/*! 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 20 Are the following statements tru... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Are the following statements true or false? Either prove the statement is true or provide a counterexample to show it is false. (a) For all integers \(a\) and \(b\), if \(a \cdot b \equiv 0(\bmod 6),\) then \(a \equiv 0(\bmod 6)\) or \(b \equiv 0(\bmod 6)\) (b) For all integers \(a\) and \(b\), if \(a \cdot b \equiv 0(\bmod 8)\), then \(a \equiv 0(\bmod 8)\) or \(b \equiv 0(\bmod 8)\) (c) For all integers \(a\) and \(b\), if \(a \cdot b=1(\bmod 6),\) then \(a=1(\bmod 6)\) or \(b \equiv 1(\bmod 6)\) (d) For all integers \(a\) and \(b\), if \(a b=7(\bmod 12),\) then either \(a=1(\bmod 12)\) or \(a \equiv 7(\bmod 12)\)

Short Answer

Expert verified
(a) True (b) False (c) False (d) False

Step by step solution

01

a) True or False: If \(a \cdot b \equiv 0 \pmod{6},\) then \(a \equiv 0 \pmod{6}\) or \(b \equiv 0 \pmod{6}\)

To prove the statement is true, we will first assume that \(a \cdot b \equiv 0 \pmod{6}\). We can rewrite this as: \(a \cdot b = 6k\), where k is an integer. Now if a is divisible by 6 (i.e., \(a = 6r\) for some integer r), then the statement is true. If a is not divisible by 6, that means it must be divisible by either 2 or 3, because 6 = 2 * 3. Let's consider both cases: 1. When a is divisible by 2 but not by 6: In this case, \(a = 2t\) for some integer t. So \(2t \cdot b = 6k\), which simplifies to \(t \cdot b = 3k\). This indicates that \(b\) must be divisible by 3, and hence, \(a \cdot b\) is divisible by 6. 2. When a is divisible by 3 but not by 6: In this case, \(a = 3s\) for some integer s. So \(3s \cdot b = 6k\), which simplifies to \(s \cdot b = 2k\). This indicates that \(b\) must be divisible by 2, and hence, \(a \cdot b\) is divisible by 6. So in both cases, \(a \cdot b \equiv 0 \pmod{6}\). Therefore, the statement is true.
02

b) True or False: If \(a \cdot b \equiv 0 \pmod{8}\), then \(a \equiv 0 \pmod{8}\) or \(b \equiv 0 \pmod{8}\)

To check if the statement is true or false, we can provide a counterexample. Let's consider \(a = 2\) and \(b = 4\). \(a \cdot b = 2 \cdot 4 = 8\), which is \(0 \pmod{8}\). However, neither \(a \equiv 0 \pmod{8}\) nor \(b \equiv 0 \pmod{8}\). Thus, the statement is false.
03

c) True or False: If \(a \cdot b \equiv 1 \pmod{6}\), then \(a \equiv 1 \pmod{6}\) or \(b \equiv 1 \pmod{6}\)

We can find a counterexample to show that the statement is false. Let \(a = 5\) and \(b = 5\). \(a \cdot b = 5 \cdot 5 = 25\), which is \(1 \pmod{6}\). However, neither \(a \equiv 1 \pmod{6}\) nor \(b \equiv 1 \pmod{6}\). Thus, the statement is false.
04

d) True or False: If \(a \cdot b \equiv 7 \pmod{12}\), then either \(a \equiv 1 \pmod{12}\) or \(a \equiv 7 \pmod{12}\)

We can find a counterexample to show that the statement is false. Let \(a = 5\) and \(b = 5\). \(a \cdot b = 5 \cdot 5 = 25\), which is \(7 \pmod{12}\). However, neither \(a \equiv 1 \pmod{12}\) nor \(a \equiv 7 \pmod{12}\). Thus, the statement is false.

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.

Integer Division
Integer division is a fundamental concept in mathematics that involves dividing one integer by another. Unlike regular division, the key here is that we're interested in whole numbers as the result. For example, dividing 10 by 3 in integer division would give 3 as the result, not 3.33, because we're only accounting for the complete wholes. This is where modular arithmetic comes into play, as it allows us to find the remainder when dividing two integers.

In the exercise, the problem statements deal with integers in the form of products, which often relate back to the idea of divisibility. For instance, when we see an expression like \(a \cdot b \equiv 0 \pmod{6}\), it implies that the product \(a \cdot b\) is divisible by 6. This doesn't necessarily mean that either \(a\) or \(b\) needs to be divisible by 6 on their own. Instead, they could both be divisible by smaller numbers that multiply to 6, such as 2 and 3, highlighting the nuanced nature of integer division in this mathematical context.
Mathematical Proof
Mathematical proofs are logical arguments that demonstrate the truth of a given statement. In math, proving a statement true often involves assuming the conditions of the statement and showing step by step how those conditions necessarily lead to a true conclusion. This step-by-step verification guarantees that the statement holds under all applicable circumstances.

