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Chapter 4: Problem 10

Use mathematical induction to prove that for each natural number \(n, 3 \mathrm{di}-\) vides \(n^{3}+23 n\). Compare this proof to the proof from Exercise (19) in Section 3.5

Short Answer

Expert verified
In short, we used mathematical induction to prove that for every natural number n, \(3|(n^3 + 23n)\). We first proved the base case for n=1, then assumed the statement was true for n=k, and finally proved it for n=k+1 using the inductive hypothesis. This proof is similar to Exercise (19) in Section 3.5 in the sense that both use mathematical induction for proving divisibility statements; however, the steps and calculations in each proof might be different. The key aspect of both proofs is the use of the inductive hypothesis to show the statement is true for n=k+1 based on the assumption that it is true for n=k.

Step by step solution

01

Proving the Base Case (n=1)

Let's prove the base case when n=1: \[ 1^3 + 23\cdot 1 = 1 + 23 = 24 \] Since 24 is divisible by 3, the base case is true.
02

Inductive Hypothesis

Now, let's assume that the statement is true for n=k, that is \(3|(k^3 + 23k)\). This can be written as: \[ k^3 + 23k = 3a \] where 'a' is some integer.
03

Proving the Statement for n=k+1

Next, we need to prove the statement is true for n=k+1. Let's compute \({(k+1)}^3 + 23(k+1)\) and simplify it: \( (k+1)^3 + 23(k+1) = (k^3 + 3k^2 + 3k + 1) + (23k + 23) \) Now, we want to show that this expression is divisible by 3, based on the inductive hypothesis. Let's substitute the assumed expression from the inductive hypothesis: \[(k^3 + 23k) + (3k^2 + 1) + 23 = 3a + 3k^2 + 1 + 23 \] Now we want to prove that the remaining terms are divisible by 3: \[3k^2 + 1 + 23 = 3k^2 + 24\] Since k is a natural number, \(3k^2\) is divisible by 3. Moreover, 24 is also divisible by 3. Therefore, our expression can be rewritten as: \[3a + 3k^2 + 24 = 2*3 + 3(a + k^2 + 8)\] Since we were able to represent the expression for n=k+1 as a multiple of 3, we have proven that \(3|({(k+1)}^3 + 23(k+1))\). By the principle of mathematical induction, we have proven that for each natural number n, \(3|(n^3 + 23n)\). Comparison to proof from Exercise (19) in Section 3.5: While this exercise and Exercise (19) in Section 3.5 both involve using mathematical induction to prove a statement regarding divisibility, the steps and calculations involved in the proofs may differ. The key aspect of both proofs is the use of the inductive hypothesis to progress from the assumption that the statement is true for n=k to showing that it is also true for n=k+1.

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Key Concepts

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

Divisibility by Natural Numbers
Divisibility is a foundational concept in mathematics where one integer can be divided by another with no remainder. For instance, we say that an integer 'a' is divisible by an integer 'b' if there exists another integer 'c' such that \(b \times c = a\).

This property is central to many areas of number theory and comes into play when discussing factors, multiples, and division algorithms among other topics. In the case of our exercise, we're interested in divisibility by the natural number 3, which arises frequently due to its prevalence in geometrical and algebraic applications. An understanding of divisibility rules, especially for small numbers like 3, can greatly simplify many mathematical endeavors and proofs.

When we state that \(3 | (n^3 + 23n)\) for all natural numbers 'n', we're asserting a universal trait about this algebraic expression, linking it with divisibility by 3.
Inductive Hypothesis
The inductive hypothesis is the pivotal assumption in the process of mathematical induction that allows us to establish a general rule. After verifying the base case (such as \(3|(1^3 + 23 \times 1)\)), we suppose that what we are trying to prove is true for a particular natural number 'k'. This assumption, termed the inductive hypothesis, is written as \(3|(k^3 + 23k)\) - effectively assuming that \(k^3 + 23k\) is a multiple of 3.

The true power of the inductive hypothesis lies in its ability to act as a 'domino effect'. By demonstrating that if the statement is valid for 'k' then it must also be valid for 'k+1', the inductive step, we set off a chain reaction that proves the statement for all natural numbers following our base case. This kind of reasoning is akin to proving that every next domino will fall if the current one does, thereby assuring the collapse of an infinitely long line of dominoes.
Proof Writing in Mathematics
In proof writing, clarity, logical flow, and thorough understanding of the mathematical principles are paramount. Mathematical induction is a classic proof technique used particularly in number theory for establishing the truth of infinitely many cases with a finite amount of work.

To carry out a proof by induction, like our provided exercise, you need to accomplish two main steps: verify the base case and perform the inductive step. A well-structured proof will cleanly separate these sections and articulate each logical transition. The goal is to establish a clear narrative that 'if it works for one, it works for the next, and therefore it works for all'. Just as a well-written story has a beginning, middle, and end, a good mathematical proof has distinct sections that guide the reader through the argument.

It is important to be explicit about each step, especially when manipulating an equation, to demonstrate that the inductive hypothesis is carried through correctly. Such detail aids in understanding, reduces the chance of error, and reinforces the universal truth of the theorem being proved.

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Most popular questions from this chapter

(a) Calculate the value of \(5^{n}-2^{n}\) for \(n=1, n=2, n=3, n=4, n=5\), and \(n=6\). (b) Based on your work in Part (a), make a conjecture about the values of \(5^{n}-2^{n}\) for each natural number \(n\). (c) Use mathematical induction to prove your conjecture in Part (b).

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=5,\) and that for each \(n \in \mathbb{N}, a_{n+1}=a_{n}+2 a_{n-1}\). Prove that for each natural number \(n\), \(a_{n}=2^{n}+(-1)^{n}\)

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=1,\) and for each \(n \in \mathbb{N}, a_{n+2}=a_{n+1}+3 a_{n} .\) Determine which terms in this sequence are divisible by 4 and prove that your answer is correct.

The quadratic formula can be used to show that \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2}\) are the two real number solutions of the quadratic equation \(x^{2}-x-1=0\). Notice that this implies that $$ \begin{array}{l} \alpha^{2}=\alpha+1, \text { and } \\ \beta^{2}=\beta+1 \end{array} $$ It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers. (a) Verify that \(f_{1}=\frac{\alpha^{1}-\beta^{1}}{\alpha-\beta}\) and that \(f_{2}=\frac{\alpha^{2}-\beta^{2}}{\alpha-\beta}\). (b) (This part is optional, but it may help with the induction proof in part (c).) Work with the relation \(f_{3}=f_{2}+f_{1}\) and substitute the expressions for \(f_{1}\) and \(f_{2}\) from part (a). Rewrite the expression as a single fraction and then in the numerator use \(\alpha^{2}+\alpha=\alpha(\alpha+1)\) and a similar equation involving \(\beta .\) Now prove that \(f_{3}=\frac{\alpha^{3}-\beta^{3}}{\alpha-\beta}\).(c) Use induction to prove that for each natural number \(n,\) if \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2},\) then \(f_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} .\) Note: This formula for the \(n^{t h}\) Fibonacci number is known as Binet's formula, named after the French mathematician Jacques Binet ( \(1786-1856\) ).

For which natural numbers \(n\) is \(n^{2}<2^{n} ?\) Justify your conclusion.
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