Math Problem Statement

Tunjukan relasi tilde merupakan relasi ekuivalensi setelah di abstaksikan, yaitu di daerah integral D

Tunjukan relasi tilde merupakan relasi ekuivalensi setelah di abstraksikan, yaitu di daerah integral D.

Solution

The image you provided contains detailed information regarding a relation, denoted by the tilde (~), being described as an equivalence relation. This is then followed by explanations, including an abstract definition based on integral domains and some discussion of rational numbers.

Key Concepts:

  1. Definition of Equivalence Relation: A relation ~ is called an equivalence relation if it satisfies the following three properties:

    • Reflexive: aaa \sim a for all aAa \in A.
    • Symmetric: If aba \sim b, then bab \sim a.
    • Transitive: If aba \sim b and bcb \sim c, then aca \sim c.
  2. Context from the Image:

    • Integral Domain: A subset DZ×ZD \subseteq Z \times Z is discussed, where ZZ represents integers. The relation \sim is defined based on rational numbers a1b1=a2b2\frac{a_1}{b_1} = \frac{a_2}{b_2}, meaning that for two pairs (a1,b1)(a_1, b_1) and (a2,b2)(a_2, b_2), they are related if and only if a1b2=a2b1a_1b_2 = a_2b_1, which is a typical way of representing equivalence of fractions.

    • Equation (4.1): This seems to describe how pairs of integers are compared to verify equivalence, with conditions that b1b_1 and b2b_2 are non-zero, indicating that this relation holds between valid rational numbers.

  3. Abstraction into an Integral Domain: The relation is applied to elements in the product set Z×ZZ \times Z, and the discussion appears to abstract this into an integral domain where the properties of equivalence (reflexivity, symmetry, and transitivity) can be verified.

Proving Equivalence:

  1. Reflexivity:

    • Any pair (p,q)Z×Z(p, q) \in Z \times Z^* is related to itself because pq=pq\frac{p}{q} = \frac{p}{q}, or equivalently pq=pqp \cdot q = p \cdot q, which satisfies the condition.
  2. Symmetry:

    • If p1q1=p2q2\frac{p_1}{q_1} = \frac{p_2}{q_2}, then p1q2=p2q1p_1 q_2 = p_2 q_1, and by symmetry, p2q1=p1q2p_2 q_1 = p_1 q_2, hence the relation is symmetric.
  3. Transitivity:

    • If p1q1=p2q2\frac{p_1}{q_1} = \frac{p_2}{q_2} and p2q2=p3q3\frac{p_2}{q_2} = \frac{p_3}{q_3}, then p1q2=p2q1p_1 q_2 = p_2 q_1 and p2q3=p3q2p_2 q_3 = p_3 q_2. From this, you can deduce p1q3=p3q1p_1 q_3 = p_3 q_1, hence transitivity holds.

Thus, the relation \sim defined in the image is indeed an equivalence relation as it satisfies all three necessary conditions.

Would you like further details on any specific part of the proof or explanation? Here are some related questions you might find interesting:

  1. What are the properties of an integral domain, and how do they relate to equivalence relations?
  2. How is the concept of fractions formalized using pairs of integers in abstract algebra?
  3. Can this definition of equivalence be extended to other mathematical structures beyond rational numbers?
  4. What are other common examples of equivalence relations in mathematics?
  5. How does this abstract definition apply to real-world problems involving proportional relationships?

Tip: When working with equivalence relations, always check the three fundamental properties: reflexivity, symmetry, and transitivity.

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Math Problem Analysis

Mathematical Concepts

Equivalence Relation
Rational Numbers
Integral Domain

Formulas

Reflexivity: a ~ a
Symmetry: if a ~ b, then b ~ a
Transitivity: if a ~ b and b ~ c, then a ~ c
Equivalence of fractions: a1/b1 = a2/b2 if and only if a1*b2 = a2*b1

Theorems

Equivalence Relation Properties

Suitable Grade Level

Undergraduate Mathematics