In the first statement from the exercise, a proof is attempted to determine whether \(a \cdot b \equiv 0 \pmod{6}\) implies \(a \equiv 0 \pmod{6}\) or \(b \equiv 0 \pmod{6}\). The proof explores different scenarios – such as when \(a\) is divisible by factors of 6 (2 or 3) – to ensure that even in separate cases, the product's divisibility by 6 can still be explained. A successful proof shows that for any integers \(a\) and \(b\) meeting the given condition, the observed pattern holds true universally. This makes mathematical proofs essential for establishing the reliability of mathematical statements.
Counterexample
A counterexample is a specific case that shows a general statement to be false. If you can find even one example where the statement doesn't hold, it becomes clear that the statement isn't universally true.

In the textbook exercise, several statements are evaluated, and for some, counterexamples are used to show their falseness. For example, consider the second statement about divisibility by 8. By choosing \(a = 2\) and \(b = 4\), we find \(a \cdot b = 8\) which is \(0 \pmod{8}\), showing the overall product is zero mod 8. Yet, neither \(a\) nor \(b\) is individually \(0 \pmod{8}\). Another sophisticated counterexample is selecting \(a = 5\) and \(b = 5\) for the condition \(a \cdot b \equiv 1 \pmod{6}\). Here, their product satisfies the given modular condition, but neither meets the proposed individual congruence condition. Thus, counterexamples are powerful tools for testing the limits and accuracy of mathematical claims.

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

We have already seen examples of Pythagorean triples, which are natural numbers \(a, b,\) and \(c\) where \(a^{2}+\) \(b^{2}=c^{2}\). For example, \(3,4,\) and 5 form a Pythagorean triple as do \(5,\) \(12,\) and \(13 .\) One of the famous mathematicians of the 17 th century was Pierre de Fermat \((1601-1665)\). Fermat made an assertion that for each natural number \(n\) with \(n \geq 3,\) there are no positive integers \(a, b,\) and \(c\) for which \(a^{n}+b^{n}=c^{n}\). This assertion was discovered in a margin of one of Fermat's books after his death, but Fermat provided no proof. He did, however, state that he had discovered a truly remarkable proof but the margin did not contain enough room for the proof. This assertion became known as Fermat's Last Theorem but it more properly should have been called Fermat's Last Conjecture. Despite the efforts of mathematicians, this "theorem" remained unproved until Andrew Wiles, a British mathematician, first announced a proof in June of \(1993 .\) However, it was soon recognized that this proof had a serious gap, but a widely accepted version of the proof was published by Wiles in \(1995 .\) Wiles' proof uses many concepts and techniques that were unknown at the time of Fermat. We cannot discuss the proof here, but we will explore and prove the following proposition, which is a (very) special case of Fermat's Last Theorem. Proposition. There do not exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\) Although Fermat's Last Theorem implies this proposition is true, we will use a proof by contradiction to prove this proposition. For a proof by contradiction, we assume that there exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\). Since 2 is the only even prime number, we will use the following cases: (1) \(a=b=2 ;\) (2) \(a\) and \(b\) are both odd; and (3) one of \(a\) and \(b\) is odd and the other one is 2 . (a) Show that the case where \(a=b=2\) leads to a contradiction and hence, this case is not possible. (b) Show that the case where \(a\) and \(b\) are both odd leads to a contradiction and hence, this case is not possible. (c) We now know that one of \(a\) or \(b\) must be equal to 2 . So we assume that \(b=2\) and that \(a\) is an odd prime. Substitute \(b=2\) into the equation \(b^{3}=c^{3}-a^{3}\) and then factor the expression \(c^{3}-a^{3}\). Use this to obtain a contradiction. (d) Write a complete proof of the proposition.

Prove that for each natural number \(n, \sqrt{3 n+2}\) is not a natural number.

In Exercise (15) in Section 3.2, we proved that there exists a real number solution to the equation \(x^{3}-4 x^{2}=7\). Prove that there is no integer \(x\) such that \(x^{3}-4 x^{2}=7\)

Prove that for all integers \(a\) and \(m,\) if \(a\) and \(m\) are the lengths of the sides of a right triangle and \(m+1\) is the length of the hypotenuse, then \(a\) is an odd integer.

In Section \(3.1,\) we defined congruence modulo \(n\) where \(n\) is a natural number. If \(a\) and \(b\) are integers, we will use the notation \(a \neq b(\bmod n)\) to mean that \(a\) is not congruent to \(b\) modulo \(n\). * (a) Write the contrapositive of the following conditional statement: For all integers \(a\) and \(b,\) if \(a \neq 0(\bmod 6)\) and \(b \neq 0(\bmod 6),\) then \(a b \not \equiv 0(\bmod 6)\). (b) Is this statement true or false? Explain.

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.