Inductive Step for Part (b)

Assume 5 divides \(k^5 - k\) for some integer \(k\). We need to show that 5 divides \((k+1)^5 - (k+1)\). Expand: \((k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 - k - 1\). Simplify to \(k^5 - k + 5(k^4 + 2k^3 + 2k^2 + k)\). Since 5 divides \(k^5 - k\) and 5 divides the entire second term due to the factor of 5, 5 divides the entire expression.

Base Case for Part (c)

For part (c), take \(n = 0\). Then, we have \(0^3 - 0 = 0\), which is divisible by 6. This establishes the base case for part (c).

Inductive Step for Part (c)

Assume 6 divides \(k^3 - k\) for some integer \(k\). We need to show that 6 divides \((k+1)^3 - (k+1)\). Expand: \((k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1\). Simplify to \(k^3 - k + 3(k^2 + k)\). Since 6 divides \(k^3 - k\) (as 3 and 2 both divide this), and clearly, any number multiplied by 3 is divisible by 3 and 2 simultaneously when considering modulo, 6 divides the total expression.

Base Case for Part (d)

For part (d), take \(n=0\). Then, we have \(0^3 + 5 \times 0 = 0\), which is divisible by 6. This establishes the base case for part (d).

Inductive Step for Part (d)

Assume 6 divides \(k^3 + 5k\) for some integer \(k\). We need to show that 6 divides \((k+1)^3 + 5(k+1)\). Expand: \((k+1)^3 + 5(k+1) = k^3 + 3k^2 + 3k + 1 + 5k + 5\). Simplify to \(k^3 + 5k + 3k^2 + 3k + 6\). Since 6 divides \(k^3 + 5k\) and the terms \(3k^2 + 3k + 6\) are divisible by 6 due to the factors included, the total expression is divisible by 6.

Key concepts

Divisibility

Divisibility is a fundamental concept in mathematics, particularly where we deal with proving specific expressions to be divisible by certain numbers. In exercises involving divisibility, like the ones presented here, our goal is to verify if an expression can be evenly divided by a number without leaving a remainder.
To test divisibility, a straightforward method is to simplify the expression and analyze if each term contributes to the divisibility of the whole. For example, if we determine that the sum of terms, or perhaps individual groups within the sum, can be divided by our number of interest, we conclude that the entire expression is divisible by it as well.
Understanding divisibility helps us solve problems more systematically using properties of numbers; it often serves as the underlying principle to ascertain results through induction.

Base Case

The base case in mathematical induction is the initial step where you verify the statement for a starting value, often denoted by a simple integer like 0 or 1. This forms the foundation for the inductive process, ensuring that the property holds true at the inception before it can be assumed for later values.

• In our examples, for the base cases of each part, we assume simple values of \(n\) (commonly 0) to evaluate the expression's divisibility by the target number.
• We substitute \(n = 0\) into our polynomial, yielding results like 0, which are easily checked for divisibility.

This initial confirmation is crucial; without a true base case, the entire inductive reasoning could falter, resulting in false conclusions. Therefore, confirming this step assures us that our logic is on solid ground from the start.

Inductive Step

The inductive step of mathematical induction is where the bulk of our logical reasoning happens. After establishing our base case, we move to this step to continue proving for all natural numbers, relying heavily on what is assumed as the inductive hypothesis.
**Inductive Hypothesis Assumption**
- Assume the statement is true for an integer \(k\), expressed as the divisibility we want to establish.- This sets the stage for the next part of our proof.
**Proving for \(k+1\)**
- Expand the expression involving \(k+1\) in terms of known expressions.- Simplify, aiming to express the new statement \((k+1)\) as a sum or product based on your hypothesis. This converts at least part of it into terms known to be divisible from the hypothesis.Once these steps are verified, you solidify the legitimacy of the inductive step, allowing you to extend the property's application to all integers beyond the base case.

Polynomial Expansion

In many induction proofs, polynomial expansion is utilized to break down expressions, particularly those carefully structured to prove divisibility.
**Basic Expansion Rules**
- Utilize formulas for binomial expansions, such as format \( (k+1)^n \), which provides terms to work with for comparison against the divisibility criteria.
**Breaking Down the Expressions**
- Separately calculate each part of the expanded polynomial in terms of \(k\) and constants.- Guide these calculations towards forming combinations that simplify with obvious factors of the divisibility target. Look for terms like \(3k^2 + 3k\) when proving with divisibility by 3.
Polynomials allow step-by-step manipulation, bringing out factors and simplifying them to ensure divisibility. By arranging expanded forms smartly, we can plainly see what needs to be true for multiplication or addition to direct our proof towards the intended outcome.