/*! 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 10 (a) Prove Part (2) of Propositio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Prove Part (2) of Proposition 3.23 . For each \(x \in \mathbb{R},|-x|=|x|\). (b) Prove Part (2) of Theorem 3.25 . For all real numbers \(x\) and \(y,|x y|=|x||y|\).

Short Answer

Expert verified
(a) Proof of Proposition 3.23 part (2): We consider two cases, \(x \ge 0\) and \(x < 0\). In both cases, we show that |-x|=|x|. (b) Proof of Theorem 3.25 part (2): We consider four cases based on the signs of x and y. In all cases, we show that |xy|=|x||y|.

Step by step solution

01

Proof of Proposition 3.23 part (2)

By definition, the absolute value of a real number x, denoted by |x|, is equal to x if x is greater than or equal to 0, and -x if x is less than 0. We'll use this definition to prove that |-x|=|x| for any real number x. There are two cases to consider: Case 1: \(x \ge 0\) For this case, |-x| = -(-x) = x. Since x is greater than or equal to 0, |x| = x by definition. Therefore, |-x|=|x| in this case. Case 2: \(x < 0\) For this case, |-x| = -(-x) = x. However, since x is less than 0, |x| = -x by definition. Therefore, |-x|=|x| in this case as well. Since we've shown that |-x|=|x| holds for both cases, we have proved Proposition 3.23 part (2).
02

Proof of Theorem 3.25 part (2)

We are given that the product of the absolute values of two real numbers x and y is equal to the absolute value of their product. In other words, we need to prove that |xy|=|x||y|. We will use the definition of absolute value and basic properties of real numbers to prove this theorem. There are four cases to consider based on the signs of x and y: Case 1: \(x \ge 0\) and \(y \ge 0\) Since both x and y are greater than or equal to 0, we have |x| = x and |y| = y. Then, |xy| = xy and |x||y| = xy. Therefore, |xy| = |x||y| in this case. Case 2: \(x \ge 0\) and \(y < 0\) In this case, |x| = x and |y| = -y. Then, |xy| = x(-y)= -(xy) and |x||y| = x(-y). Therefore, |xy| = |x||y| in this case as well. Case 3: \(x < 0\) and \(y \ge 0\) Here, |x| = -x and |y| = y. Then, |xy| = (-x)y = -(xy) and |x||y| = (-x)y. Therefore, |xy| = |x||y| in this case. Case 4: \(x < 0\) and \(y < 0\) In this case, both x and y are less than 0, so |x| = -x and |y| = -y. Then, |xy| = (-x)(-y) = xy and |x||y| = (-x)(-y) = xy. Therefore, |xy| = |x||y| in this case as well. Since we've shown that |xy|=|x||y| holds for all four cases, we have proved Theorem 3.25 part (2).

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 91Ó°ÊÓ!

Key Concepts

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

Absolute Value Properties
Understanding the absolute value of a number is fundamental in math. It refers to the non-negative value of a number without regard to its sign. In simple terms, it's the distance of a number from zero on a number line.

Let's explore some key properties of absolute values that underpin the proofs we're dealing with. One essential property is that the absolute value of a number is always non-negative, meaning \( |x| \geq 0 \) for any real number \( x \). Another property is the one we see in our exercise: \( |-x| = |x| \), indicating that the absolute value of a number is the same regardless of its sign.

These properties play a pivotal role in solving equations and inequalities. They also help predict the behavior of functions that involve absolute values and lay the groundwork for more advanced mathematical concepts like complex numbers.
Proof Writing in Mathematics
Proof writing is a crucial skill in mathematics, enabling mathematicians to verify the truth of a statement beyond any doubt. A well-written proof follows a logical progression, is clear, and conveys the reasoning behind each step.

When writing a proof, start by understanding the statement you want to prove. Identify the assumptions and definitions relevant to the problem. Then, plan your proof by breaking it down into smaller, manageable steps, as demonstrated in the step-by-step solution to our exercise. Each step should follow naturally from the previous one, maintaining a coherent and logical argument throughout.

Using definitions, like the definition of absolute value in our exercise, is often the starting point for a proof. By considering various cases—such as the different signs of the numbers involved—you can address all possible scenarios and ensure your proof is complete and robust.
Real Numbers
The real numbers system is the set of all rational and irrational numbers, denoted by \( \mathbb{R} \). It includes positive and negative numbers, zero, fractions, and non-repeating, non-terminating decimals. Understanding the behavior of real numbers is decisive in various mathematical proofs.

