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Chapter 7: Problem 10

Let \(\sim\) and \(\approx\) be relations on \(\mathbb{Z}\) defined as follows: \- For \(a, b \in \mathbb{Z}, a \sim b\) if and only if 2 divides \(a+b\). \- For \(a, b \in \mathbb{Z}, a \approx b\) if and only if 3 divides \(a+b\). (a) Is \(\sim\) an equivalence relation on \(\mathbb{Z} ?\) If not, is this relation reflexive, symmetric, or transitive? (b) Is \(\approx\) an equivalence relation on \(\mathbb{Z} ?\) If not, is this relation reflexive, symmetric, or transitive?

Short Answer

Expert verified
The relation \(\sim\) is an equivalence relation on \(\mathbb{Z}\) since it is reflexive, symmetric, and transitive. The relation \(\approx\) is not an equivalence relation on \(\mathbb{Z}\) since it is not reflexive, but it is symmetric and transitive.

Step by step solution

01

Reflexivity of \(\sim\)#

Consider any integer \(a \in \mathbb{Z}\). The condition for \(a \sim a\) is that 2 divides \(a+a\). Since \(2a\) is always an even number, 2 must always divide \(a+a\). Therefore, the reflexivity property holds for the \(\sim\) relation.
02

Symmetry of \(\sim\)#

Let \(a, b \in \mathbb{Z}\). If \(a \sim b\), then 2 divides \(a+b\). The sum \(a+b\) is equal to the sum \(b+a\), so if 2 divides \(a+b\), it also divides \(b+a\). This implies \(b \sim a\). So, the symmetry property holds for the \(\sim\) relation.
03

Transitivity of \(\sim\)#

Suppose \(a, b, c \in \mathbb{Z}\). If \(a \sim b\) and \(b \sim c\), then 2 divides \(a+b\) and 2 divides \(b+c\). These relations can be written as \(a+b=2m\) and \(b+c=2n\) for some integers \(m\) and \(n\). Adding these equations, we get \(a+2b+c=2(m+n)\). Now, subtracting \(2b\) from both sides, we find that \(a+c=2(m+n-b)\). So, 2 divides \(a+c\), and we conclude that \(a \sim c\). The transitivity property holds for the \(\sim\) relation. All three properties 鈥� reflexivity, symmetry, and transitivity 鈥� hold for the relation \(\sim\). Therefore, \(\sim\) is an equivalence relation on \(\mathbb{Z}\). (b) Is \(\approx\) an equivalence relation on \(\mathbb{Z}?\) If not, is this relation reflexive, symmetric, or transitive?
04

Reflexivity of \(\approx\)#

Consider any integer \(a \in \mathbb{Z}\). The condition for \(a \approx a\) is that 3 divides \(a+a\). Since \(2a\) is always an even number, 3 does not necessarily divide \(a+a\). In particular, if \(a=1\), then \(2a=2\), and 3 does not divide 2. Therefore, the reflexivity property does not hold for the \(\approx\) relation. Since reflexivity is not satisfied for the relation \(\approx\), it cannot be an equivalence relation on \(\mathbb{Z}\). Nevertheless, we will analyze if the relation is symmetric and transitive.
05

Symmetry of \(\approx\)#

Let \(a, b \in \mathbb{Z}\). If \(a \approx b\), then 3 divides \(a+b\). The sum \(a+b\) is equal to the sum \(b+a\), so if 3 divides \(a+b\), it also divides \(b+a\). This implies \(b \approx a\). So, the symmetry property holds for the \(\approx\) relation.
06

Transitivity of \(\approx\)#

Suppose \(a, b, c \in \mathbb{Z}\). If \(a \approx b\) and \(b \approx c\), then 3 divides \(a+b\) and 3 divides \(b+c\). These relations can be written as \(a+b=3m\) and \(b+c=3n\) for some integers \(m\) and \(n\). Adding these equations, we get \(a+2b+c=3(m+n)\). Now, subtracting \(3b\) from both sides, we find that \(a+c=3(m+n-b)\). So, 3 divides \(a+c\), and we conclude that \(a \approx c\). The transitivity property holds for the \(\approx\) relation. In conclusion, the relation \(\approx\) is not an equivalence relation on \(\mathbb{Z}\) since it is not reflexive. However, it is symmetric and transitive.

