/*! 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 8 (a) Prove that for each real num... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 8

(a) Prove that for each real number \(x,(x+\sqrt{2})\) is irrational or \((-x+\sqrt{2})\) is irrational. (b) Generalize the proposition in Part (a) for any irrational number (instead of just \(\sqrt{2}\) ) and then prove the new proposition.

Short Answer

Expert verified
In part (a), we assume that \(x + \sqrt{2}\) is rational, which leads to the expression \(-x + \sqrt{2} = -\frac{p}{q} + 2\sqrt{2}\). By contradiction, we show that \(-x + \sqrt{2}\) must be irrational. In part (b), we replace \(\sqrt{2}\) with any irrational number y and follow the same logic. We assume that \(x + y\) is rational, which leads to the expression \(-x + y = -\frac{p}{q} + 2y\). By contradiction, we show that \(-x + y\) must be irrational.

Step by step solution

01

Assume (x + 鈭�2) is rational

Let x be any real number. Now, let's assume that (x + 鈭�2) is rational. This means that there exist two integers, say p and q, such that (x + 鈭�2) = p/q where q is not equal to 0.
02

Rearrange the equation

Now, let's try to express x in terms of p, q, and 鈭�2. We can write x as: \[x = \frac{p}{q} - \sqrt{2}\]
03

Consider (-x + 鈭�2) and show it's irrational

We'll now consider (-x + 鈭�2) and show that it is irrational. We can rewrite the expression (-x + 鈭�2) using the expression for x from Step 2: \[-x + \sqrt{2} = -\left(\frac{p}{q} - \sqrt{2}\right) + \sqrt{2}\] \[=-\frac{p}{q} + 2\sqrt{2}\] Now, suppose (-x + 鈭�2) is also rational. Then there exist two integers, say m and n, such that (-x + 鈭�2) = m/n where n is not equal to 0. This would lead to: \[-\frac{p}{q} + 2\sqrt{2} = \frac{m}{n}\] Since p/q and m/n are both rationals, their sum or difference should also be rational. Therefore, \[2\sqrt{2} = \frac{m}{n} + \frac{p}{q}\] This implies that 2鈭�2 is a rational number, which is a contradiction because 2鈭�2 is irrational. Thus, if (x + 鈭�2) is rational, then (-x + 鈭�2) must be irrational.
04

Generalize the proposition for any irrational number

Let y be any irrational number instead of 鈭�2. We can follow the same logic as for the part (a) of the exercise. Suppose (x + y) is a rational number. Then there exist two integers, say p and q, such that (x + y) = p/q where q is not equal to 0. We can rewrite x as: \[x = \frac{p}{q} - y\] Now, we consider (-x + y) and try to show that it's irrational. We rewrite (-x + y) using the expression for x: \[-x + y = -\left(\frac{p}{q} - y\right) + y\] \[=-\frac{p}{q} + 2y\] Suppose (-x + y) is also rational. Then there exist two integers, say m and n, such that (-x + y) = m/n where n is not equal to 0. This would lead to: \[-\frac{p}{q} + 2y = \frac{m}{n}\] Thus, we would have: \[2y = \frac{m}{n} + \frac{p}{q}\] Since p/q and m/n are both rationals, their sum or difference should also be rational, meaning that 2y would be rational, which is a contradiction because y is irrational. Therefore, for any irrational number y, if (x + y) is rational, then (-x + y) must be irrational.

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.

Rational and Irrational Numbers
Understanding the difference between rational and irrational numbers is fundamental in mathematics. A rational number is any number that can be expressed as the quotient or fraction \frac{p}{q} of two integers, a numerator p and a non-zero denominator q. For example, 1/2, 3, and -4.5 are all rational because they can be written as fractions.

In contrast, an irrational number cannot be expressed as a simple fraction. This means that their decimal expansion is non-repeating and non-terminating. Classic examples include \(\sqrt{2}\), \(\pi\), and the mathematical constant e. These numbers have decimal representations that go on forever without repeating a pattern. The exercise's reference to \(\sqrt{2}\) as an irrational number underlines this concept. It's essential to grasp that the sum or difference of a rational number and an irrational number is always irrational, which is a pivotal point in proving our textbook exercise.
Proof by Contradiction
Proof by contradiction is a powerful technique in mathematical reasoning. It involves assuming the opposite of what you want to prove and showing that this assumption leads to a contradiction - an impossibility or an instance where something is both true and false at the same time, which breaks logical rules.

