Math Problem Statement

Let ∼ be a relation defined on Z by a∼b if 3a≡7b(mod 7). is it an equivalent relation? If so, find the distinct equivalent classes

Solution

To determine if the given relation \sim defined on Z\mathbb{Z} by 3a7b(mod7)3a \equiv 7b \pmod{7} is an equivalence relation, we must check the three properties: reflexivity, symmetry, and transitivity. If all hold, we will find the distinct equivalence classes.

1. Reflexivity

For any aZa \in \mathbb{Z}, we must show aaa \sim a, i.e., 3a7a(mod7)3a \equiv 7a \pmod{7}.
Since 7a0(mod7)7a \equiv 0 \pmod{7}, it follows that: 3a \equiv 7a \pmod{7} \quad \text{(true for all a)}. Thus, the relation is reflexive.

2. Symmetry

For any a,bZa, b \in \mathbb{Z}, if aba \sim b, then bab \sim a.
Assume 3a7b(mod7)3a \equiv 7b \pmod{7}. Rearrange to: 3a7b0(mod7).3a - 7b \equiv 0 \pmod{7}. Since 7b0(mod7)7b \equiv 0 \pmod{7}, this simplifies to: 3a7b    3b7a(mod7).3a \equiv 7b \implies 3b \equiv 7a \pmod{7}. Thus, bab \sim a, and the relation is symmetric.

3. Transitivity

For any a,b,cZa, b, c \in \mathbb{Z}, if aba \sim b and bcb \sim c, then aca \sim c.
Assume: 3a7b(mod7)and3b7c(mod7).3a \equiv 7b \pmod{7} \quad \text{and} \quad 3b \equiv 7c \pmod{7}. From 3a7b3a \equiv 7b, we know: 3a7b0(mod7).3a - 7b \equiv 0 \pmod{7}. From 3b7c3b \equiv 7c, we know: 3b7c0(mod7).3b - 7c \equiv 0 \pmod{7}. Adding these two equations gives: (3a7b)+(3b7c)0(mod7).(3a - 7b) + (3b - 7c) \equiv 0 \pmod{7}. This simplifies to: 3a7c0(mod7).3a - 7c \equiv 0 \pmod{7}. Thus, 3a7c(mod7)3a \equiv 7c \pmod{7}, and aca \sim c. The relation is transitive.

Conclusion

Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.


Distinct Equivalence Classes

The equivalence class of aZa \in \mathbb{Z} is: [a]={bZ3a7b(mod7)}.[a] = \{ b \in \mathbb{Z} \mid 3a \equiv 7b \pmod{7} \}. Simplify 3a7b(mod7)3a \equiv 7b \pmod{7}. Since 7b0(mod7)7b \equiv 0 \pmod{7}, this reduces to: 3a0(mod7).3a \equiv 0 \pmod{7}. Thus, the equivalence classes depend only on 3amod73a \mod{7}. Since 33 and 77 are coprime, there are 77 residue classes: [ [0], [1], [2], [3], [4], [5], [6]. ]

Each equivalence class contains integers aa such that ak(mod7)a \equiv k \pmod{7}, where k=0,1,,6k = 0, 1, \dots, 6.


Summary

  1. The relation \sim is an equivalence relation.
  2. There are 7 distinct equivalence classes: [ [0], [1], [2], [3], [4], [5], [6]. ]

Do you want further clarification on the steps?
Here are related questions for deeper exploration:

  1. How do modular arithmetic properties ensure reflexivity in relations like this?
  2. Why do the equivalence classes correspond to the residue classes modulo 7?
  3. How can the coprimality of 33 and 77 affect the distinct equivalence classes?
  4. What happens if the modulus changes to a non-prime like 8 or 9?
  5. How do equivalence relations apply to quotient groups in abstract algebra?

Tip: When dealing with equivalence relations in modular arithmetic, always reduce coefficients to simplify proofs.

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Equivalence Relations
Number Theory

Formulas

3a ≡ 7b (mod 7)
Equivalence Class Definition: [a] = {b ∈ Z | 3a ≡ 7b (mod 7)}

Theorems

Properties of Equivalence Relations (Reflexivity, Symmetry, Transitivity)
Residue Classes in Modular Arithmetic

Suitable Grade Level

Grades 10-12