Real numbers have an orderly structure that allows us to perform arithmetic operations like addition, subtraction, multiplication, and division (except by zero). The properties of these operations, especially multiplication, are what we relied on in proving Theorem 3.25 part (2) in our exercise.

Recognizing that multiplication of real numbers is commutative (\( xy = yx \)) and associative (\( (xy)z = x(yz) \)) helps simplify proofs involving products of absolute values. It is this deep understanding of the characteristics of real numbers that builds the foundation for higher-level math.
Mathematical Reasoning
Mathematical reasoning is the process of drawing logical conclusions based on premises and known facts. It encompasses various types of reasoning, including inductive and deductive reasoning. For example, deductive reasoning involves starting with general principles and deriving specific conclusions, which is often how proof writing unfolds.

In our proofs, we applied mathematical reasoning by considering all possible cases for the values of \( x \) and \( y \)—whether they are positive, negative, or zero—and then applying the properties of real numbers and absolute value to draw conclusions. This step-by-step logical thinking ensures that no scenario is left unchecked, making the proof complete and sound.

Moreover, mathematical reasoning isn't just about getting the correct answer; it's about understanding why the answer is correct. This deeper insight is what empowers students to tackle more challenging problems and advance their math skills.

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 \(n\) be a natural number. Prove each of the following: (a) For every integer \(a, a \equiv a(\bmod n)\). This is called the reflexive property of congruence modulo \(n\). (b) For all integers \(a\) and \(b,\) if \(a \equiv b(\bmod n),\) then \(b \equiv a(\bmod n)\). This is called the symmetric property of congruence modulo \(n\). (c) For all integers \(a, b,\) and \(c,\) if \(a \equiv b(\bmod n)\) and \(b \equiv c(\bmod n),\) then \(a \equiv c(\bmod n)\) This is called the transitive property of congruence modulo \(n\).

(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 \(|x+y| \leq|x|+|y|\)

The purpose of this exploration is to investigate the possibilities for which integers cannot be the sum of the cubes of two or three integers. (a) If \(x\) is an integer, what are the possible values (between 0 and 8 , inclusive) for \(x^{3}\) modulo \(9 ?\) (b) If \(x\) and \(y\) are integers, what are the possible values for \(x^{3}+y^{3}\) (between 0 and 8 , inclusive) modulo \(9 ?\) (c) If \(k\) is an integer and \(k \equiv 3(\bmod 9), \operatorname{can} k\) be equal to the sum of the cubes of two integers? Explain. (d) If \(k\) is an integer and \(k \equiv 4(\bmod 9), \operatorname{can} k\) be equal to the sum of the cubes of two integers? Explain. (e) State and prove a theorem of the following form: For each integer \(k\), if (conditions on \(k\) ), then \(k\) cannot be written as the sum of the cubes of two integers. Be as complete with the conditions on \(k\) as possible based on the explorations in Part (b). (f) If \(x, y,\) and \(z\) are integers, what are the possible values (between 0 and 8 , inclusive) for \(x^{3}+y^{3}+z^{3}\) modulo \(9 ?\) (g) If \(k\) is an integer and \(k \equiv 4(\bmod 9),\) can \(k\) be equal to the sum of the cubes of three integers? Explain. (h) State and prove a theorem of the following form: For each integer \(k\), if (conditions on \(k\) ), then \(k\) cannot be written as the sum of the cubes of three integers. Be as complete with the conditions on \(k\) as possible based on the explorations in Part (f).

Two prime numbers that differ by 2 are called twin primes. For example, 3 and 5 are twin primes, 5 and 7 are twin primes, and 11 and 13 are twin primes. Determine at least two other pairs of twin primes. Is the following proposition true or false? Justify your conclusion. For all natural numbers \(p\) and \(q\) if \(p\) and \(q\) are twin primes other than 3 and \(5,\) then \(p q+1\) is a perfect square and 36 divides \(p q+1\).

Are the following propositions true or false? Justify all your conclusions. If a biconditional statement is found to be false, you should clearly determine if one of the conditional statements within it is true. In that case, you should state an appropriate theorem for this conditional statement and prove it. (a) For all integers \(m\) and \(n, m\) and \(n\) are consecutive integers if and only if 4 divides \(\left(m^{2}+n^{2}-1\right)\) (b) For all integers \(m\) and \(n, 4\) divides \(\left(m^{2}-n^{2}\right)\) if and only if \(m\) and \(n\) are both even or \(m\) and \(n\) are both odd.

See all solutions

Recommended explanations on Math Textbooks

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.