Here's how it works in simple terms:
  • Start by assuming that the statement you want to prove is false.
  • Deductively arrive at a contradiction, something that we know is not true or violates established truths.
  • Conclude that because the assumption leads to an impossible situation, the original statement must be true.
In our textbook example, we utilize proof by contradiction to show that for every real number \(x\), at least one of the numbers \(x + \sqrt{2}\) or \( - x + \sqrt{2}\) must be irrational. By demonstrating the impossibility of both numbers being rational, the exercise exemplifies this proof technique elegantly.
Mathematical Reasoning
Mathematical reasoning is the logical thought process behind solving mathematical problems. It allows us to justify why certain mathematical statements are true and to construct proofs, like the one in our textbook exercise.

There are two main types of mathematical reasoning: deductive and inductive. Deductive reasoning involves beginning with assumed truths (like axioms or previously proven theorems) and logically deriving conclusions from them. Inductive reasoning, on the other hand, involves looking at specific cases and extrapolating a general rule.

The textbook exercise puts deductive reasoning into action. Starting from known properties of rational and irrational numbers, it deduces that the sum of a rational and an irrational number must be irrational. The ability to logically step from known facts to new conclusions is at the heart of effective mathematical reasoning and problem-solving.

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

Consider the following proposition: Proposition. For all integers \(m\) and \(n,\) if \(n\) is odd, then the equation $$ x^{2}+2 m x+2 n=0 $$ has no integer solution for \(x\). (a) What are the solutions of the equation when \(m=1\) and \(n=-1 ?\) That is, what are the solutions of the equation \(x^{2}+2 x-2=0 ?\) (b) What are the solutions of the equation when \(m=2\) and \(n=3\) ? That is, what are the solutions of the equation \(x^{2}+4 x+6=0 ?\) (c) Solve the resulting quadratic equation for at least two more examples using values of \(m\) and \(n\) that satisfy the hypothesis of the proposition. (d) For this proposition, why does it seem reasonable to try a proof by contradiction? (e) For this proposition, state clearly the assumptions that need to be made at the beginning of a proof by contradiction. (f) Use a proof by contradiction to prove this proposition.

Determine if each of the following statements is true or false. Provide a counterexample for statements that are false and provide a complete proof for those that are true. (a) For all real numbers \(x\) and \(y, \sqrt{x y} \leq \frac{x+y}{2}\). (b) For all real numbers \(x\) and \(y, x y \leq\left(\frac{x+y}{2}\right)^{2}\). (c) For all nonnegative real numbers \(x\) and \(y, \sqrt{x y} \leq \frac{x+y}{2}\).

Consider the following proposition: For each integer \(a, a \equiv 2(\bmod 8)\) if and only if \(\left(a^{2}+4 a\right) \equiv 4(\bmod 8)\) (a) Write the proposition as the conjunction of two conditional statements. (b) Determine if the two conditional statements in Part (a) are true or false. If a conditional statement is true, write a proof, and if it is false, provide a counterexample. (c) Is the given proposition true or false? Explain.

(a) Verify that the triangle inequality is true for several different real numbers \(x\) and \(y .\) Be sure to have some examples where the real numbers are negative. (b) Explain why the following proposition is true: For each real number \(r\), \(-|r| \leq r \leq|r|\) (c) Now let \(x\) and \(y\) be real numbers. Apply the result in Part (14b) to both \(x\) and \(y\). Then add the corresponding parts of the two inequalities to obtain another inequality. Use this to prove that \(|x+y| \leq|x|+|y|\)

(a) Give an example that shows that the sum of two irrational numbers can be a rational number. (b) Now explain why the following proof that \((\sqrt{2}+\sqrt{5})\) is an irrational number is not a valid proof: Since \(\sqrt{2}\) and \(\sqrt{5}\) are both irrational numbers, their sum is an irrational number. Therefore, \((\sqrt{2}+\sqrt{5})\) is an irrational number. Note: You may even assume that we have proven that \(\sqrt{5}\) is an irrational number. (We have not proven this.) (c) Is the real number \(\sqrt{2}+\sqrt{5}\) a rational number or an irrational number? Justify your conclusion.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Decision 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.