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Key Concepts

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

Reflexive Property
When discussing relations on integers and equivalence relations, one crucial part is the reflexive property. This property implies that every element is related to itself. For instance, in the context of integer divisibility, for a relation to be reflexive, a number must meet the criteria when paired with itself.In our textbook exercise, the relation \(\sim\) showcases reflexivity because any integer \(a\), when added to itself (yielding \(2a\)), results in a number divisible by 2. Hence, every integer is in relation with itself under the \(\sim\) operation. However, the relation \(\approx\) does not fulfill this criteria, as not every integer, when doubled, results in a number divisible by 3. This illustrates a pivotal principle: a relation cannot be termed an equivalence relation without being reflexive.
Symmetric Property
The symmetric property is another fundamental component of an equivalence relation, implying that if an element \(a\) is related to \(b\), then \(b\) is also related to \(a\). This mutual relationship must hold true for every pair of elements within the set.Our textbook scenario with the relation \(\sim\) demonstrates symmetry, because if 2 divides \(a+b\), it naturally divides \(b+a\) since addition is commutative. The relation \(\approx\), though not reflexive, also possesses symmetry. For students, grasping the symmetric property can be aided by visualizing the relationship as a two-way street: if a path exists from \(a\) to \(b\), symmetry ensures an identical path back from \(b\) to \(a\).
Transitive Property
The core aspect of an equivalence relation is the transitive property, which states that if \(a\) is related to \(b\), and \(b\) to \(c\), then \(a\) must be related to \(c\). This transitive nature links elements together in a chain of relations, defining an interconnected network within the set.In our textbook solution, both relations \(\sim\) and \(\approx\) satisfy transitivity. This is demonstrated through the algebraic manipulation provided in the original solutions. Whenever students examine such properties within an equivalence relation, the transitive property should be seen as establishing a 鈥榙omino effect鈥欌�攐nce the initial relationships are set, the subsequent ones will naturally follow.
Integer Divisibility
Integer divisibility plays a critical role when determining the characteristics of relations like \(\sim\) and \(\approx\). It involves understanding whether one integer can be divided by another without leaving a remainder. The concepts of reflexive, symmetric, and transitive properties all hinge upon integer divisibility.In our relations on \(\mathbb{Z}\), divisibility by 2 and 3 forms the basis for determining the validity of equivalence. Students exploring these relations gain insight into divisibility rules, which are key for mathematical reasoning and the analysis of integer relations.
Mathematical Reasoning
Mathematical reasoning is paramount when dissecting relations and their properties. It encompasses logical thought processes, which helps students understand and apply the definitions of reflexive, symmetric, and transitive properties systematically.Using the textbook exercise as a learning tool, one can apply mathematical reasoning to deduce that while the relation \(\sim\) on \(\mathbb{Z}\) satisfies all three properties making it an equivalence relation, \(\approx\) fails to be reflexive, thus is not an equivalence relation. Encouraging the development of mathematical reasoning is beneficial not only for this topic but also for a wider scope of mathematical theory and applications.
Relation on Integers
A relation on integers is a way to pair numbers from the set of integers \(\mathbb{Z}\) based on a specific rule or condition. In equivalence relations such as \(\sim\) and \(\approx\), the rule is related to the divisibility of the sum of pairs, providing a practical understanding of such relationships.Using the relations defined in our exercise, we see that they can vary in their adherence to the properties that make up an equivalence relation. Understanding these relations provides a strong foundation for concepts in number theory and broader mathematical contexts. It's through exercises like this that students can apply concrete reasoning to abstract mathematical concepts.

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Most popular questions from this chapter

The set \(\mathbb{Z}_{n}\) is a finite set, and hence one way to prove things about \(\mathbb{Z}_{n}\) is to simply use the \(n\) elements in \(\mathbb{Z}_{n}\) as the \(n\) cases for a proof using cases. For example, if \(n \in \mathbb{Z},\) then in \(\mathbb{Z}_{4},[n]=[0]\), \([n]=[1],[n]=[2],\) or \([n]=[3]\) (a) Prove that if \(n \in \mathbb{Z},\) then in \(\mathbb{Z}_{4},[n]^{2}=[0]\) or \([n]^{2}=[1] .\) Use this to conclude that in \(\mathbb{Z}_{4},\left[n^{2}\right]=[0]\) or \(\left[n^{2}\right]=[1]\) (b) Translate the equations \(\left[n^{2}\right]=[0]\) and \(\left[n^{2}\right]=[1]\) in \(\mathbb{Z}_{4}\) into congruences modulo 4 . (c) Use a result in Exercise (12) to determine the value of \(r\) so that \(r \in \mathbb{Z}\) \(0 \leq r<3,\) and $$ 104257833259 \equiv r(\bmod 4) $$ That is, \([104257833259]=[r]\) in \(\mathbb{Z}_{4}\). (d) Is the natural number 104257833259 a perfect square? Justify your conclusion.

Let \(A=\mathbb{Z} \times(\mathbb{Z}-\\{0\\}) .\) That is, \(A=\\{(a, b) \in \mathbb{Z} \times \mathbb{Z} \mid b \neq 0\\} .\) Define the relation \(\approx\) on \(A\) as follows: For \((a, b),(c, d) \in A,(a, b) \approx(c, d)\) if and only if \(a d=b c\) (a) Prove that \(\approx\) is an equivalence relation on \(A\). (b) Why was it necessary to include the restriction that \(b \neq 0\) in the definition of the set \(A ?\) - (c) Determine an equation that gives a relation between \(a\) and \(b\) if \((a, b) \in\) \(A\) and \((a, b) \approx(2,3)\) (d) Determine at least four different elements in [(2,3)] , the equivalence class of (2,3) . (e) Use set builder notation to describe \([(2,3)],\) the equivalence class of (2,3)

Let \(A\) be a nonempty set and let \(\sim\) be an equivalence relation on \(A\). Prove each of the following: (a) For each \(a, b \in A, a \approx b\) if and only if \([a] \cap[b]=\emptyset\). (b) For each \(a, b \in A,\) if \([a] \neq[b],\) then \([a] \cap[b]=\emptyset\). (c) For each \(a, b \in A,\) if \([a] \cap[b] \neq \emptyset\) then \([a]=[b]\).

Let \(n \in \mathbb{N}\) and assume $$ n=\left(a_{k} \times 10^{k}\right)+\left(a_{k-1} \times 10^{k-1}\right)+\cdots+\left(a_{1} \times 10^{1}\right)+\left(a_{0} \times 10^{0}\right) $$ Use the result in Exercise (15) to help prove each of the following: (a) \(n \equiv \sum_{j=0}^{k}(-1)^{j} a_{j}(\bmod 11)\). (b) \([n]=\left[\sum_{j=0}^{k}(-1)^{j} a_{j}\right],\) using congruence classes modulo 11 . (c) 11 divides \(n\) if and only if 11 divides \(\sum_{j=0}^{k}(-1)^{j} a_{j}\).

Let \(R_{<}=\\{(x, y) \in \mathbb{R} \times \mathbb{R} \mid x
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