/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 \(9-16 .\) Using induction, veri... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 14

\(9-16 .\) Using induction, verify the solutions to Exercises \(1-8.\)

Short Answer

Expert verified
Using mathematical induction, we have verified that the formula \(n^2 - n + 41\) generates a prime number for \(n = 1\) to \(n = 8\). We showed that it holds true for the base case (n = 1), and if it is true for n = k, then it is true for n = k + 1. Therefore, we conclude that the statement is valid for the given range.

Step by step solution

01

Base Case (n = 1)

Verify the statement for the base case, which is when \(n = 1\). Substitute n = 1 into the formula: \(P(1) = (1)^2 - (1) + 41 = 41\) Since 41 is a prime number, the statement holds true for the base case.
02

Inductive Hypothesis

Assume that the statement is true for \(n = k\), which means that \(P(k) = k^2 - k + 41\) is prime for some integer k.
03

Inductive Step (n = k + 1)

Now, we must prove that the statement is true for \(n = k + 1\): \(P(k+1) = (k+1)^2 - (k+1) + 41\) Simplify the expression: \(P(k+1) = (k^2 + 2k + 1) - k - 1 + 41\) Combine like terms: \(P(k+1) = k^2 + k + 41\) Now, we must show that the expression is either a prime number or can be reduced to a prime number. \(P(k+1) = k^2 + k + 41 = (k^2 - k + 41) + 2k\) We know that \(k^2 - k + 41\) generates a prime by our inductive hypothesis. We have to show that adding \(2k\) will also result in a prime number. If there is a common divisor of more than one between \(k^2 - k + 41\) and \(2k\), then \(P(k+1)\) will not be prime. However, since \(2k\) is even, its factors must either be \(1\) or \(2\). The factors of \(k^2 - k + 41\) are either \(1\) or odd, due to the inductive hypothesis stating it generates a prime number. As a result, the largest common divisor that \(k^2 - k + 41\) and \(2k\) can share is 1, so the sum of these two expressions will result in a prime number.
04

Conclusion

We have shown that the expression holds true for the base case (n = 1), and by induction, we have also shown that if it is true for n = k, it is true for n = k + 1. Therefore, we can conclude that the formula \(n^2 - n + 41\) generates a prime number for \(n = 1\) to \(n = 8\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Base Case
The base case in mathematical induction is like the foundation of a house. It provides the initial step upon which the rest of the structure is built. In our exercise, we start with the first value, which is usually the smallest integer, such as \(n = 1\).
We substitute \(n = 1\) into the formula to see if the statement holds. Let's evaluate:
  • \(P(1) = 1^2 - 1 + 41 = 41\)
Number 41 is a prime number. Therefore, our base case stands firm. Demonstrating that the statement is accurate at this initial step is crucial because this validity is carried through each subsequent step of the induction process.
Inductive Hypothesis
The inductive hypothesis is like a well-educated guess based on the base case. Let's assume the statement holds true for a particular case \(n = k\). This assumption allows us to use it as a stepping stone to prove further cases.

Mathematically written, the inductive hypothesis for our problem is:
  • \(P(k) = k^2 - k + 41\)
Assume this expression results in a prime number for a specific integer \(k\). This assumption is crucial because it's used in the next step to prove the statement for \(n = k + 1\).
Inductive Step
In the inductive step, you take what you assumed to be true in the inductive hypothesis and use it to prove the next case. It鈥檚 like using the end of one textbook chapter to build up to the next.For \(n = k+1\), plug \(k+1\) into the expression:
  • \((k+1)^2 - (k+1) + 41\)
When you simplify this, it becomes \(k^2 + k + 41\). Next, we leverage our inductive hypothesis, stating that \(k^2 - k + 41\) is prime. The task is to show adding \(2k\) does not disrupt its primeness.In our proof:
  • The terms \(2k\) and \(k^2 - k + 41\) have no common factors other than 1.
This reinforcement connects the hypothesis to the newer case, solidifying that the expression remains prime.
Primes
A prime number is one more fascinating than it seems: it can only be divided by 1 and itself, and this indivisibility makes it fundamental in mathematics. In our exercise, we've been consistently dealing with expressions that yield prime numbers, verified through induction.Prime numbers are crucial, like the atoms of arithmetic, breaking into no smaller parts other than themselves. They keep showing up in the sequence generated by our formula, ensuring our expressions keep their prime nature.Understanding how primes behave, especially in sequences or formulas like \(n^2 - n + 41\), can enhance a student鈥檚 perception of mathematical patterns and the underlying structure of numbers in mathematics. It鈥檚 remarkable how one check at a base case can lead to many evidences of prime outputs.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(a, b, k \in \mathbb{N}, b \geq 2,\) and \(n=b^{k} .\) Consider the function \(f\) defined by \(f(n)=a f(n / b)+g(n) .\) Show that \(f(n)=a^{k} f(1)+\sum_{i=0}^{k-1} a^{i} g\left(n / b^{i}\right)\)

Algorithm 5.10 computes the \(n\)th power of a positive real number \(x,\) where \(n \geq 0 .\) Use it to answer Exercises 18-24 . Algorithm exponentiation(x,n) (* This algorithm computes the nth power of \(x\) using recursion and returns the value in the variable answer.*) 0\. Begin (* algorithm *) 1\. if \(n=0\) then 2\. answer \(\leftarrow 1\) 3\. else 1 if \(n=1\) then 4\. answer \(\leftarrow x\) 5\. else 6\. begin (* else *) 7\. value \(\leftarrow\) exponentiation \((x,\lfloor n / 2\rfloor)\) 8\. answer \(\leftarrow\) value \(\cdot\) value 9\. If \(n\) is odd then 10\. answer \(\leftarrow\) answer \(\cdot\) \(x\) 11\. endelse 12\. End (* algorithm *) Let \(a_{n}\) denote the number of multiplications (lines \(7-10 )\) required by the algorithm to compute \(x^{n}\) . Compute each. $$a_{5}$$

The number \(h_{n}=\sum_{i=1}^{n}\left(\frac{1}{i}\right)\) called the harmonic number, occurs frequently in the analysis of algorithms. Compute \(h_{4}\) and \(h_{5}\)

Consider the recurrence relation \(c_{n}=c_{[n / 2]}+c_{[L n+1 / 2]}+2,\) where \(c_{1}=0\) Find the order of magnitude of \(c_{n}\) when \(n\) is a power of \(2 .\)

Express each quotient as a sum of partial fractions. $$\frac{x^{3}+x^{2}+5 x-2}{x^{4}-x^{2}+x-1}$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.