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Chapter 5: Problem 32

Use mathematical induction in Exercises \(31-37\) to prove divisibility facts. Prove that 3 divides \(n^{3}+2 n\) whenever \(n\) is a positive integer.

Short Answer

Expert verified
By mathematical induction, 3 divides \(n^3 + 2n\) for all positive integers n.

Step by step solution

01

- Base Case

To prove by induction, start with the base case where the smallest value of the positive integer is considered. For this problem, take n = 1. Calculate the expression: \[ f(1) = 1^3 + 2 \times 1 = 1 + 2 = 3 \] As 3 divides 3, the base case holds true.
02

- Induction Hypothesis

Assume that the statement is true for some positive integer k, i.e., 3 divides k鲁 + 2k. This means: \[ k^3 + 2k = 3m \] for some integer m.
03

- Inductive Step

Next, prove that if the statement holds for n = k, it also holds for n = k + 1. Consider the expression for n = k + 1: \[ f(k+1) = (k+1)^3 + 2(k+1) \] Expand the expression: \[ f(k+1) = k^3 + 3k^2 + 3k + 1 + 2k + 2 \] Combine like terms: \[ f(k+1) = k^3 + 3k^2 + 5k + 3 \]
04

- Using Induction Hypothesis

Rewrite the expression by separating the part we assumed to be divisible by 3: \[ f(k+1) = (k^3 + 2k) + 3(k^2 + k + 1) \] By the induction hypothesis, we know that \(k^3 + 2k\) is divisible by 3 (let's call it 3m). Hence: \[ f(k+1) = 3m + 3(k^2 + k + 1) \]
05

- Conclusion

Factor out the 3 from the expression: \[ f(k+1) = 3(m + k^2 + k + 1) \] Since \(m + k^2 + k + 1\) is an integer, this shows that \(f(k+1)\) is divisible by 3. Thus, by mathematical induction, 3 divides \(n^3 + 2n\) for all positive integers n.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divisibility
Divisibility is a fundamental concept in number theory. It means that one number can be divided by another without leaving a remainder. For example, if we say '3 divides 9', it means when 9 is divided by 3, the result is 3, with no remainder.

Understanding divisibility helps simplify many mathematical problems. When we talk about proving divisibility, as in the exercise where we show 3 divides the expression (n^3 + 2n) for all positive integers, we need to demonstrate it consistently through mathematical methods.

In this case, mathematical induction is used to verify that the expression (n^3 + 2n) is divisible by 3 for every positive integer n.
Base Case
The base case is the starting point of the induction process. It's where we prove that a statement is true for the very first number in our domain, typically the smallest positive integer. In our exercise, we start with n = 1.

For n = 1: 1^3 + 2 * 1 = 3. Clearly, 3 divides 3 with no remainder, establishing the base case.

This first step is crucial because it gives the foundation upon which we build our inductive argument. If the base case holds true, we can proceed to further steps.
Induction Hypothesis
In the induction hypothesis, we assume that the statement we want to prove is true for some arbitrary positive integer k. This step helps us set up the framework to prove it for the next integer, k + 1.

In our exercise, we assume: 3 divides k^3 + 2k. Mathematically, this means: k^3 + 2k = 3m for some integer m.

This assumption leverages the power of mathematical induction, helping us bridge known information to unknown scenarios. Once we have this assumption in place, we move on to the next critical step.
Inductive Step
The inductive step is where we demonstrate that if our statement is true for a positive integer k, it must also be true for k + 1. This step logically ensures that our hypothesis can extend to all subsequent integers.

For our problem, we need to show that: f(k + 1) = (k+1)^3 + 2(k+1) is divisible by 3.

By expanding and simplifying: (k+1)^3 + 2(k+1) becomes k^3 + 3k^2 + 5k + 3 . Using the hypothesis, we rewrite: k^3 + 2k = 3m and combine it with remaining terms to get: 3(m + k^2 + k + 1) .

Since m + k^2 + k + 1 is an integer, the entire expression is divisible by 3, proving our original statement true for k + 1. This completes the proof by induction.

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Most popular questions from this chapter

When does a string belong to the set \(A\) of bit strings defined recursively by $$ \begin{array}{l}{\lambda \in A} \\ {0 x 1 \in A \text { if } x \in A}\end{array} $$ where \(\lambda\) is the empty string?

Suppose that \(P\) is a simple polygon with vertices \(v_{1}, v_{2}, \ldots, v_{n}\) listed so that consecutive vertices are con- nected by an edge, and \(v_{1}\) and \(v_{n}\) are connected by an edge. A vertex \(v_{i}\) is called an ear if the line segment connecting the two vertices adjacent to \(v_{i}\) is an interior diagonal of the simple polygon. Two ears \(v_{i}\) and \(v_{j}\) are called nonoverlapping if the interiors of the triangles with vertices \(v_{i}\) and its two adjacent vertices and \(v_{j}\) and its two adjacent vertices do not intersect. Prove that every simple polygon with at least four vertices has at least two nonoverlapping ears.

Pick's theorem says that the area of a simple polygon \(P\) in the plane with vertices that are all lattice points (that is, points with integer coordinates) equals \(I(P)+B(P) / 2-1\) where \(I(P)\) and \(B(P)\) are the number of lattice points in the interior of \(P\) and on the boundary of \(P,\) respectively. Use strong induction on the number of vertices of \(P\) to prove Pick's theorem. [Hint: For the basis step, first prove the theorem for rectangles, then for right triangles, and finally for all triangles by noting that the area of a tri- angle is the area of a larger rectangle containing it with the areas of at most three triangles subtracted. For the inductive step, take advantage of Lemma \(1 . ]\)

a) Determine which amounts of postage can be formed using just 3 -cent and 10 -cent stamps. b) Prove your answer to (a) using the principle of math- ematical induction. Be sure to state explicitly your inductive hypothesis in the inductive step. c) Prove your answer to (a) using strong induction. How does the inductive hypothesis in this proof differ from that in the inductive hypothesis for a proof using mathematical induction?

Give a recursive definition of \(w^{i},\) where \(w\) is a string and \(i\) is a nonnegative integer. (Here \(w^{i}\) represents the concatenation of \(i\) copies of the string \(w . )\)
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